{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0001","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"090ca8f2f025d9eaecbccc0363be2c9bf994017021b1e56c21c6a6b5362df514","family_id":"qf-ftc-accumulation-derivative","seed":202610091001,"semantic_template":"qf-ftc-accumulation-derivative/[[0]]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [0], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(-2). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 0/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0002","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ab9776ee7ec830917fa2b19c33ebd59097253dbf5bee19ae14c34e35e2ce03d","family_id":"qf-ftc-accumulation-derivative","seed":202610091002,"semantic_template":"qf-ftc-accumulation-derivative/[[3]]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [3], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(-1). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"3/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 3/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0003","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8337a5617bae37f6383568258acf86d50f2f94bcefe40187da2062b25b74584","family_id":"qf-ftc-accumulation-derivative","seed":202610091003,"semantic_template":"qf-ftc-accumulation-derivative/[[-10,3,1,3],0]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [-10, 3, 1, 3], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(0). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-10/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -10/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0004","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0cc630a8ebc52af497b286bf181e501a1fadb926cc035b77f68cc96bf529f721","family_id":"qf-ftc-accumulation-derivative","seed":202610091004,"semantic_template":"qf-ftc-accumulation-derivative/[[-9,4,-1,4],1]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [-9, 4, -1, 4], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(1). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-2/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -2/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0005","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b7b35fd944c5811dbc2bdb5d3e7264fd4cebd9944e221b435b4d4922894f7833","family_id":"qf-ftc-accumulation-derivative","seed":202610091005,"semantic_template":"qf-ftc-accumulation-derivative/[[-8,5,0,1],2]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [-8, 5, 0, 1], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(2). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"10/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 10/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0006","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ab15645db1765fda46c954cf5070edf9f54a1c39ecd3f6f71d35cd6173c38e8","family_id":"qf-ftc-accumulation-derivative","seed":202610091006,"semantic_template":"qf-ftc-accumulation-derivative/[[-7,6,1,2],-2]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [-7, 6, 1, 2], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(-2). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-31/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -31/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0007","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe1f8f49dc544b7e7fae32e1097b1ad446c724f3d47aec8769c37b9daeb4297a","family_id":"qf-ftc-accumulation-derivative","seed":202610091007,"semantic_template":"qf-ftc-accumulation-derivative/[[-6,7,-1,3],-1]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [-6, 7, -1, 3], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(-1). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-17/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -17/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0008","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"29268c274a69e60343035524f5204a8370b52614ef42c5759d718c07cc8a6773","family_id":"qf-ftc-accumulation-derivative","seed":202610091008,"semantic_template":"qf-ftc-accumulation-derivative/[[-5,8,0,4],0]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [-5, 8, 0, 4], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(0). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-5/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -5/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0009","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"882fd8b4d6546c21ff3532719af966e4b6967e344ac968dd9a54157667583049","family_id":"qf-ftc-accumulation-derivative","seed":202610091009,"semantic_template":"qf-ftc-accumulation-derivative/[[-4,9,1,1],1]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [-4, 9, 1, 1], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(1). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"7/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 7/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0010","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ccd02cdbe8f747b34ec8cf6283da6a87b8f1ab0d11aed6aa1217d52a2189b3d0","family_id":"qf-ftc-accumulation-derivative","seed":202610091010,"semantic_template":"qf-ftc-accumulation-derivative/[[-3,10,-1,2],2]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [-3, 10, -1, 2], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(2). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"29/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 29/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0011","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2184d41fedd9c75e31f682180bed4ecba14e1aeeeaf3d2feb951565257ac3464","family_id":"qf-ftc-accumulation-derivative","seed":202610091011,"semantic_template":"qf-ftc-accumulation-derivative/[[-2,11,0,3],-2]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [-2, 11, 0, 3], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(-2). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-48/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -48/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0012","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4cd3192366cdb10d25684b0cc25192838600e0471675423d1145d775c4fca928","family_id":"qf-ftc-accumulation-derivative","seed":202610091012,"semantic_template":"qf-ftc-accumulation-derivative/[[-1,12,1,4],-1]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [-1, 12, 1, 4], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(-1). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-16/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -16/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0013","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"836aa901426e547b9f87ba9dbf824394dde4bc74b1761ef028189088f91d41cd","family_id":"qf-ftc-accumulation-derivative","seed":202610091013,"semantic_template":"qf-ftc-accumulation-derivative/[[0,13,-1,1],0]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [0, 13, -1, 1], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(0). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 0/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0014","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cb4b2dbeac8647d29e58e318d5b615d0d0d07f759fe0ace8444f5fa28109d4d0","family_id":"qf-ftc-accumulation-derivative","seed":202610091014,"semantic_template":"qf-ftc-accumulation-derivative/[[1,14,0,2],1]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [1, 14, 0, 2], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(1). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"17/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 17/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0015","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91037f5ddfa99ed765e28347ead6c83c04bcd73089ff44d974a8bc4c341b5e92","family_id":"qf-ftc-accumulation-derivative","seed":202610091015,"semantic_template":"qf-ftc-accumulation-derivative/[[2,15,1,3],2]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [2, 15, 1, 3], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(2). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"60/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 60/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0016","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a7318915fede767a1001bbd0f37e2de1abd2990b5b22f2678a19d126bbe21ea","family_id":"qf-ftc-accumulation-derivative","seed":202610091016,"semantic_template":"qf-ftc-accumulation-derivative/[[3,16,-1,4],-2]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [3, 16, -1, 4], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(-2). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-65/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -65/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0017","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14cfc5bcd7a764d30e95b248fb150b60802c3ea81fb2832114b14d5e5d415e8d","family_id":"qf-ftc-accumulation-derivative","seed":202610091017,"semantic_template":"qf-ftc-accumulation-derivative/[[4,17,0,1],-1]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [4, 17, 0, 1], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(-1). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-14/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -14/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0018","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"feb697c158ce6cb00689f663d45125b832ff4996d3cc859f39e252c5651d6360","family_id":"qf-ftc-accumulation-derivative","seed":202610091018,"semantic_template":"qf-ftc-accumulation-derivative/[[5,18,1,2],0]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [5, 18, 1, 2], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(0). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"5/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 5/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0019","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"62cdbfe3c3d7887196b264137ec803a00398709116806b41abe8b8846ccfa2a2","family_id":"qf-ftc-accumulation-derivative","seed":202610091019,"semantic_template":"qf-ftc-accumulation-derivative/[[6,19,-1,3],1]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [6, 19, -1, 3], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(1). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"27/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 27/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0020","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"886ab6fe470cf486f6aa08521ae9df524f73e58dc7a7dad359fc7ff89c04e537","family_id":"qf-ftc-accumulation-derivative","seed":202610091020,"semantic_template":"qf-ftc-accumulation-derivative/[[7,20,0,4],2]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [7, 20, 0, 4], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(2). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"79/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 79/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0021","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11c23893cdf1581eddaf5034a3c809b18763d6219ef8522cfb08eaa628c8c05a","family_id":"qf-ftc-accumulation-derivative","seed":202610091021,"semantic_template":"qf-ftc-accumulation-derivative/[[8,21,1,1],-2]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [8, 21, 1, 1], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(-2). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-38/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -38/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0022","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e555bcdf828dc46d8104d3a15b994ec4decfa7410e1849bfcc751c31c5b8aa89","family_id":"qf-ftc-accumulation-derivative","seed":202610091022,"semantic_template":"qf-ftc-accumulation-derivative/[[9,22,-1,2],-1]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [9, 22, -1, 2], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(-1). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-16/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving -16/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0023","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7bd1a85ff1c107ef928a8308865e7d73c5f4acb3c53a643a6878d8987e7ead53","family_id":"qf-ftc-accumulation-derivative","seed":202610091023,"semantic_template":"qf-ftc-accumulation-derivative/[[10,23,0,3],0]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [10, 23, 0, 3], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(0). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"10/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 10/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0024","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b3cbac2454b1469aad81af1bde9617a02d0ec95096aefdbd2d1aae5f49627b3","family_id":"qf-ftc-accumulation-derivative","seed":202610091024,"semantic_template":"qf-ftc-accumulation-derivative/[[11,24,1,4],1]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [11, 24, 1, 4], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(1). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"40/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 40/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0025","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"70d57ee44e2b4e2b00f50bf5649ed6dae8851e654760f59fe393bfbf0ece1d03","family_id":"qf-ftc-accumulation-derivative","seed":202610091025,"semantic_template":"qf-ftc-accumulation-derivative/[[12,25,-1,1],2]"},"fields":{"title":"Point Derivative of a Continuous Accumulating Integral","stem":"On all R, the polynomial f is continuous and f has ascending coefficients [12, 25, -1, 1], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(2). Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"66/1","explanation":"Under the stated full-real continuity premise, G′(x)=f(x), giving 66/1. This exact polynomial instance is not a new general FTC formal proof."},"skills":["Point Derivative of a Continuous Accumulating Integral"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0026","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9f68e27cce2ef8c1f508f318e6967a8245c031790ff00305f5e760dd93a0ddd","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091026,"semantic_template":"qf-ftc-moving-bounds-derivative/[[1],1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [1], constant term first. Let lower l(x)=(1)x+(0) and upper u(x)=(2)x+(0). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"1/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 1/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0027","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ed0fdb526cf7cfa8b9d04f068d219c9ae6327664b45f1a471611105be17c9ec","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091027,"semantic_template":"qf-ftc-moving-bounds-derivative/[\"zero_integrand\"]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [0], constant term first. Let lower l(x)=(1)x+(0) and upper u(x)=(3)x+(0). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 0/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0028","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"896c7568fb7c1d714bd6477f1b25bc6b4e0e342d001bf92197d55eea1d5a8488","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091028,"semantic_template":"qf-ftc-moving-bounds-derivative/[\"identical_bounds\"]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [3, 1], constant term first. Let lower l(x)=(2)x+(1) and upper u(x)=(2)x+(1). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 0/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0029","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d22a1f60e989e8da59cda7a174fb38630baaa0aa6d832654852a7f74df3ea035","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091029,"semantic_template":"qf-ftc-moving-bounds-derivative/[[-9,4,-1,4],-1,-3,0,-9,1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [-9, 4, -1, 4], constant term first. Let lower l(x)=(-1)x+(-3) and upper u(x)=(0)x+(-9). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-297/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -297/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0030","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"974ad2213e23fa786d5334673b65130b1c567a326bd382fcb4f7fde47da011f5","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091030,"semantic_template":"qf-ftc-moving-bounds-derivative/[[-8,5,0,1],2,0,4,-8,-2]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [-8, 5, 0, 1], constant term first. Let lower l(x)=(2)x+(0) and upper u(x)=(4)x+(-8). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-16552/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -16552/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0031","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8f9de086fe733a17bbcb114b4cfd5e313878e1aff5d91e73ee9f47f19f47530","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091031,"semantic_template":"qf-ftc-moving-bounds-derivative/[[-7,6,1,2],-3,-1,0,-7,-1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [-7, 6, 1, 2], constant term first. Let lower l(x)=(-3)x+(-1) and upper u(x)=(0)x+(-7). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"75/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 75/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0032","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b31bba6527f9f3546867557782fe6771ea082da2c7b9888f9104f6c3026d2aa","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091032,"semantic_template":"qf-ftc-moving-bounds-derivative/[[-6,7,-1,3],1,-2,2,-6,0]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [-6, 7, -1, 3], constant term first. Let lower l(x)=(1)x+(-2) and upper u(x)=(2)x+(-6). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-1416/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -1416/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0033","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c329cd5c38fed273547e8383efe6b0b0e266ee00051112ab8004cf5a976bea6","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091033,"semantic_template":"qf-ftc-moving-bounds-derivative/[[-5,8,0,4],-2,-3,0,-5,1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [-5, 8, 0, 4], constant term first. Let lower l(x)=(-2)x+(-3) and upper u(x)=(0)x+(-5). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-1090/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -1090/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0034","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64e0048371805d347b687ee9a6f3282c7533127c8d4c4e5291ab60b8976ac6ce","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091034,"semantic_template":"qf-ftc-moving-bounds-derivative/[[-4,9,1,1],3,0,6,-4,-2]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [-4, 9, 1, 1], constant term first. Let lower l(x)=(3)x+(0) and upper u(x)=(6)x+(-4). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-23214/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -23214/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0035","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d2221f189032eb07e991c38e8bb2608741ee1a16cb2eed1c68526239c640bbc","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091035,"semantic_template":"qf-ftc-moving-bounds-derivative/[[-3,10,-1,2],-1,-1,0,-3,-1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [-3, 10, -1, 2], constant term first. Let lower l(x)=(-1)x+(-1) and upper u(x)=(0)x+(-3). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-3/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -3/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0036","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36915f78a86b377f1d619ccf352c492081b64c92c178e936eca87a8c9022967f","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091036,"semantic_template":"qf-ftc-moving-bounds-derivative/[[-2,11,0,3],2,-2,4,-2,0]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [-2, 11, 0, 3], constant term first. Let lower l(x)=(2)x+(-2) and upper u(x)=(4)x+(-2). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-96/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -96/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0037","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8932d2d79cc19596376a1c09a9d1d61cb032c63ac0d29b71a4b2cdad575cf036","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091037,"semantic_template":"qf-ftc-moving-bounds-derivative/[[-1,12,1,4],-3,-3,0,-1,1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [-1, 12, 1, 4], constant term first. Let lower l(x)=(-3)x+(-3) and upper u(x)=(0)x+(-1). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-2703/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -2703/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0038","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95549c248a948df882eb00570f2c2c8bcc1ab6b9e0414a8c4e5371a2eebac52d","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091038,"semantic_template":"qf-ftc-moving-bounds-derivative/[[0,13,-1,1],1,0,2,0,-2]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [0, 13, -1, 1], constant term first. Let lower l(x)=(1)x+(0) and upper u(x)=(2)x+(0). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-226/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -226/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0039","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"270be42ff515b2ca976b5ab19fdb6b62b761b289a220a297eaa67f37949ed0b7","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091039,"semantic_template":"qf-ftc-moving-bounds-derivative/[[1,14,0,2],-2,-1,0,1,-1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [1, 14, 0, 2], constant term first. Let lower l(x)=(-2)x+(-1) and upper u(x)=(0)x+(1). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"34/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 34/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0040","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a644043ac7e94071c2ba12a6e90187381eafd1fd3cad59027a76dfe8b678dfe3","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091040,"semantic_template":"qf-ftc-moving-bounds-derivative/[[2,15,1,3],3,-2,6,2,0]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [2, 15, 1, 3], constant term first. Let lower l(x)=(3)x+(-2) and upper u(x)=(6)x+(2). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"504/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 504/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0041","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"195966ec173876d339e0b6177684900506bfa9c9fcadc083da6bbe38e905de2f","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091041,"semantic_template":"qf-ftc-moving-bounds-derivative/[[3,16,-1,4],-1,-3,0,3,1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [3, 16, -1, 4], constant term first. Let lower l(x)=(-1)x+(-3) and upper u(x)=(0)x+(3). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-333/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -333/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0042","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d88aaa671c2d111e86a0a5629250d736b69b9cb273bb644306b80e2faa05f35e","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091042,"semantic_template":"qf-ftc-moving-bounds-derivative/[[4,17,0,1],2,0,4,4,-2]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [4, 17, 0, 1], constant term first. Let lower l(x)=(2)x+(0) and upper u(x)=(4)x+(4). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-256/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -256/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0043","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7584241e56529c09d26112085dadc1072d1a430dd01185cb78f52555f9b690a7","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091043,"semantic_template":"qf-ftc-moving-bounds-derivative/[[5,18,1,2],-3,-1,0,5,-1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [5, 18, 1, 2], constant term first. Let lower l(x)=(-3)x+(-1) and upper u(x)=(0)x+(5). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"183/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 183/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0044","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f940bf759e6d196651c4e764376b585377e733e91d394574115fd8e24bafcb07","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091044,"semantic_template":"qf-ftc-moving-bounds-derivative/[[6,19,-1,3],1,-2,2,6,0]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [6, 19, -1, 3], constant term first. Let lower l(x)=(1)x+(-2) and upper u(x)=(2)x+(6). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"1524/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 1524/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0045","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e5a6c722bd6e6d7c224568f4e68b773617058496f87a3dfaa4452c80a4d511c2","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091045,"semantic_template":"qf-ftc-moving-bounds-derivative/[[7,20,0,4],-2,-3,0,7,1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [7, 20, 0, 4], constant term first. Let lower l(x)=(-2)x+(-3) and upper u(x)=(0)x+(7). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-1186/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -1186/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0046","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aebc006f98be339748c316792a43b76ac4c84760b7d5f865acb63611efe4d15b","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091046,"semantic_template":"qf-ftc-moving-bounds-derivative/[[8,21,1,1],3,0,6,8,-2]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [8, 21, 1, 1], constant term first. Let lower l(x)=(3)x+(0) and upper u(x)=(6)x+(8). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"150/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 150/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0047","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e0702f0ac3069c43f3fb763be0ea0b5ce7a98302bbe57975893963a5f5495097","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091047,"semantic_template":"qf-ftc-moving-bounds-derivative/[[9,22,-1,2],-1,-1,0,9,-1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [9, 22, -1, 2], constant term first. Let lower l(x)=(-1)x+(-1) and upper u(x)=(0)x+(9). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"9/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 9/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0048","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0713731974d062d042588331495bab041e6adb05260ca824418023d4adb8c48b","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091048,"semantic_template":"qf-ftc-moving-bounds-derivative/[[10,23,0,3],2,-2,4,10,0]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [10, 23, 0, 3], constant term first. Let lower l(x)=(2)x+(-2) and upper u(x)=(4)x+(10). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"13080/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 13080/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0049","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b72bec082a34d405a68318eeaf1f963a2a09381b6d7a0489257474c748218c20","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091049,"semantic_template":"qf-ftc-moving-bounds-derivative/[[11,24,1,4],-3,-3,0,11,1]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [11, 24, 1, 4], constant term first. Let lower l(x)=(-3)x+(-3) and upper u(x)=(0)x+(11). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-2883/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -2883/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0050","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3c17c8c62e73bdf699042df8c787f64d69a8737f24ae713bf706592b9bdf5e3","family_id":"qf-ftc-moving-bounds-derivative","seed":202610091050,"semantic_template":"qf-ftc-moving-bounds-derivative/[[12,25,-1,1],1,0,2,12,-2]"},"fields":{"title":"Derivative of an Oriented Integral with Moving Bounds","stem":"On all R, continuous f has ascending coefficients [12, 25, -1, 1], constant term first. Let lower l(x)=(1)x+(0) and upper u(x)=(2)x+(12). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"1370/1","explanation":"Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 1370/1. Bounds coinciding at this point do not force equal velocities or a zero derivative."},"skills":["Derivative of an Oriented Integral with Moving Bounds"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0051","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d579807c5ed4161ba75673c43537670c805a574c37a7c08b9a094327a1a8db1d","family_id":"qf-ftc-anchor-value-derivative","seed":202610091051,"semantic_template":"qf-ftc-anchor-value-derivative/[[0]]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [0], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 0/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 0/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0052","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4cdf22690c61bb318cb2431a579449ae4493eb580026c9776027d2276f0185f4","family_id":"qf-ftc-anchor-value-derivative","seed":202610091052,"semantic_template":"qf-ftc-anchor-value-derivative/[[3]]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [3], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 3/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 3/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0053","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4304fd9248d7a7293bd54ff77d5f7d3caff4ef60a6cceb2f3983b715770c2cf6","family_id":"qf-ftc-anchor-value-derivative","seed":202610091053,"semantic_template":"qf-ftc-anchor-value-derivative/[[-10,3,1,3],1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [-10, 3, 1, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, -3/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -3/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0054","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d92d37afc7e39cdbf5df0e3ae314c7f3799599bf283f3afc3c5f7ae0108dc48f","family_id":"qf-ftc-anchor-value-derivative","seed":202610091054,"semantic_template":"qf-ftc-anchor-value-derivative/[[-9,4,-1,4],2]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [-9, 4, -1, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 27/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 27/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0055","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e028eaa1a0484fb96459f7287542b048f3c7286edb2a80fbd24de1cfa13aebbf","family_id":"qf-ftc-anchor-value-derivative","seed":202610091055,"semantic_template":"qf-ftc-anchor-value-derivative/[[-8,5,0,1],-1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [-8, 5, 0, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, -14/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -14/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0056","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01c3dddde3d26d8b0714e60a51f345d26457ca9d8e4a8bebe5a7471d6e6202fc","family_id":"qf-ftc-anchor-value-derivative","seed":202610091056,"semantic_template":"qf-ftc-anchor-value-derivative/[[-7,6,1,2],0]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [-7, 6, 1, 2], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, -7/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -7/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0057","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd9e0191b22757315b4f24ffab7e1015d38d4f1caa25bf71a250f31394518cd2","family_id":"qf-ftc-anchor-value-derivative","seed":202610091057,"semantic_template":"qf-ftc-anchor-value-derivative/[[-6,7,-1,3],1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [-6, 7, -1, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 3/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 3/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0058","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5001d794d5140ea0f8441c939adcc39a66d22bb4cce102aa2866f7c27bfcae17","family_id":"qf-ftc-anchor-value-derivative","seed":202610091058,"semantic_template":"qf-ftc-anchor-value-derivative/[[-5,8,0,4],2]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [-5, 8, 0, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 43/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 43/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0059","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9849fbba931198304de632be72811fbcbbc12113b0ddac24ff8f5727ca403d69","family_id":"qf-ftc-anchor-value-derivative","seed":202610091059,"semantic_template":"qf-ftc-anchor-value-derivative/[[-4,9,1,1],-1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [-4, 9, 1, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, -13/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -13/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0060","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5e0cda7f6f658e92f8755311a303631a6940cad380fe82e78cce3695f49e916","family_id":"qf-ftc-anchor-value-derivative","seed":202610091060,"semantic_template":"qf-ftc-anchor-value-derivative/[[-3,10,-1,2],0]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [-3, 10, -1, 2], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, -3/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -3/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0061","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d78d1f494a063c1599fe520f49191e50da8a29f10783f6a60adfb9dd13fbca8","family_id":"qf-ftc-anchor-value-derivative","seed":202610091061,"semantic_template":"qf-ftc-anchor-value-derivative/[[-2,11,0,3],1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [-2, 11, 0, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 12/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 12/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0062","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e88db4b9fef3300d20393e6e238f2f1526ff979430f46452503eb22d8950b733","family_id":"qf-ftc-anchor-value-derivative","seed":202610091062,"semantic_template":"qf-ftc-anchor-value-derivative/[[-1,12,1,4],2]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [-1, 