{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r02-mean-value-theorem-0001","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19fad7c1495a65ad9b15fa02431d849a7a5a3b525b890f54c3f71687c9b7d046","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102001,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":1,\"B\":-10,\"C\":1,\"a\":-2,\"b\":3}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=-10, C=1, a=-2, b=3 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-2, 3, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-2, 3, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0002","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b19dcabe062740a53bb1127b22471274787da5c4376daf6a9449155a4d9f48f","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102002,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":2,\"B\":-9,\"C\":2,\"a\":-3,\"b\":4}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=-9, C=2, a=-3, b=4 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-3, 4, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-3, 4, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0003","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"411f8340e5d0d147d2a20eba36e48029e0ad0f05cdc58fe8a1f9abe10f5e9f80","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102003,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":3,\"B\":-8,\"C\":5,\"a\":-4,\"b\":5}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=-8, C=5, a=-4, b=5 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-4, 5, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-4, 5, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0004","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e43df3b2d501dad69bd3581f9baf8d1c99fb1c4fb7c3e8e2d4d3f9b3652c0fe1","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102004,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":4,\"B\":-7,\"C\":10,\"a\":-5,\"b\":6}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=-7, C=10, a=-5, b=6 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-5, 6, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-5, 6, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0005","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba9fc27b6456c8a67969e25b843c4618bca55e80b125ce0bd49e2c6c67b5361a","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102005,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":5,\"B\":-6,\"C\":17,\"a\":-6,\"b\":7}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=-6, C=17, a=-6, b=7 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-6, 7, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-6, 7, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0006","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2c0399417cac2ae6ff35d4e06b3b0b3d170bce588be544c0d189f24695a4cdc","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102006,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":1,\"B\":-5,\"C\":26,\"a\":-7,\"b\":8}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=-5, C=26, a=-7, b=8 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-7, 8, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-7, 8, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0007","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d2560c0874c64902b89f946bae1feba85664ba9d3c076419b5e004bce5d06fc","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102007,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":2,\"B\":-4,\"C\":37,\"a\":-8,\"b\":9}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=-4, C=37, a=-8, b=9 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-8, 9, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-8, 9, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0008","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a497c22d69acf13ec2f0529b16eb300a94fc4349d54e1bd5b41dfdcb24c7a4f","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102008,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":3,\"B\":-3,\"C\":50,\"a\":-9,\"b\":10}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=-3, C=50, a=-9, b=10 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-9, 10, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-9, 10, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0009","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e79c75271f386dbfd690ddf6bb8fedbd6d128f596301d3cb24b13a94c4fba8ec","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102009,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":4,\"B\":-2,\"C\":65,\"a\":-10,\"b\":11}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=-2, C=65, a=-10, b=11 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-10, 11, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-10, 11, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0010","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a975adbfec09a3b2f625c3fd4db1263ce681481da566c9cc91981a664e422bf9","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102010,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":5,\"B\":-1,\"C\":82,\"a\":-11,\"b\":12}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=-1, C=82, a=-11, b=12 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-11, 12, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-11, 12, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0011","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"47f981b63df2483634b6888e0efd4b74e31c5c27ab1f1b466814f256d126c91b","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102011,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":1,\"B\":0,\"C\":101,\"a\":-12,\"b\":13}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=0, C=101, a=-12, b=13 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-12, 13, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-12, 13, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0012","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"573fe9bc4aacdc2bb5e3e90002ca40063ce27e9345a77945172081145d9dfc9f","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102012,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":2,\"B\":1,\"C\":122,\"a\":-13,\"b\":14}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=1, C=122, a=-13, b=14 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-13, 14, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-13, 14, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0013","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eed4c6ed047611abcf66067b83c917afafb194518007ae5a2a599f0416a05d46","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102013,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":3,\"B\":2,\"C\":145,\"a\":-14,\"b\":15}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=2, C=145, a=-14, b=15 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-14, 15, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-14, 15, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0014","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78e9a4aa1441521e266b3f95f727f6dfc1e936ba5c086bc9ab30283aa3e0ea42","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102014,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":4,\"B\":3,\"C\":170,\"a\":-15,\"b\":16}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=3, C=170, a=-15, b=16 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-15, 16, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-15, 16, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0015","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fc9f53103af11eb58ccedc5b53cfc0177ef85a2f899c99d53ccb39384e5bdcc3","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102015,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":5,\"B\":4,\"C\":197,\"a\":-16,\"b\":17}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=4, C=197, a=-16, b=17 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-16, 17, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-16, 17, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0016","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"327300dff16a3d59de88d2c5b2862fb29628e1ba100c36d64051307d9895963b","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102016,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":1,\"B\":5,\"C\":226,\"a\":-17,\"b\":18}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=5, C=226, a=-17, b=18 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-17, 18, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-17, 18, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0017","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f08e7c92d78105eb729bd550c2e7b1ed6bff080b132f87bb3a335b669f09e001","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102017,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":2,\"B\":6,\"C\":257,\"a\":-18,\"b\":19}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=6, C=257, a=-18, b=19 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-18, 19, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-18, 19, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0018","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"946b571d081a979e0eb86c112b389afa98423e945168f99147da1bab4c3b3f1b","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102018,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":3,\"B\":7,\"C\":290,\"a\":-19,\"b\":20}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=7, C=290, a=-19, b=20 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-19, 20, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-19, 20, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0019","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ac7858e4b27e391af61e052090863532afdc3ac6a4cf534380f2cfce854ca45","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102019,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":4,\"B\":8,\"C\":325,\"a\":-20,\"b\":21}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=8, C=325, a=-20, b=21 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-20, 21, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-20, 21, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0020","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6ec04c545cc601987d0bf3cbb2e0370b72df4a8e1ad27242aa744e8c17535b16","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102020,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":5,\"B\":9,\"C\":362,\"a\":-21,\"b\":22}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=9, C=362, a=-21, b=22 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-21, 22, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-21, 22, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0021","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a535e8d5a5cbb8f7a3ca044eb0925a28aee86824850d64e7fa1781d698e54fb","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102021,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":1,\"B\":10,\"C\":401,\"a\":-22,\"b\":23}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=10, C=401, a=-22, b=23 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-22, 23, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-22, 23, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0022","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cfc8626e22ed8b5f300bbf0a574d858108f8aba290703a27569caccdab4653af","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102022,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":2,\"B\":11,\"C\":442,\"a\":-23,\"b\":24}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=11, C=442, a=-23, b=24 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-23, 24, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-23, 24, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0023","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca0e36384e756a41495d393ca0f6c1689977e913ebedabd1ecc3e756ce1792d7","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102023,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":3,\"B\":12,\"C\":485,\"a\":-24,\"b\":25}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=12, C=485, a=-24, b=25 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-24, 25, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-24, 25, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0024","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3251c8096714d70a5d04ef9319640e67c9aba02abf86b92d1a55561ac00f2a4c","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102024,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":4,\"B\":13,\"C\":530,\"a\":-25,\"b\":26}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=13, C=530, a=-25, b=26 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-25, 26, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-25, 26, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0025","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e7d49f5d1a797c57f9d0ac4896f3557d8b97b4941ed30dfa1816e10499f7d8dd","family_id":"qf-mod41-mvt-hypothesis-audit","seed":202611102025,"semantic_template":"qf-mod41-mvt-hypothesis-audit/{\"A\":5,\"B\":14,\"C\":577,\"a\":-26,\"b\":27}"},"fields":{"title":"Continuity and Differentiability Apply on Different Interval Parts","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=14, C=577, a=-26, b=27 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (a, b, ordered_endpoints, closed_interval_continuity, open_interval_differentiability, certificate).","answer":{"kind":"expression","value":"(-26, 27, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)","explanation":"Audit a<b, continuity on the closed interval and differentiability on the open interval. A theorem witness must lie strictly inside (a,b); existence does not imply uniqueness. The exact result, retaining the requested order, is (-26, 27, Yes, Yes, Yes, polynomial_continuous_on_closed_interval_and_differentiable_on_open_interval)."