{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0001","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"661c4c6beb23e4d0d77783bcb4872298c23851039d1a7f6767bfb5309c920b93","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253001,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[-1,-12,-3],\"interval\":[-2,3]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[-1, -12, -3], not ascending-power order. The declared interval [lower,upper]=[-2, 3] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-1, -12, -3], [-2, 3], No, Yes, No, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-1, -12, -3], [-2, 3], No, Yes, No, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0002","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"182cf4135072e7f83e7d6981b201c53ede850ee240af81b6c97894cd37c75510","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253002,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[1,-11,-2],\"interval\":[-3,4]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, -11, -2], not ascending-power order. The declared interval [lower,upper]=[-3, 4] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, -11, -2], [-3, 4], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, -11, -2], [-3, 4], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0003","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"625ef78c3da23538df485165a1e9ff687b54bd289db8f38c266ace65178b08a1","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253003,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[2,-10,-1],\"interval\":[-4,5]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, -10, -1], not ascending-power order. The declared interval [lower,upper]=[-4, 5] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, -10, -1], [-4, 5], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, -10, -1], [-4, 5], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0004","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d883d79c4ed7dcaa86ce66e5cfbb6ebac0e983541d2e91c08dcb69ced9ee9e1","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253004,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[3,-9,0],\"interval\":[-5,6]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, -9, 0], not ascending-power order. The declared interval [lower,upper]=[-5, 6] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, -9, 0], [-5, 6], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -9, 0], [-5, 6], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0005","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56d534f093f879bc7c201c61b1fa7ae75efd981a123d211f8c1c3f86d973d3c8","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253005,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[0,-8,1],\"interval\":[-6,7]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, -8, 1], not ascending-power order. The declared interval [lower,upper]=[-6, 7] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, -8, 1], [-6, 7], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, -8, 1], [-6, 7], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0006","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef88d1a255963efc56749013c889e405836369f6b71e4e02e633a3151103f28c","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253006,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[-1,-7,2],\"interval\":[-7,8]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[-1, -7, 2], not ascending-power order. The declared interval [lower,upper]=[-7, 8] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-1, -7, 2], [-7, 8], No, Yes, No, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-1, -7, 2], [-7, 8], No, Yes, No, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0007","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d817992081a32429a3ae87392d042efd8ab0baf9e9669a9547d9742368d4d3dc","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253007,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[2,-6,3],\"interval\":[-8,9]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, -6, 3], not ascending-power order. The declared interval [lower,upper]=[-8, 9] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, -6, 3], [-8, 9], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, -6, 3], [-8, 9], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0008","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"320bdb751360699e86d91040b533f1152309ccc2e24f83894629a7765653f526","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253008,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[3,-5,-3],\"interval\":[-9,10]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, -5, -3], not ascending-power order. The declared interval [lower,upper]=[-9, 10] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, -5, -3], [-9, 10], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -5, -3], [-9, 10], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0009","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f40dbf51196bcfe3dc6ae84764d8a90a38792e848421daedc2c5419f63f2cbe8","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253009,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[0,-4,-2],\"interval\":[-10,11]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, -4, -2], not ascending-power order. The declared interval [lower,upper]=[-10, 11] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, -4, -2], [-10, 11], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, -4, -2], [-10, 11], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0010","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8972c16df75c7f796570dd0be15c79481bbd904ba9e39c1fc265e2f234d90ba3","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253010,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[1,-3,-1],\"interval\":[-11,12]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, -3, -1], not ascending-power order. The declared interval [lower,upper]=[-11, 12] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, -3, -1], [-11, 12], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, -3, -1], [-11, 12], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0011","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5afd4079ccd7022637921e025b58f27a9946271104ee6e63334501b28172446d","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253011,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[-1,-2,0],\"interval\":[-12,13]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[-1, -2, 0], not ascending-power order. The declared interval [lower,upper]=[-12, 13] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-1, -2, 0], [-12, 13], No, Yes, No, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-1, -2, 0], [-12, 13], No, Yes, No, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0012","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f368ead1966c16646adf33b11a655538d580eb2069ee85fb5dd142110ae160f","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253012,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[3,-1,1],\"interval\":[-13,14]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, -1, 1], not ascending-power order. The declared interval [lower,upper]=[-13, 14] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, -1, 1], [-13, 14], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -1, 1], [-13, 14], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0013","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"137936d99275f7f9606aa78758e6249372197da5db38af3b523f5e04026aee5e","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253013,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[0,0,2],\"interval\":[-14,15]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 0, 2], not ascending-power order. The declared interval [lower,upper]=[-14, 15] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 0, 2], [-14, 15], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 0, 2], [-14, 15], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0014","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f838ef81ff9062240258d21a228fb122cd01ba70ee158bf6933f2219d1198d3e","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253014,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[1,1,3],\"interval\":[-15,16]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, 1, 3], not ascending-power order. The declared