12, 1, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 59/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 59/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0063","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dc142d3ad0fb82e16ad7d25081a499b0f0e4f4e7006e8ff60fe74ffc8c1e4c0","family_id":"qf-ftc-anchor-value-derivative","seed":202610091063,"semantic_template":"qf-ftc-anchor-value-derivative/[[0,13,-1,1],-1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [0, 13, -1, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, -15/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -15/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0064","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d36b749336378dfd8d0d0609f68aa483a83b8f4c6f61c569eb38dd2c3ecade08","family_id":"qf-ftc-anchor-value-derivative","seed":202610091064,"semantic_template":"qf-ftc-anchor-value-derivative/[[1,14,0,2],0]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [1, 14, 0, 2], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 1/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 1/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0065","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"355984f0edd6c05562a547cf2a77f72dd59a1480916b725b104b50a21d621365","family_id":"qf-ftc-anchor-value-derivative","seed":202610091065,"semantic_template":"qf-ftc-anchor-value-derivative/[[2,15,1,3],1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [2, 15, 1, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 21/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 21/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0066","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d0f6c4336074a25c65dbde44f3000aa74808d8ff1232d9d66d856590d0ab3c6b","family_id":"qf-ftc-anchor-value-derivative","seed":202610091066,"semantic_template":"qf-ftc-anchor-value-derivative/[[3,16,-1,4],2]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [3, 16, -1, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 63/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 63/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0067","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e58df36278fe2587616cf912ca94bad7c60db3a192dfcfdcc0e8420f362df870","family_id":"qf-ftc-anchor-value-derivative","seed":202610091067,"semantic_template":"qf-ftc-anchor-value-derivative/[[4,17,0,1],-1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [4, 17, 0, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, -14/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -14/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0068","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66b4a3051a579ae8a84b3da291aa70852e99ed2397878b1b5b5bda95beb0766a","family_id":"qf-ftc-anchor-value-derivative","seed":202610091068,"semantic_template":"qf-ftc-anchor-value-derivative/[[5,18,1,2],0]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [5, 18, 1, 2], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 5/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 5/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0069","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be38aa3ec58411a26a7c92a49e6f7edc85efdd0450582cd9816ae5aba7d23720","family_id":"qf-ftc-anchor-value-derivative","seed":202610091069,"semantic_template":"qf-ftc-anchor-value-derivative/[[6,19,-1,3],1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [6, 19, -1, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 27/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 27/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0070","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"29e98960293dd3440eab32106dfae9aa3f55c969a35282e54187a0e968bdda2c","family_id":"qf-ftc-anchor-value-derivative","seed":202610091070,"semantic_template":"qf-ftc-anchor-value-derivative/[[7,20,0,4],2]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [7, 20, 0, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 79/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 79/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0071","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"040c96b12303e6a863be83d8d86fed08e43e34a46a923f7d8bedda06101bbf68","family_id":"qf-ftc-anchor-value-derivative","seed":202610091071,"semantic_template":"qf-ftc-anchor-value-derivative/[[8,21,1,1],-1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [8, 21, 1, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, -13/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -13/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0072","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d7bf4313e432130c6d9a910bee302dd5d6f327a8f70dd64488a12a9ac9aee63","family_id":"qf-ftc-anchor-value-derivative","seed":202610091072,"semantic_template":"qf-ftc-anchor-value-derivative/[[9,22,-1,2],0]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [9, 22, -1, 2], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 9/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 9/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0073","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"331c2faa00efc006d710998883b0ee35de0c6a06accbffd65f6e42f8b03ec4d0","family_id":"qf-ftc-anchor-value-derivative","seed":202610091073,"semantic_template":"qf-ftc-anchor-value-derivative/[[10,23,0,3],1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [10, 23, 0, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 36/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 36/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0074","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"409da18b9a614771b6efdbc6b76bfa384d01804e4f0b13f7a32dd89bb828b275","family_id":"qf-ftc-anchor-value-derivative","seed":202610091074,"semantic_template":"qf-ftc-anchor-value-derivative/[[11,24,1,4],2]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [11, 24, 1, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 95/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 95/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0075","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a010310e17913e2ad8d8f328cadfed7db4d20b23162429f59f4a5119c6c6f545","family_id":"qf-ftc-anchor-value-derivative","seed":202610091075,"semantic_template":"qf-ftc-anchor-value-derivative/[[12,25,-1,1],-1]"},"fields":{"title":"Accumulation Value and Derivative at the Anchor","stem":"On all R, continuous f has ascending coefficients [12, 25, -1, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, -15/1)","explanation":"The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -15/1). A zero integral value does not imply a zero instantaneous derivative."