},"skills":["Continuity and Differentiability Apply on Different Interval Parts"],"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-20261014-advance-20261110-r02-mean-value-theorem-0026","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa6a8a2cad1afbe246f2520c1490a6d39df6dc399209502be70c1022e2d3649a","family_id":"qf-mod41-mvt-secant-slope","seed":202611102026,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":1,\"B\":-10,\"C\":1,\"a\":-2,\"b\":3}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=-10, C=1, a=-2, b=3 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(25, -20, 5, -9, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (25, -20, 5, -9, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0027","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e4f0d3c379d7a8eb570fde9ba7a4ac855c5a35937254606cc485ac7c527267c","family_id":"qf-mod41-mvt-secant-slope","seed":202611102027,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":2,\"B\":-9,\"C\":2,\"a\":-3,\"b\":4}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=-9, C=2, a=-3, b=4 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(47, -2, 7, -7, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (47, -2, 7, -7, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0028","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"955d2d9470d0ef02cf437cfb6670619ea16bda632b34a0fb48a2e40766260aa1","family_id":"qf-mod41-mvt-secant-slope","seed":202611102028,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":3,\"B\":-8,\"C\":5,\"a\":-4,\"b\":5}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=-8, C=5, a=-4, b=5 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(85, 40, 9, -5, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (85, 40, 9, -5, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0029","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b1ed5fcee8a63c67d3cfd77c8c99b0d36e278dbca8070ec51f264080dc0d656","family_id":"qf-mod41-mvt-secant-slope","seed":202611102029,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":4,\"B\":-7,\"C\":10,\"a\":-5,\"b\":6}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=-7, C=10, a=-5, b=6 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(145, 112, 11, -3, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (145, 112, 11, -3, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0030","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5098df19339d7942d77f1b438d9ec1b12c6dcc49efab7e9981f7b783d0735543","family_id":"qf-mod41-mvt-secant-slope","seed":202611102030,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":5,\"B\":-6,\"C\":17,\"a\":-6,\"b\":7}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=-6, C=17, a=-6, b=7 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(233, 220, 13, -1, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (233, 220, 13, -1, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0031","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4f1d5ba950a951aac8ccf5b350ac5d13e4af1702a84f974119869f41d8c4f88","family_id":"qf-mod41-mvt-secant-slope","seed":202611102031,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":1,\"B\":-5,\"C\":26,\"a\":-7,\"b\":8}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=-5, C=26, a=-7, b=8 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(110, 50, 15, -4, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (110, 50, 15, -4, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0032","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1b95dc857ca5860aa2a5aedfa2dac9efa27dddc613dd11427b89771059b567a","family_id":"qf-mod41-mvt-secant-slope","seed":202611102032,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":2,\"B\":-4,\"C\":37,\"a\":-8,\"b\":9}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=-4, C=37, a=-8, b=9 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(197, 163, 17, -2, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (197, 163, 17, -2, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0033","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0bf65d70c7060f5b4b68f4fbe17a307eda3bbf932c2d1e357d9a187dec731821","family_id":"qf-mod41-mvt-secant-slope","seed":202611102033,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":3,\"B\":-3,\"C\":50,\"a\":-9,\"b\":10}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=-3, C=50, a=-9, b=10 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(320, 320, 19, 0, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (320, 320, 19, 0, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0034","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4057d8083992ed8c4cc569381d7eaf98adb1be589b0534d97976b7eb89926bfc","family_id":"qf-mod41-mvt-secant-slope","seed":202611102034,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":4,\"B\":-2,\"C\":65,\"a\":-10,\"b\":11}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=-2, C=65, a=-10, b=11 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(485, 527, 21, 2, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (485, 527, 21, 2, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0035","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d1dcfd7383f57ababe566e0762c70d44b81623776b695dbf6c44b07339d0e85","family_id":"qf-mod41-mvt-secant-slope","seed":202611102035,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":5,\"B\":-1,\"C\":82,\"a\":-11,\"b\":12}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=-1, C=82, a=-11, b=12 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(698, 790, 23, 4, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (698, 790, 23, 4, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0036","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5bbe19b0ddfeed83cb67afc8d022621757c82e5428d7fa549559ac4b5dc04f82","family_id":"qf-mod41-mvt-secant-slope","seed":202611102036,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":1,\"B\":0,\"C\":101,\"a\":-12,\"b\":13}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=0, C=101, a=-12, b=13 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(245, 270, 25, 1, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (245, 270, 25, 1, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0037","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"023e5d3645180804c73a52ef4d3e2caf66b2cea734ae12fdbb3e79c50bd7c275","family_id":"qf-mod41-mvt-secant-slope","seed":202611102037,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":2,\"B\":1,\"C\":122,\"a\":-13,\"b\":14}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=1, C=122, a=-13, b=14 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(447, 528, 27, 3, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (447, 528, 27, 3, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0038","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"05adab6e8af59641a31ece4c3f840f3b80f01b4f6d1ce7a7049c45df29813677","family_id":"qf-mod41-mvt-secant-slope","seed":202611102038,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":3,\"B\":2,\"C\":145,\"a\":-14,\"b\":15}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=2, C=145, a=-14, b=15 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(705, 850, 29, 5, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (705, 850, 29, 5, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0039","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e447f6d394a1b303051331991873473a6859febecaef641d6acde5931803a58","family_id":"qf-mod41-mvt-secant-slope","seed":202611102039,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":4,\"B\":3,\"C\":170,\"a\":-15,\"b\":16}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=3, C=170, a=-15, b=16 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(1025, 1242, 31, 7, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (1025, 1242, 31, 7, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0040","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1bd388f29eeacf12a24a7925377ecbd2603318e23d73b6bc7735d08fc79798b","family_id":"qf-mod41-mvt-secant-slope","seed":202611102040,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":5,\"B\":4,\"C\":197,\"a\":-16,\"b\":17}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=4, C=197, a=-16, b=17 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(1413, 1710, 33, 9, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (1413, 1710, 33, 9, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0041","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5717f32a50230043fd8e5d8e8f81212fc8cbb13120b58c708a892682e6b4ea41","family_id":"qf-mod41-mvt-secant-slope","seed":202611102041,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":1,\"B\":5,\"C\":226,\"a\":-17,\"b\":18}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=5, C=226, a=-17, b=18 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(430, 640, 35, 6, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (430, 640, 35, 6, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0042","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc5ae1cf13feb730dbdcd1a153444752f9752c8e786d6cb7c9be2bc051263efc","family_id":"qf-mod41-mvt-secant-slope","seed":202611102042,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":2,\"B\":6,\"C\":257,\"a\":-18,\"b\":19}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=6, C=257, a=-18, b=19 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(797, 1093, 37, 8, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (797, 1093, 37, 8, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0043","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d370931d9ad35151cf70dd11ab2705a74a1374ee22c1c7fe88dcabe4f65e024","family_id":"qf-mod41-mvt-secant-slope","seed":202611102043,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":3,\"B\":7,\"C\":290,\"a\":-19,\"b\":20}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=7, C=290, a=-19, b=20 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(1240, 1630, 39, 10, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (1240, 1630, 39, 10, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0044","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8d26f08bf819c97cb5618812f7eb9c2d4eeca38236b8c8eebdad8395cf2153c","family_id":"qf-mod41-mvt-secant-slope","seed":202611102044,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":4,\"B\":8,\"C\":325,\"a\":-20,\"b\":21}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=8, C=325, a=-20, b=21 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(1765, 2257, 41, 12, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (1765, 2257, 41, 12, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0045","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0bce3e9f5c13f3a03d5d4e21127e15b7ebb004d9f5d409fac3498d883105ceb8","family_id":"qf-mod41-mvt-secant-slope","seed":202611102045,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":5,\"B\":9,\"C\":362,\"a\":-21,\"b\":22}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=9, C=362, a=-21, b=22 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(2378, 2980, 43, 14, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (2378, 2980, 43, 14, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0046","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f200ef6f1feddb34c768afe63c30186c4d237a3273c79d066361aceee67a9942","family_id":"qf-mod41-mvt-secant-slope","seed":202611102046,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":1,\"B\":10,\"C\":401,\"a\":-22,\"b\":23}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=10, C=401, a=-22, b=23 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(665, 1160, 45, 11, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (665, 1160, 45, 11, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0047","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a133627e16eda87974801f29f7201b69f5fa82571e1fe7601adb44e68208269","family_id":"qf-mod41-mvt-secant-slope","seed":202611102047,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":2,\"B\":11,\"C\":442,\"a\":-23,\"b\":24}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=11, C=442, a=-23, b=24 