interval [lower,upper]=[-15, 16] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, 1, 3], [-15, 16], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 1, 3], [-15, 16], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0015","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"74ceab958d31600890e3b1b6730b5559b87b3bb021dea50a13cd8d878475f2d9","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253015,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[2,2,-3],\"interval\":[-16,17]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, 2, -3], not ascending-power order. The declared interval [lower,upper]=[-16, 17] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, 2, -3], [-16, 17], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 2, -3], [-16, 17], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0016","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"084f8e7e9c55aafc0aff12a9031194f29ad4d4c28aba527f42da9c10a7a78a8c","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253016,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[-1,3,-2],\"interval\":[-17,18]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[-1, 3, -2], not ascending-power order. The declared interval [lower,upper]=[-17, 18] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-1, 3, -2], [-17, 18], No, Yes, No, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-1, 3, -2], [-17, 18], No, Yes, No, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0017","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02daf3bfb89ffd28eaee2cde85983f593a7ad2cb6e450239e601c8cd0e3e74eb","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253017,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[0,4,-1],\"interval\":[-18,19]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 4, -1], not ascending-power order. The declared interval [lower,upper]=[-18, 19] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 4, -1], [-18, 19], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 4, -1], [-18, 19], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0018","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7144e5434495888543e17a45154953f4fb50e02f7575f5d9b0ef573366c4fe7","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253018,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[1,5,0],\"interval\":[-19,20]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, 5, 0], not ascending-power order. The declared interval [lower,upper]=[-19, 20] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, 5, 0], [-19, 20], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 5, 0], [-19, 20], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0019","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54008ee9aeea72cc317941c13403b3e6dadaf2657ebf26f7feaf69946495f86b","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253019,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[2,6,1],\"interval\":[-20,21]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, 6, 1], not ascending-power order. The declared interval [lower,upper]=[-20, 21] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, 6, 1], [-20, 21], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 6, 1], [-20, 21], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0020","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2cd8378a05ac8350ac5276ceace0d02868d9ae31e8f28e845c57b3b87825119","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253020,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[3,7,2],\"interval\":[-21,22]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, 7, 2], not ascending-power order. The declared interval [lower,upper]=[-21, 22] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, 7, 2], [-21, 22], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, 7, 2], [-21, 22], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0021","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1eaa6d7b23c942e66e67b3f16c1850b45942efdb7595625a4f61de6974b3d051","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253021,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[-1,8,3],\"interval\":[-22,23]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[-1, 8, 3], not ascending-power order. The declared interval [lower,upper]=[-22, 23] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-1, 8, 3], [-22, 23], No, Yes, No, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-1, 8, 3], [-22, 23], No, Yes, No, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0022","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e91347ac4eb59023d614ce046c9cc34707b8093e1e51b76a1ac8a7dedc39fa1","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253022,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[1,9,-3],\"interval\":[-23,24]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, 9, -3], not ascending-power order. The declared interval [lower,upper]=[-23, 24] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, 9, -3], [-23, 24], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 9, -3], [-23, 24], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0023","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a69510335b6e5074ddced16070e60479d84c04df73421869e8afd86bfc62324","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253023,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[2,10,-2],\"interval\":[-24,25]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, 10, -2], not ascending-power order. The declared interval [lower,upper]=[-24, 25] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, 10, -2], [-24, 25], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 10, -2], [-24, 25], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0024","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15226ce23bdfefe1213a5833666af3b53238b59963089089efa7fcd7732e9178","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253024,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[3,11,-1],\"interval\":[-25,26]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, 11, -1], not ascending-power order. The declared interval [lower,upper]=[-25, 26] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, 11, -1], [-25, 26], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, 11, -1], [-25, 26], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0025","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb618c8b8d1a417d5ee34e2fcf7f49bb1c6f02cc0d47cbdd476184befe3e2b23","family_id":"qf-mod25-convexopt-problem-qualification","seed":202610253025,"semantic_template":"qf-mod25-convexopt-problem-qualification/{\"coefficients\":[0,12,0],\"interval\":[-26,27]}"},"fields":{"title":"Separate a Quadratic Objective Certificate from Feasible-Interval Qualification","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 12, 0], not ascending-power order. The declared interval [lower,upper]=[-26, 27] belongs to the function data. Give (coefficients,interval,global quadratic certificate a≥0?,qualified nonempty ordered feasible interval?,both qualifications?,qualification_only_no_solution_claim). The objective flag uses the named global quadratic criterion; failure does not decide every possible restricted singleton case. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 12, 0], [-26, 27], Yes, Yes, Yes, qualification_only_no_solution_claim)","explanation":"Check the two declared qualifications separately. A reversed interval record is outside this nonempty-domain protocol. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 12, 0], [-26, 27], Yes, Yes, Yes, qualification_only_no_solution_claim)."