},"skills":["Accumulation Value and Derivative at the Anchor"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0076","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0bd483b3f9d2f5837fd37bccac028f86f251834f15db8f8245cc18a289600583","family_id":"qf-ftc-primitive-hypothesis","seed":202610091076,"semantic_template":"qf-ftc-primitive-hypothesis/[[0],[]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [0], constant term first. Candidate F has ascending rational coefficients [5/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0077","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"efd136d6c99b782b4cf6c1351dcc8a8fc815199a0a48a645b91fe982f933abac","family_id":"qf-ftc-primitive-hypothesis","seed":202610091077,"semantic_template":"qf-ftc-primitive-hypothesis/[[1],[\"1/1\",\"0/1\",\"0/1\",\"1/1\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [1], constant term first. Candidate F has ascending rational coefficients [0/1, 1/1, 0/1, 0/1, 1/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"No","explanation":"Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0078","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"728bdd781b633df5a492f131e963271cf7ec6694249852a60379c9cc8722b13a","family_id":"qf-ftc-primitive-hypothesis","seed":202610091078,"semantic_template":"qf-ftc-primitive-hypothesis/[[-10,3,1,3],[\"-10/1\",\"3/2\",\"1/3\",\"3/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [-10, 3, 1, 3], constant term first. Candidate F has ascending rational coefficients [-10/1, -10/1, 3/2, 1/3, 3/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0079","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e9bd89f81a8a921a767571a93e1139fb91087b0c0a74e96596127ea7348f922","family_id":"qf-ftc-primitive-hypothesis","seed":202610091079,"semantic_template":"qf-ftc-primitive-hypothesis/[[-9,4,-1,4],[\"-9/1\",\"2/1\",\"-1/3\",\"1/1\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [-9, 4, -1, 4], constant term first. Candidate F has ascending rational coefficients [-9/1, -9/1, 2/1, -1/3, 1/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0080","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7217046bd0b13946a7342b29fe02c693905abe1c4b957ddc640495fe28f90747","family_id":"qf-ftc-primitive-hypothesis","seed":202610091080,"semantic_template":"qf-ftc-primitive-hypothesis/[[-8,5,0,1],[\"-8/1\",\"5/2\",\"0/1\",\"5/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [-8, 5, 0, 1], constant term first. Candidate F has ascending rational coefficients [-8/1, -8/1, 5/2, 0/1, 5/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"No","explanation":"Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0081","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa54e86eacf92d36d9e6505692856bfb93a7298b239d6ef5e2ddfff08588ded0","family_id":"qf-ftc-primitive-hypothesis","seed":202610091081,"semantic_template":"qf-ftc-primitive-hypothesis/[[-7,6,1,2],[\"-7/1\",\"3/1\",\"1/3\",\"1/2\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [-7, 6, 1, 2], constant term first. Candidate F has ascending rational coefficients [-7/1, -7/1, 3/1, 1/3, 1/2]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0082","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4a2f528ad92aab40c2bc1c14cbc7c68098418e20908b8757b64debc72816cff3","family_id":"qf-ftc-primitive-hypothesis","seed":202610091082,"semantic_template":"qf-ftc-primitive-hypothesis/[[-6,7,-1,3],[\"-6/1\",\"7/2\",\"-1/3\",\"3/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [-6, 7, -1, 3], constant term first. Candidate F has ascending rational coefficients [-6/1, -6/1, 7/2, -1/3, 3/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0083","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c6de228940060de02f968097b3cd6ee30e50bb517a979b0e8f54e289b378306","family_id":"qf-ftc-primitive-hypothesis","seed":202610091083,"semantic_template":"qf-ftc-primitive-hypothesis/[[-5,8,0,4],[\"-5/1\",\"4/1\",\"0/1\",\"2/1\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [-5, 8, 0, 4], constant term first. Candidate F has ascending rational coefficients [-5/1, -5/1, 4/1, 0/1, 2/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"No","explanation":"Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0084","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1da696c51507bc10e90dbf023b83361b34bff1c80707b85f0d3aabd60046ee88","family_id":"qf-ftc-primitive-hypothesis","seed":202610091084,"semantic_template":"qf-ftc-primitive-hypothesis/[[-4,9,1,1],[\"-4/1\",\"9/2\",\"1/3\",\"1/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [-4, 9, 1, 1], constant term first. Candidate F has ascending rational coefficients [-4/1, -4/1, 9/2, 1/3, 1/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0085","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc2b93319771afc7757d4e92a967c8c1ac71449c027613b1ec84907ef8ca9e24","family_id":"qf-ftc-primitive-hypothesis","seed":202610091085,"semantic_template":"qf-ftc-primitive-hypothesis/[[-3,10,-1,2],[\"-3/1\",\"5/1\",\"-1/3\",\"1/2\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [-3, 10, -1, 2], constant term first. Candidate F has ascending rational coefficients [-3/1, -3/1, 5/1, -1/3, 1/2]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0086","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e37e5714d00d31396c42ce50d7b0a36fe13c31cc99aacfeb34be3fb0e07bda5","family_id":"qf-ftc-primitive-hypothesis","seed":202610091086,"semantic_template":"qf-ftc-primitive-hypothesis/[[-2,11,0,3],[\"-2/1\",\"11/2\",\"0/1\",\"7/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [-2, 11, 0, 3], constant term first. Candidate F has ascending rational coefficients [-2/1, -2/1, 11/2, 0/1, 7/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"No","explanation":"Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0087","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"38dc3d1c9c96480f2d7429a10576016e6ae35f502ae331ef641ba6758b349ebd","family_id":"qf-ftc-primitive-hypothesis","seed":202610091087,"semantic_template":"qf-ftc-primitive-hypothesis/[[-1,12,1,4],[\"-1/1\",\"6/1\",\"1/3\",\"1/1\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [-1, 12, 1, 4], constant term first. Candidate F has ascending rational coefficients [-1/1, -1/1, 6/1, 1/3, 1/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0088","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f62039dafbde632fb34e0ba475b36c11d576f7bfa4753af454b1562217590f6c","family_id":"qf-ftc-primitive-hypothesis","seed":202610091088,"semantic_template":"qf-ftc-primitive-hypothesis/[[0,13,-1,1],[\"0/1\",\"13/2\",\"-1/3\",\"1/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [0, 13, -1, 1], constant term first. Candidate F has ascending rational coefficients [0/1, 0/1, 13/2, -1/3, 