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(1247, 1858, 47, 13, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (1247, 1858, 47, 13, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0048","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f74d94be7f497bcada73771de040043b7551ec20f6cb6be6792250c9234bcc1","family_id":"qf-mod41-mvt-secant-slope","seed":202611102048,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":3,\"B\":12,\"C\":485,\"a\":-24,\"b\":25}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=12, C=485, a=-24, b=25 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(1925, 2660, 49, 15, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (1925, 2660, 49, 15, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0049","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd63cfa87c7d2b6589722d7f719d2ec2f63a94d21fb4bae7d819902150a6c970","family_id":"qf-mod41-mvt-secant-slope","seed":202611102049,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":4,\"B\":13,\"C\":530,\"a\":-25,\"b\":26}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=13, C=530, a=-25, b=26 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(2705, 3572, 51, 17, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (2705, 3572, 51, 17, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0050","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6533fdb23f2b1f0b4d8c97cc88108f6bcb74100adb21a9ddce5c9f4deab49b1a","family_id":"qf-mod41-mvt-secant-slope","seed":202611102050,"semantic_template":"qf-mod41-mvt-secant-slope/{\"A\":5,\"B\":14,\"C\":577,\"a\":-26,\"b\":27}"},"fields":{"title":"The Secant Uses Exact Endpoint Values and a Nonzero Denominator","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=14, C=577, a=-26, b=27 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (value_at_a, value_at_b, interval_length, secant_slope, certificate).","answer":{"kind":"expression","value":"(3593, 4600, 53, 19, exact_endpoint_secant_slope)","explanation":"Compute f(a), f(b), b−a and (f(b)−f(a))/(b−a) exactly. This is an average slope, not an asserted endpoint derivative or finite-sampling differentiability proof. The exact result, retaining the requested order, is (3593, 4600, 53, 19, exact_endpoint_secant_slope)."},"skills":["The Secant Uses Exact Endpoint Values and a Nonzero Denominator"],"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-20261014-advance-20261110-r02-mean-value-theorem-0051","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3757fb7e9dc24eefa723be26573f9dcbac669445bb93c7c820086250ed31aa6d","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102051,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":1,\"B\":-10,\"C\":1,\"a\":-2,\"b\":3}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=-10, C=1, a=-2, b=3 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, -9, -9, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, -9, -9, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0052","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e980393e1a9bc336496e008ae4581e18ee78d896f7ff5423849314c3e2809b43","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102052,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":2,\"B\":-9,\"C\":2,\"a\":-3,\"b\":4}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=-9, C=2, a=-3, b=4 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, -7, -7, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, -7, -7, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0053","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a019e9a7ee007d1c392e46eb0fc92402da38b36179bea05564903dd2df3d0a07","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102053,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":3,\"B\":-8,\"C\":5,\"a\":-4,\"b\":5}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=-8, C=5, a=-4, b=5 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, -5, -5, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, -5, -5, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0054","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"544fe5feb8f638b73c507f33d97856ae60fd4fba40f1b47a8988326d1ec74e09","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102054,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":4,\"B\":-7,\"C\":10,\"a\":-5,\"b\":6}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=-7, C=10, a=-5, b=6 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, -3, -3, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, -3, -3, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0055","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"257fac0f296fb38468694730553064b118ef74338e4d43f49c564989df279df9","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102055,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":5,\"B\":-6,\"C\":17,\"a\":-6,\"b\":7}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=-6, C=17, a=-6, b=7 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, -1, -1, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, -1, -1, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0056","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"972c0009bd64fcea438324d709bcea8c21c5241a7abb69207f527917a8659208","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102056,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":1,\"B\":-5,\"C\":26,\"a\":-7,\"b\":8}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=-5, C=26, a=-7, b=8 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, -4, -4, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, -4, -4, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0057","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ded7e75c762e990bcabd85565df19a594dffc37e0f522cee28f174d66da3aa7","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102057,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":2,\"B\":-4,\"C\":37,\"a\":-8,\"b\":9}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=-4, C=37, a=-8, b=9 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, -2, -2, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, -2, -2, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0058","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6da476e88b895311b1a7f1e1a992bdc3aa23a6eb770ec00a876fcf8cf85cc20c","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102058,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":3,\"B\":-3,\"C\":50,\"a\":-9,\"b\":10}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=-3, C=50, a=-9, b=10 