},"skills":["Separate a Quadratic Objective Certificate from Feasible-Interval Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0026","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14f2f37b78355ee830aaf05f65309f3d725cef2bff2e099bfc67bffcf34d6803","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253026,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-2,3],\"point\":[-3,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-2, 3], candidate=-3/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-2, 3], -3, No, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-2, 3], -3, No, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0027","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27f2181d56f02bdad02affef46a3669279dfdd7f380f9312ab4afb3deafed16c","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253027,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-3,4],\"point\":[-2,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-3, 4], candidate=-2/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-3, 4], -2, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-3, 4], -2, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0028","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a72d10febc7ed8bd189067ef6ff3585ad372a966b34ead38d896c47fda962ef","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253028,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-4,5],\"point\":[-1,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-4, 5], candidate=-1/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-4, 5], -1, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-4, 5], -1, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0029","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4544289de801af2061859d4ebee26624c8b01be3e4d8dec09696510fd994471c","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253029,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-5,6],\"point\":[0,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-5, 6], candidate=0/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-5, 6], 0, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-5, 6], 0, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0030","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"20cd5cbcc2f47236d0063522a2aeee78b8c32dd1341f8b9d92537a5d1869b553","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253030,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-6,7],\"point\":[1,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-6, 7], candidate=1/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-6, 7], 1, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-6, 7], 1, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0031","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65be5fa7217a19f4394a5d5a2b3a5a125cbb927fd20f36072f16574e4589eef3","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253031,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-7,8],\"point\":[2,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-7, 8], candidate=2/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-7, 8], 2, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-7, 8], 2, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0032","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3bc079fdf3cb8633756f2984ccbe444de1ff9ccf3566fcfd45912bef6045f6e4","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253032,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-8,9],\"point\":[3,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-8, 9], candidate=3/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-8, 9], 3, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-8, 9], 3, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0033","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"12c5bfb621b8cc4997f1c99ddf6cd9efbc61aa04e70eb7d3acaeb79f535ef418","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253033,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-9,10],\"point\":[-3,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-9, 10], candidate=-3/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-9, 10], -3, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-9, 10], -3, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0034","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce07aeda6badce1fac54f5fe4fd0acbb41ba1ef6158b0a390b0c817a3850e423","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253034,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-10,11],\"point\":[-2,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-10, 11], candidate=-2/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-10, 11], -2, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-10, 11], -2, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0035","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2c7a37276f5ae13503cff745cf1b2bb8c56c0d16be6f4fa6f566b2ebedd9746c","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253035,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-11,12],\"point\":[-1,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-11, 12], candidate=-1/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-11, 12], -1, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-11, 12], -1, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0036","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dbeeff2cf934b22794a39900f92951b5739e84325c50c4a4c14b2a47a6d132d","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253036,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-12,13],\"point\":[0,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-12, 13], candidate=0/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-12, 13], 0, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-12, 13], 0, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0037","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43587a64b7682d36d9dfa805043425fc7dd1a58bfc88ece45e307da5f95d05ea","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253037,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-13,14],\"point\":[1,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-13, 14], candidate=1/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-13, 14], 1, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-13, 14], 1, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0038","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af89ce133bdf56bebfe39e8fc12480f304f2a76ab5c07193eaec792c8fb8d24a","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253038,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-14,15],\"point\":[2,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-14, 15], candidate=2/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-14, 15], 2, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-14, 15], 2, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0039","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa2e3e4485334b4bf853d801c15f9c71910659c23a78516b5e1c8425a3ed37b0","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253039,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-15,16],\"point\":[3,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-15, 16], candidate=3/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-15, 16], 3, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-15, 16], 3, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0040","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db23add2755ac849456fe1da29adb050eb9694498b3ac6fce71c250cd22c1235","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253040,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-16,17],\"point\":[-3,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-16, 17], candidate=-3/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-16, 17], -3, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-16, 17], -3, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0041","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"107a0e43718847f4c219af407017eca88a146b59aa6e7f0218c7603f0b0b9339","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253041,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-17,18],\"point\":[-2,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-17, 18], candidate=-2/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-17, 18], -2, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-17, 18], -2, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0042","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"144f7eb3a8983f891a10348b9b9d72d7dba767c198b67610bd2b882ea230f9cb","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253042,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-18,19],\"point\":[-1,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-18, 19], candidate=-1/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-18, 19], -1, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-18, 19], -1, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0043","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e3f9cbcdc5d4c61d83119500e8e17858ca96f055351ea2486813446f51cbc2f","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253043,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-19,20],\"point\":[0,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-19, 