1/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0089","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fedf63db083c98c73f771df4dd481a325e9feef0983d0517641e2ea3e556d4dd","family_id":"qf-ftc-primitive-hypothesis","seed":202610091089,"semantic_template":"qf-ftc-primitive-hypothesis/[[1,14,0,2],[\"1/1\",\"7/1\",\"0/1\",\"3/2\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [1, 14, 0, 2], constant term first. Candidate F has ascending rational coefficients [1/1, 1/1, 7/1, 0/1, 3/2]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"No","explanation":"Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0090","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45b03d8bd033ca9401f1c7151ce86a88e59cf6201ab3da677b08e5007210cc82","family_id":"qf-ftc-primitive-hypothesis","seed":202610091090,"semantic_template":"qf-ftc-primitive-hypothesis/[[2,15,1,3],[\"2/1\",\"15/2\",\"1/3\",\"3/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [2, 15, 1, 3], constant term first. Candidate F has ascending rational coefficients [2/1, 2/1, 15/2, 1/3, 3/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0091","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"22a9c3eb50474bec8be06474bd2c930ca787ca5c67e24c06c8337b859ed111d1","family_id":"qf-ftc-primitive-hypothesis","seed":202610091091,"semantic_template":"qf-ftc-primitive-hypothesis/[[3,16,-1,4],[\"3/1\",\"8/1\",\"-1/3\",\"1/1\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [3, 16, -1, 4], constant term first. Candidate F has ascending rational coefficients [3/1, 3/1, 8/1, -1/3, 1/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0092","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ede51bffe9716fbf376dc2a84477f2b5e9ce632eeaef7382795d3d299754dd8a","family_id":"qf-ftc-primitive-hypothesis","seed":202610091092,"semantic_template":"qf-ftc-primitive-hypothesis/[[4,17,0,1],[\"4/1\",\"17/2\",\"0/1\",\"5/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [4, 17, 0, 1], constant term first. Candidate F has ascending rational coefficients [4/1, 4/1, 17/2, 0/1, 5/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"No","explanation":"Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0093","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95296abfcdc8ea931cd3495557115bc7cd4af7430676a8c35a6d2b20709437b1","family_id":"qf-ftc-primitive-hypothesis","seed":202610091093,"semantic_template":"qf-ftc-primitive-hypothesis/[[5,18,1,2],[\"5/1\",\"9/1\",\"1/3\",\"1/2\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [5, 18, 1, 2], constant term first. Candidate F has ascending rational coefficients [5/1, 5/1, 9/1, 1/3, 1/2]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0094","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ebd049dca764c19fd25f389c3bcd82318f8efe72b5799761189637502bace364","family_id":"qf-ftc-primitive-hypothesis","seed":202610091094,"semantic_template":"qf-ftc-primitive-hypothesis/[[6,19,-1,3],[\"6/1\",\"19/2\",\"-1/3\",\"3/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [6, 19, -1, 3], constant term first. Candidate F has ascending rational coefficients [6/1, 6/1, 19/2, -1/3, 3/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0095","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"547dca706d19b2ea344338bab964174e39dd399241f95a1e7b0ad2f398443f63","family_id":"qf-ftc-primitive-hypothesis","seed":202610091095,"semantic_template":"qf-ftc-primitive-hypothesis/[[7,20,0,4],[\"7/1\",\"10/1\",\"0/1\",\"2/1\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [7, 20, 0, 4], constant term first. Candidate F has ascending rational coefficients [7/1, 7/1, 10/1, 0/1, 2/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"No","explanation":"Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0096","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b97c4b6eb99d161c437a10f8afd774350f951b26c82334b7b9cefd40bf9e9742","family_id":"qf-ftc-primitive-hypothesis","seed":202610091096,"semantic_template":"qf-ftc-primitive-hypothesis/[[8,21,1,1],[\"8/1\",\"21/2\",\"1/3\",\"1/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [8, 21, 1, 1], constant term first. Candidate F has ascending rational coefficients [8/1, 8/1, 21/2, 1/3, 1/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0097","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e3723881fbe57b54aa503bbf22dbbf1334dfa5f0c4433363ccec3a1105e77a25","family_id":"qf-ftc-primitive-hypothesis","seed":202610091097,"semantic_template":"qf-ftc-primitive-hypothesis/[[9,22,-1,2],[\"9/1\",\"11/1\",\"-1/3\",\"1/2\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [9, 22, -1, 2], constant term first. Candidate F has ascending rational coefficients [9/1, 9/1, 11/1, -1/3, 1/2]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0098","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7513a4cdfa03cc624f99c5adddb6bc3d209699ba3ea7ae205a2ccc9938ee9f7d","family_id":"qf-ftc-primitive-hypothesis","seed":202610091098,"semantic_template":"qf-ftc-primitive-hypothesis/[[10,23,0,3],[\"10/1\",\"23/2\",\"0/1\",\"7/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [10, 23, 0, 3], constant term first. Candidate F has ascending rational coefficients [10/1, 10/1, 23/2, 0/1, 7/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"No","explanation":"Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0099","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c178944021687bc8a31467f20e246d2951bc565ab58040e327fb48a7518603a","family_id":"qf-ftc-primitive-hypothesis","seed":202610091099,"semantic_template":"qf-ftc-primitive-hypothesis/[[11,24,1,4],[\"11/1\",\"12/1\",\"1/3\",\"1/1\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [11, 24, 1, 4], constant term first. Candidate F has ascending rational coefficients [11/1, 11/1, 12/1, 1/3, 1/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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-20261007-advance-20261009-r01-fundamental-theorem-calculus-0100","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf5006c13d1c040ecc3d489529d7a242824c1cdf46accdc307bb531d3ede01b5","family_id":"qf-ftc-primitive-hypothesis","seed":202610091100,"semantic_template":"qf-ftc-primitive-hypothesis/[[12,25,-1,1],[\"12/1\",\"25/2\",\"-1/3\",\"1/4\"]]"},"fields":{"title":"Candidate Antiderivative Premise for the Endpoint Formula","stem":"On all R, continuous f has ascending coefficients [12, 25, -1, 1], constant term first. Candidate F has ascending rational coefficients [12/1, 12/1, 25/2, -1/3, 1/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?","answer":{"kind":"short_text","value":"Yes","explanation":"Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility."},"skills":["Candidate Antiderivative Premise for the Endpoint Formula"],"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."}}