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 0, 0, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 0, 0, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0059","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0976cfaeb872e3f85612fd6e43391b758f5bc94edd260898dc47c4a529de2f80","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102059,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":4,\"B\":-2,\"C\":65,\"a\":-10,\"b\":11}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=-2, C=65, a=-10, b=11 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 2, 2, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 2, 2, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0060","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88ac4dae17ca773c205cb7aa94787969dfafa807bc57cbf3abb5d5dc8c77599f","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102060,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":5,\"B\":-1,\"C\":82,\"a\":-11,\"b\":12}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=-1, C=82, a=-11, b=12 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 4, 4, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 4, 4, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0061","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6102d7c31485451af062f6e1aea1934c63699612451d75df5e8cf7fbe7135395","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102061,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":1,\"B\":0,\"C\":101,\"a\":-12,\"b\":13}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=0, C=101, a=-12, b=13 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 1, 1, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 1, 1, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0062","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8745b9001f9eeb3f8f4815e7c38957f26241e6a994b4f94beb368812b382546","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102062,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":2,\"B\":1,\"C\":122,\"a\":-13,\"b\":14}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=1, C=122, a=-13, b=14 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 3, 3, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 3, 3, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0063","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03afd4d35b974771b044d02d28514703fa0345057bed6233f101bafb91768e47","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102063,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":3,\"B\":2,\"C\":145,\"a\":-14,\"b\":15}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=2, C=145, a=-14, b=15 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 5, 5, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 5, 5, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0064","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b680a8a8442426d801d8da3623a71dfed2b44076ac44a81080f5a46c0c19b007","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102064,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":4,\"B\":3,\"C\":170,\"a\":-15,\"b\":16}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=3, C=170, a=-15, b=16 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 7, 7, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 7, 7, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0065","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"67d2e62c0a87cdb71cd4c4d164f01bd864017c70210b42448d69d00e4a8d9624","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102065,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":5,\"B\":4,\"C\":197,\"a\":-16,\"b\":17}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=4, C=197, a=-16, b=17 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 9, 9, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 9, 9, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0066","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c193a576fa5d4f5e8e80a169c796a56a90cc9da2cadce32af46c1ea86fe83903","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102066,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":1,\"B\":5,\"C\":226,\"a\":-17,\"b\":18}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=5, C=226, a=-17, b=18 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 6, 6, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 6, 6, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0067","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e4ddc80e4b3104051b5dcc02fad1725f7124dee7926f649dc0c8de8ce82a770","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102067,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":2,\"B\":6,\"C\":257,\"a\":-18,\"b\":19}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=6, C=257, a=-18, b=19 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 8, 8, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 8, 8, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0068","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a660b81c602f4c7f4b2cf86bf8526e994c7e7418e1888f336d163e123893c5ee","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102068,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":3,\"B\":7,\"C\":290,\"a\":-19,\"b\":20}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=7, C=290, a=-19, b=20 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 10, 10, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 10, 10, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0069","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ac6eb86f20fe2e1bf26634140603737ee3bb64c3549e962a13d3d1772b12acb","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102069,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":4,\"B\":8,\"C\":325,\"a\":-20,\"b\":21}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=8, C=325, a=-20, b=21 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 12, 12, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 12, 12, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0070","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e03627aaebc72896a4464b5727b492206f275f587fad13b82334ea306394333a","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102070,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":5,\"B\":9,\"C\":362,\"a\":-21,\"b\":22}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=9, C=362, a=-21, b=22 