20], candidate=0/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-19, 20], 0, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-19, 20], 0, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0044","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f39938b219f52da964d58584fc16898e25233da9e47e405dbd5b2ce63a90f9f1","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253044,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-20,21],\"point\":[1,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-20, 21], candidate=1/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-20, 21], 1, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-20, 21], 1, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0045","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b3a5adb25e038db09e229c1866cbefcb7966686b4602c856c50c71d641a3fca8","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253045,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-21,22],\"point\":[2,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-21, 22], candidate=2/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-21, 22], 2, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-21, 22], 2, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0046","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca86dd0cd7a348ae1471bde0acd8240c871dd2bbbbcf05a3fd18455879ba0408","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253046,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-22,23],\"point\":[3,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-22, 23], candidate=3/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-22, 23], 3, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-22, 23], 3, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0047","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f1bbedcc06d2d48c574b69e654f62fa437d8753fb80535e425b1d274e8735d8","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253047,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-23,24],\"point\":[-3,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-23, 24], candidate=-3/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-23, 24], -3, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-23, 24], -3, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0048","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51e8cdbf345be39e3c6f35c09fadab3261779c721c5e53621e651741a1897a6e","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253048,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-24,25],\"point\":[-2,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-24, 25], candidate=-2/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-24, 25], -2, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-24, 25], -2, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0049","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b070376475add98319e9cff0f56312daaf641303b273972895bee801c74c79a5","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253049,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-25,26],\"point\":[-1,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-25, 26], candidate=-1/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-25, 26], -1, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-25, 26], -1, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0050","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5eb976c01d1d3f8836580d8b5b278be43e33fd0794b72b67d417b730bfe1355e","family_id":"qf-mod25-convexopt-feasible-point-audit","seed":202610253050,"semantic_template":"qf-mod25-convexopt-feasible-point-audit/{\"interval\":[-26,27],\"point\":[0,1]}"},"fields":{"title":"Feasibility Alone Supplies No Optimum","stem":"Declared nonempty interval record=[-26, 27], candidate=0/1. Give (interval,normalized candidate,within both inclusive bounds?,feasibility_only). A reversed record has no qualified feasible point in this protocol. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([-26, 27], 0, Yes, feasibility_only)","explanation":"Compare the exact point with both declared endpoints, retaining equality at a boundary. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-26, 27], 0, Yes, feasibility_only)."},"skills":["Feasibility Alone Supplies No Optimum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0051","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c91cc144f4094c7e971d7a7186d68e3ec2468af714648f6a803553f6dcdb2c9e","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253051,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[0,-12,-3],\"interval\":[-2,3],\"point\":[-3,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, -12, -3], not ascending-power order. The declared interval [lower,upper]=[-2, 3] belongs to the function data. point=-3/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, -12, -3], [-2, 3], -3, No, 33, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, -12, -3], [-2, 3], -3, No, 33, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0052","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24c70c22451af8e05ebf215b2c47a3247ea33e48ab3d8c38b0a43fac7319de39","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253052,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[1,-11,-2],\"interval\":[-3,4],\"point\":[-2,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, -11, -2], not ascending-power order. The declared interval [lower,upper]=[-3, 4] belongs to the function data. point=-2/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, -11, -2], [-3, 4], -2, Yes, 24, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, -11, -2], [-3, 4], -2, Yes, 24, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0053","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"133517413d624f4ee04dd6467bc14afbd451080c8be811e2aecfa0676de19aad","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253053,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[2,-10,-1],\"interval\":[-4,5],\"point\":[-1,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, -10, -1], not ascending-power order. The declared interval [lower,upper]=[-4, 5] belongs to the function data. point=-1/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, -10, -1], [-4, 5], -1, Yes, 11, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, -10, -1], [-4, 5], -1, Yes, 11, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0054","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf81427e708819793d73c9358abdf4fedec1f5aea6daacd97182be011aaff19f","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253054,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[3,-9,0],\"interval\":[-5,6],\"point\":[0,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, -9, 0], not ascending-power order. The declared interval [lower,upper]=[-5, 6] belongs to the function data. point=0/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, -9, 0], [-5, 6], 0, Yes, 0, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -9, 0], [-5, 6], 0, Yes, 0, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0055","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"63ad10ecb9a09dba6870fff096d4e9f5d7ceaebbfa0188625afdd76bddd4da72","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253055,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[0,-8,1],\"interval\":[-6,7],\"point\":[1,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, -8, 1], not ascending-power order. The declared interval [lower,upper]=[-6, 7] belongs to the function data. point=1/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, -8, 1], [-6, 7], 1, Yes, -7, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, -8, 1], [-6, 7], 1, Yes, -7, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0056","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d73778f818cc9ccac516a821556a9fc815fd905cb37389e4f5a87434530f27e","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253056,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[1,-7,2],\"interval\":[-7,8],\"point\":[2,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, -7, 2], not ascending-power order. The declared interval [lower,upper]=[-7, 8] belongs to the function data. point=2/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, -7, 2], [-7, 8], 2, Yes, -8, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, -7, 2], [-7, 8], 2, Yes, -8, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0057","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"872664b6fba33128e3141d64a343467ac7878b62bb6d80eb86fbd0df10c081d0","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253057,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[2,-6,3],\"interval\":[-8,9],\"point\":[3,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, -6, 3], not ascending-power order. The declared interval [lower,upper]=[-8, 9] belongs to the function data. point=3/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, -6, 3], [-8, 9], 3, Yes, 3, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, -6, 3], [-8, 9], 3, Yes, 3, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0058","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"febbdad23295b209dd08a72b372de6cdcf66691844c281166cb28733299f485b","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253058,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[3,-5,-3],\"interval\":[-9,10],\"point\":[-3,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, -5, -3], not ascending-power order. The declared interval [lower,upper]=[-9, 10] belongs to the function data. point=-3/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, -5, -3], [-9, 10], -3, Yes, 39, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -5, -3], [-9, 10], -3, Yes, 39, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0059","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6310630b06fb289d3d34f8e201eecbb5a0bb687c888f8db8e5326e8e86e2571f","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253059,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[0,-4,-2],\"interval\":[-10,11],\"point\":[-2,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, -4, -2], not ascending-power order. The declared interval [lower,upper]=[-10, 11] belongs to the function data. point=-2/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, -4, -2], [-10, 11], -2, Yes, 6, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, -4, -2], [-10, 11], -2, Yes, 6, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0060","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"965033d88690601927097501459f5123ce05bb345c6b696ed1e52ceb35848982","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253060,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[1,-3,-1],\"interval\":[-11,12],\"point\":[-1,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, -3, -1], not ascending-power order. The declared interval [lower,upper]=[-11, 12] belongs to the function data. point=-1/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, -3, -1], [-11, 12], -1, Yes, 3, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, -3, -1], [-11, 12], -1, Yes, 3, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0061","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ccdd70d1e37849222db1de3ffc0a45cdacb41b7fabb4702df921ddfdc040736c","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253061,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[2,-2,0],\"interval\":[-12,13],\"point\":[0,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, -2, 0], not ascending-power order. The declared interval [lower,upper]=[-12, 13] belongs to the function data. point=0/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, -2, 0], [-12, 13], 0, Yes, 0, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, -2, 0], [-12, 13], 0, Yes, 0, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0062","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"649a6b44a55830642422292a2ff1d398c545795787da4b138f4dc105fe523e45","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253062,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[3,-1,1],\"interval\":[-13,14],\"point\":[1,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, -1, 1], not ascending-power order. The declared interval [lower,upper]=[-13, 14] belongs to the function data. point=1/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, -1, 1], [-13, 14], 1, Yes, 3, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -1, 1], [-13, 14], 1, Yes, 3, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0063","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af05d52ba9b02919d2fae2b1fcfbd9a864b3a234221175faf70033260d3d19e0","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253063,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[0,0,2],\"interval\":[-14,15],\"point\":[2,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 0, 2], not ascending-power order. The declared interval [lower,upper]=[-14, 15] belongs to the function data. point=2/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 0, 2], [-14, 15], 2, Yes, 2, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 0, 2], [-14, 15], 2, Yes, 2, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0064","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b2ab6a43e00b54ecf0e030a2fa8583c921b26fe412cc1d3dda67d48b3bc6a94","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253064,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[1,1,3],\"interval\":[-15,16],\"point\":[3,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, 1, 3], not ascending-power order. The declared interval [lower,upper]=[-15, 16] belongs to the function data. point=3/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, 1, 3], [-15, 16], 3, Yes, 15, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 1, 3], [-15, 16], 3, Yes, 15, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0065","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4a464a970352878b1aab89018bdfe51590aa78ade5607a4637f327a63d353575","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253065,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[2,2,-3],\"interval\":[-16,17],\"point\":[-3,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, 2, -3], not ascending-power order. The declared interval [lower,upper]=[-16, 17] belongs to the function data. point=-3/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, 2, -3], [-16, 17], -3, Yes, 9, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 2, -3], [-16, 17], -3, Yes, 9, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0066","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"222c9d9efe995c780435974563d8664a57c07990562c06fb980896f5f875ef48","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253066,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[3,3,-2],\"interval\":[-17,18],\"point\":[-2,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, 3, -2], not ascending-power order. The declared interval [lower,upper]=[-17, 18] belongs to the function data. point=-2/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, 3, -2], [-17, 18], -2, Yes, 4, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, 3, -2], [-17, 18], -2, Yes, 4, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0067","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cf94d17b3db0544f5ac79c45bba70ac8d9402f1a3b3098c940912e94808328e","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253067,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[0,4,-1],\"interval\":[-18,19],\"point\":[-1,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 4, -1], not ascending-power order. The declared interval [lower,upper]=[-18, 19] belongs to the function data. point=-1/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 4, -1], [-18, 19], -1, Yes, -5, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 4, -1], [-18, 19], -1, Yes, -5, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0068","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87feeb2205803e03617023f7c98fdb294446ba245de0f99a6ed2f5cd9b0f6f7f","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253068,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[1,5,0],\"interval\":[-19,20],\"point\":[0,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, 5, 0], not ascending-power order. The declared interval [lower,upper]=[-19, 20] belongs to the function data. point=0/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, 5, 0], [-19, 20], 0, Yes, 0, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 5, 0], [-19, 20], 0, Yes, 0, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0069","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e3c5b4ebba89840d4076f99d16151ac24f1417bbb4ffee6f6e2b559c5f206ce8","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253069,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[2,6,1],\"interval\":[-20,21],\"point\":[1,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, 6, 1], not ascending-power order. The declared interval [lower,upper]=[-20, 21] belongs to the function data. point=1/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, 6, 1], [-20, 21], 1, Yes, 9, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 6, 1], [-20, 21], 1, Yes, 9, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0070","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec012e99a80d306ed3f81dc4b15258f4958ab87ca3e295e8fcca8eeebf8e44e5","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253070,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[3,7,2],\"interval\":[-21,22],\"point\":[2,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, 7, 2], not ascending-power order. The declared interval [lower,upper]=[-21, 22] belongs to the function data. point=2/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, 7, 2], [-21, 22], 2, Yes, 28, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, 7, 2], [-21, 22], 2, Yes, 28, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0071","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fcaf6ebd9a75a800778938f11d90260787d4b332da0f174db9bc4e7274a6a7ef","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253071,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[0,8,3],\"interval\":[-22,23],\"point\":[3,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 8, 3], not ascending-power order. The declared interval [lower,upper]=[-22, 23] belongs to the function data. point=3/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 8, 3], [-22, 23], 3, Yes, 27, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 8, 3], [-22, 23], 3, Yes, 27, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0072","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"becc611d154b77389ed6b53ec999b2b794eec1472b8a9758994e0f7b15d30533","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253072,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[1,9,-3],\"interval\":[-23,24],\"point\":[-3,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, 9, -3], not ascending-power order. The declared interval [lower,upper]=[-23, 24] belongs to the function data. point=-3/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, 9, -3], [-23, 24], -3, Yes, -21, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 9, -3], [-23, 24], -3, Yes, -21, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0073","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ed61f67a97fc7f5f41352a173a0976f6331aa08785935825a3068dfab332c4f","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253073,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[2,10,-2],\"interval\":[-24,25],\"point\":[-2,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, 10, -2], not ascending-power order. The declared interval [lower,upper]=[-24, 25] belongs to the function data. point=-2/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, 10, -2], [-24, 25], -2, Yes, -14, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 10, -2], [-24, 25], -2, Yes, -14, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0074","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54c5f28746952c1487f6cce71d53295ccd10415d6f75da43af087d8cc7a17f7a","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253074,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[3,11,-1],\"interval\":[-25,26],\"point\":[-1,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, 11, -1], not ascending-power order. The declared interval [lower,upper]=[-25, 26] belongs to the function data. point=-1/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, 11, -1], [-25, 26], -1, Yes, -9, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, 11, -1], [-25, 26], -1, Yes, -9, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0075","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d48b2c691864f9fd2bc45c99f5944aeb89707eb4a04aa3639fb6b42fdf65a90a","family_id":"qf-mod25-convexopt-objective-at-feasible-point","seed":202610253075,"semantic_template":"qf-mod25-convexopt-objective-at-feasible-point/{\"coefficients\":[0,12,0],\"interval\":[-26,27],\"point\":[0,1]}"},"fields":{"title":"Evaluate the Polynomial while Separately Reporting Feasibility","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 12, 0], not ascending-power order. The declared interval [lower,upper]=[-26, 27] belongs to the function data. point=0/1. Give (coefficients,interval,normalized point,feasible?,polynomial value,evaluation_only). Retain the polynomial value even for an infeasible candidate; its numerical value does not make it eligible. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 12, 0], [-26, 27], 0, Yes, 0, evaluation_only)","explanation":"Recompute both the inclusive feasibility predicate and objective value. Do not promote evaluation to optimality. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 12, 0], [-26, 27], 0, Yes, 0, evaluation_only)."},"skills":["Evaluate the Polynomial while Separately Reporting Feasibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0076","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea3648122dfd14c18cc460d82c4e7ccb781062436f63c55731168a64a7666691","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253076,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[0,-12,-3],\"interval\":[-2,3],\"left\":[-2,1],\"right\":[-3,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, -12, -3], not ascending-power order. The declared interval [lower,upper]=[-2, 3] belongs to the function data. left=-2/1, right=-3/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, -12, -3], [-2, 3], -2, Yes, 21, -3, No, 33, not_comparable, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, -12, -3], [-2, 3], -2, Yes, 21, -3, No, 33, not_comparable, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0077","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69e7d8e609f8c9cec12dfeb9c922470ba5aa3f86498e037ab43d96089bd31cfd","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253077,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[1,-11,-2],\"interval\":[-3,4],\"left\":[-1,1],\"right\":[-2,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, -11, -2], not ascending-power order. The declared interval [lower,upper]=[-3, 4] belongs to the function data. left=-1/1, right=-2/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, -11, -2], [-3, 4], -1, Yes, 10, -2, Yes, 24, left_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, -11, -2], [-3, 4], -1, Yes, 10, -2, Yes, 24, left_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0078","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5c1ad80ed0699bf0a9501b1eb57123cda2fe59ea22b0d1dab207eff08c18a63","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253078,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[2,-10,-1],\"interval\":[-4,5],\"left\":[0,1],\"right\":[-1,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, -10, -1], not ascending-power order. The declared interval [lower,upper]=[-4, 5] belongs to the function data. left=0/1, right=-1/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, -10, -1], [-4, 5], 0, Yes, -1, -1, Yes, 11, left_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, -10, -1], [-4, 5], 0, Yes, -1, -1, Yes, 11, left_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0079","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b576190f971f27771acb6a7b2536fea35a05d35f2b5f98559b9b0c926e74e42b","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253079,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[3,-9,0],\"interval\":[-5,6],\"left\":[1,1],\"right\":[0,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, -9, 0], not ascending-power order. The declared interval [lower,upper]=[-5, 6] belongs to the function data. left=1/1, right=0/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, -9, 0], [-5, 6], 1, Yes, -6, 0, Yes, 0, left_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -9, 0], [-5, 6], 1, Yes, -6, 0, Yes, 0, left_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0080","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f40a12f25dbd7583125275d39ed4f781bc5edc38d2c2cfdeb8b64cca554510b8","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253080,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[0,-8,1],\"interval\":[-6,7],\"left\":[2,1],\"right\":[1,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, -8, 1], not ascending-power order. The declared interval [lower,upper]=[-6, 7] belongs to the function data. left=2/1, right=1/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, -8, 1], [-6, 7], 2, Yes, -15, 1, Yes, -7, left_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, -8, 1], [-6, 7], 2, Yes, -15, 1, Yes, -7, left_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0081","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93062a4d5d48551e3c7ec3274eba59e470252d6768d668183ad90c0259e30835","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253081,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[1,-7,2],\"interval\":[-7,8],\"left\":[-2,1],\"right\":[2,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, -7, 2], not ascending-power order. The declared interval [lower,upper]=[-7, 8] belongs to the function data. left=-2/1, right=2/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, -7, 2], [-7, 8], -2, Yes, 20, 2, Yes, -8, right_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, -7, 2], [-7, 8], -2, Yes, 20, 2, Yes, -8, right_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0082","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2bee8d1d1ed50c0bf3ea900b575b5b2db8cf55d87fec82ffd5683092659a57f2","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253082,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[2,-6,3],\"interval\":[-8,9],\"left\":[-1,1],\"right\":[3,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, -6, 3], not ascending-power order. The declared interval [lower,upper]=[-8, 9] belongs to the function data. left=-1/1, right=3/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, -6, 3], [-8, 9], -1, Yes, 11, 3, Yes, 3, right_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, -6, 3], [-8, 9], -1, Yes, 11, 3, Yes, 3, right_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0083","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b0837be5200a0f56d3e17af25c574da0207837aa5022afea2525aafd5020eab","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253083,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[3,-5,-3],\"interval\":[-9,10],\"left\":[0,1],\"right\":[-3,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, -5, -3], not ascending-power order. The declared interval [lower,upper]=[-9, 10] belongs to the function data. left=0/1, right=-3/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, -5, -3], [-9, 10], 0, Yes, -3, -3, Yes, 39, left_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -5, -3], [-9, 10], 0, Yes, -3, -3, Yes, 39, left_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0084","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a46c76979be4672958be320b5b37dc1edccf6f9aa3206f64e0939d3c1b5f30e","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253084,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[0,-4,-2],\"interval\":[-10,11],\"left\":[1,1],\"right\":[-2,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, -4, -2], not ascending-power order. The declared interval [lower,upper]=[-10, 11] belongs to the function data. left=1/1, right=-2/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, -4, -2], [-10, 11], 1, Yes, -6, -2, Yes, 6, left_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, -4, -2], [-10, 11], 1, Yes, -6, -2, Yes, 6, left_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0085","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80878debe6b7845e31459a41e1a52e943d6351cae0c1127ee29dc0fee5b8c888","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253085,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[1,-3,-1],\"interval\":[-11,12],\"left\":[2,1],\"right\":[-1,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, -3, -1], not ascending-power order. The declared interval [lower,upper]=[-11, 12] belongs to the function data. left=2/1, right=-1/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, -3, -1], [-11, 12], 2, Yes, -3, -1, Yes, 3, left_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, -3, -1], [-11, 12], 2, Yes, -3, -1, Yes, 3, left_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0086","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4572497b3450fcf79fda61c7fe762d69886bda1936ef4958110815d88599868a","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253086,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[2,-2,0],\"interval\":[-12,13],\"left\":[-2,1],\"right\":[0,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, -2, 0], not ascending-power order. The declared interval [lower,upper]=[-12, 13] belongs to the function data. left=-2/1, right=0/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, -2, 0], [-12, 13], -2, Yes, 12, 0, Yes, 0, right_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, -2, 0], [-12, 13], -2, Yes, 12, 0, Yes, 0, right_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0087","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"672e3ea9c81b79270dc258d64260f20d7cb5461c34b95103c97b7f4ff5fd7d63","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253087,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[3,-1,1],\"interval\":[-13,14],\"left\":[-1,1],\"right\":[1,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, -1, 1], not ascending-power order. The declared interval [lower,upper]=[-13, 14] belongs to the function data. left=-1/1, right=1/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, -1, 1], [-13, 14], -1, Yes, 5, 1, Yes, 3, right_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -1, 1], [-13, 14], -1, Yes, 5, 1, Yes, 3, right_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0088","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2719b6de8c288b8d8e68d0abf322857b0064c49940202c5ce8555044c14d3928","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253088,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[0,0,2],\"interval\":[-14,15],\"left\":[0,1],\"right\":[2,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 0, 2], not ascending-power order. The declared interval [lower,upper]=[-14, 15] belongs to the function data. left=0/1, right=2/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 0, 2], [-14, 15], 0, Yes, 2, 2, Yes, 2, equal, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 0, 2], [-14, 15], 0, Yes, 2, 2, Yes, 2, equal, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0089","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba397318dd5406b5b13ecdf888b83788749115db99af8ee99bb19d762ccada46","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253089,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[1,1,3],\"interval\":[-15,16],\"left\":[1,1],\"right\":[3,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, 1, 3], not ascending-power order. The declared interval [lower,upper]=[-15, 16] belongs to the function data. left=1/1, right=3/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, 1, 3], [-15, 16], 1, Yes, 5, 3, Yes, 15, left_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 1, 3], [-15, 16], 1, Yes, 5, 3, Yes, 15, left_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0090","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a41b4e0f5df545fa7a2aee0eae5baafaac476981f55fd38f1f425be78125bc15","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253090,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[2,2,-3],\"interval\":[-16,17],\"left\":[2,1],\"right\":[-3,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, 2, -3], not ascending-power order. The declared interval [lower,upper]=[-16, 17] belongs to the function data. left=2/1, right=-3/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, 2, -3], [-16, 17], 2, Yes, 9, -3, Yes, 9, equal, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 2, -3], [-16, 17], 2, Yes, 9, -3, Yes, 9, equal, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0091","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d23d8eab31c3be53e4d06cb88b39647b1c40e8668e57ae0d4fafc2e511489fb8","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253091,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[3,3,-2],\"interval\":[-17,18],\"left\":[-2,1],\"right\":[-2,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, 3, -2], not ascending-power order. The declared interval [lower,upper]=[-17, 18] belongs to the function data. left=-2/1, right=-2/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, 3, -2], [-17, 18], -2, Yes, 4, -2, Yes, 4, equal, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, 3, -2], [-17, 18], -2, Yes, 4, -2, Yes, 4, equal, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0092","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69fe6ad2a392b94dd81b937f378e2172d4875f5c14eeeeff3f217bd6591fdc32","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253092,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[0,4,-1],\"interval\":[-18,19],\"left\":[-1,1],\"right\":[-1,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 4, -1], not ascending-power order. The declared interval [lower,upper]=[-18, 19] belongs to the function data. left=-1/1, right=-1/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 4, -1], [-18, 19], -1, Yes, -5, -1, Yes, -5, equal, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 4, -1], [-18, 19], -1, Yes, -5, -1, Yes, -5, equal, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0093","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89df9ec580945514d47cb04fdcb1f7b5c44bb5b017687f6763f116bc2f759ba2","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253093,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[1,5,0],\"interval\":[-19,20],\"left\":[0,1],\"right\":[0,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, 5, 0], not ascending-power order. The declared interval [lower,upper]=[-19, 20] belongs to the function data. left=0/1, right=0/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, 5, 0], [-19, 20], 0, Yes, 0, 0, Yes, 0, equal, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 5, 0], [-19, 20], 0, Yes, 0, 0, Yes, 0, equal, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0094","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95dc8a40af78b4165c92ffb903cfc9d714904fc878389dcf817957982095f118","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253094,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[2,6,1],\"interval\":[-20,21],\"left\":[1,1],\"right\":[1,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, 6, 1], not ascending-power order. The declared interval [lower,upper]=[-20, 21] belongs to the function data. left=1/1, right=1/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, 6, 1], [-20, 21], 1, Yes, 9, 1, Yes, 9, equal, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 6, 1], [-20, 21], 1, Yes, 9, 1, Yes, 9, equal, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0095","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c51c1ff24acd1de76b2e8f037dd4f24ed9f1fe4e33c2ad34f61cd4cf3a09e0c0","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253095,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[3,7,2],\"interval\":[-21,22],\"left\":[2,1],\"right\":[2,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, 7, 2], not ascending-power order. The declared interval [lower,upper]=[-21, 22] belongs to the function data. left=2/1, right=2/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, 7, 2], [-21, 22], 2, Yes, 28, 2, Yes, 28, equal, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, 7, 2], [-21, 22], 2, Yes, 28, 2, Yes, 28, equal, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0096","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d936782304ec4ab58e36f02441d14d883d19deece5105626591df8386e329be","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253096,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[0,8,3],\"interval\":[-22,23],\"left\":[-2,1],\"right\":[3,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 8, 3], not ascending-power order. The declared interval [lower,upper]=[-22, 23] belongs to the function data. left=-2/1, right=3/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 8, 3], [-22, 23], -2, Yes, -13, 3, Yes, 27, left_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 8, 3], [-22, 23], -2, Yes, -13, 3, Yes, 27, left_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0097","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ae66021e8c8f6cb7943a498112f9514a0e29884c4a9c1cc63ec7b806dcda4dd","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253097,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[1,9,-3],\"interval\":[-23,24],\"left\":[-1,1],\"right\":[-3,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[1, 9, -3], not ascending-power order. The declared interval [lower,upper]=[-23, 24] belongs to the function data. left=-1/1, right=-3/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([1, 9, -3], [-23, 24], -1, Yes, -11, -3, Yes, -21, right_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 9, -3], [-23, 24], -1, Yes, -11, -3, Yes, -21, right_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0098","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db12fa7ec6e8f6ebc9dc3935784e06e16364a01255b07147056cfbe1de3b6e47","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253098,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[2,10,-2],\"interval\":[-24,25],\"left\":[0,1],\"right\":[-2,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[2, 10, -2], not ascending-power order. The declared interval [lower,upper]=[-24, 25] belongs to the function data. left=0/1, right=-2/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([2, 10, -2], [-24, 25], 0, Yes, -2, -2, Yes, -14, right_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 10, -2], [-24, 25], 0, Yes, -2, -2, Yes, -14, right_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0099","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"712d0d34fd170bd5772e4b5ff7f2f683d80079a7a67b9beb5656eabfe9148bd8","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253099,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[3,11,-1],\"interval\":[-25,26],\"left\":[1,1],\"right\":[-1,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[3, 11, -1], not ascending-power order. The declared interval [lower,upper]=[-25, 26] belongs to the function data. left=1/1, right=-1/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([3, 11, -1], [-25, 26], 1, Yes, 13, -1, Yes, -9, right_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, 11, -1], [-25, 26], 1, Yes, 13, -1, Yes, -9, right_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r03-convex-optimization-problem-0100","batch_id":"batch-20261013-advance-20261025-r03-convex-optimization-problem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e403c8553d4110613d1f9c1b7f236e63c33f8cd02d41f266a7d8c709b4dde0bc","family_id":"qf-mod25-convexopt-pairwise-feasible-comparison","seed":202610253100,"semantic_template":"qf-mod25-convexopt-pairwise-feasible-comparison/{\"coefficients\":[0,12,0],\"interval\":[-26,27],\"left\":[2,1],\"right\":[0,1]}"},"fields":{"title":"Compare Only Two Eligible Candidates, Not the Whole Problem","stem":"Use the one-variable quadratic f(t)=a*t²+b*t+c with coefficients [a,b,c]=[0, 12, 0], not ascending-power order. The declared interval [lower,upper]=[-26, 27] belongs to the function data. left=2/1, right=0/1. Give (coefficients,interval,left,left feasible?,left value,right,right feasible?,right value,relation,pairwise_only_no_global_optimum_claim). The relation is not_comparable if either point is infeasible; otherwise left_lower, right_lower or equal. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Keep convex-objective qualification, convex feasible-domain qualification, point feasibility, evaluation and comparison separate. These checks supply no general solver, global optimum, existence, uniqueness, KKT or convergence claim.","answer":{"kind":"expression","value":"([0, 12, 0], [-26, 27], 2, Yes, 24, 0, Yes, 0, right_lower, pairwise_only_no_global_optimum_claim)","explanation":"Compute values even for ineligible points but compare only when both are feasible. A lower value among these two is not a global optimum certificate. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 12, 0], [-26, 27], 2, Yes, 24, 0, Yes, 0, right_lower, pairwise_only_no_global_optimum_claim)."},"skills":["Compare Only Two Eligible Candidates, Not the Whole Problem"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}