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 14, 14, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 14, 14, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0071","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9fd9ad38d9405385e21b60fe3a01a4cb185a30ccc8faba639f0a0d86549cf4f7","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102071,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":1,\"B\":10,\"C\":401,\"a\":-22,\"b\":23}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=1, B=10, C=401, a=-22, b=23 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 11, 11, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 11, 11, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0072","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"997f90aa5e2fc5693a6ff543d6088d98d647040a30e6f371f6741eb23c636aec","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102072,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":2,\"B\":11,\"C\":442,\"a\":-23,\"b\":24}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=2, B=11, C=442, a=-23, b=24 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 13, 13, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 13, 13, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0073","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4997d068dd53600a1ad48d93cb5ca8997086c99a880b2ae6dde0d3ea4f380cdb","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102073,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":3,\"B\":12,\"C\":485,\"a\":-24,\"b\":25}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=3, B=12, C=485, a=-24, b=25 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 15, 15, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 15, 15, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0074","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d31876ba06864af54b4e216c4754e7d0eb25f81530675ac1be1d5f2ace33b74","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102074,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":4,\"B\":13,\"C\":530,\"a\":-25,\"b\":26}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=4, B=13, C=530, a=-25, b=26 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 17, 17, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 17, 17, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0075","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92e68982942c23809dcf6715476e5d75e4f8b47fad045013f119de0d6d83f508","family_id":"qf-mod41-mvt-polynomial-witness","seed":202611102075,"semantic_template":"qf-mod41-mvt-polynomial-witness/{\"A\":5,\"B\":14,\"C\":577,\"a\":-26,\"b\":27}"},"fields":{"title":"The Quadratic Midpoint Is a Strictly Interior MVT Witness","stem":"Use the whole real polynomial f(t)=A*t^2+B*t+C with A=5, B=14, C=577, a=-26, b=27 and a<b. Coefficients are quadratic, linear and constant in that order. This declared polynomial is continuous on [a,b] and differentiable on (a,b); finite samples do not certify those hypotheses for an arbitrary function. Give (c, strictly_interior, derivative_at_c, secant_slope, equal, certificate).","answer":{"kind":"expression","value":"(1/2, Yes, 19, 19, Yes, quadratic_midpoint_witness)","explanation":"Take c=(a+b)/2, verify a<c<b, and compare 2*A*c+B with the exact secant slope. For A=0 this affine/constant case has many witnesses; the midpoint is not a uniqueness claim. The exact result, retaining the requested order, is (1/2, Yes, 19, 19, Yes, quadratic_midpoint_witness)."},"skills":["The Quadratic Midpoint Is a Strictly Interior MVT Witness"],"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-20261014-advance-20261110-r02-mean-value-theorem-0076","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b2f81506ea2a70dc86f3c826522feb0b8c3940b38c090f22a59682e901fbfb4","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102076,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":0,\"b\":0,\"mode\":\"degenerate_interval\",\"offset\":0,\"scale\":1}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=degenerate_interval, a=0, b=0, scale=1, offset=0. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, ordered_endpoints, secant_slope_or_null, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(degenerate_interval, 0, 0, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (degenerate_interval, 0, 0, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0077","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44f43873e60ec3efd8ecc74920885d3b5460e7a0f53c0502580af1c33c192bf5","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102077,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-2,\"b\":3,\"mode\":\"jump_at_zero\",\"offset\":1,\"scale\":2}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=jump_at_zero, a=-2, b=3, scale=2, offset=1. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(jump_at_zero, -2, 3, 2, 1, No, No, No, endpoint_interval_continuity_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (jump_at_zero, -2, 3, 2, 1, No, No, No, endpoint_interval_continuity_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0078","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1e38ac0cec06171a82af6ef3d74745c46e88c0a2c3f78afc37f88dd363bb396","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102078,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-3,\"b\":4,\"mode\":\"cusp_at_zero\",\"offset\":2,\"scale\":3}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=cusp_at_zero, a=-3, b=4, scale=3, offset=2. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(cusp_at_zero, -3, 4, 3, 2, Yes, No, No, open_interval_differentiability_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (cusp_at_zero, -3, 4, 3, 2, Yes, No, No, open_interval_differentiability_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0079","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"57e1a7b28c428acbb7b5748109d1b5d13a51cad22c6b791b143ad41c54eec262","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102079,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":3,\"b\":3,\"mode\":\"degenerate_interval\",\"offset\":-3,\"scale\":4}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=degenerate_interval, a=3, b=3, scale=4, offset=-3. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, ordered_endpoints, secant_slope_or_null, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(degenerate_interval, 3, 3, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (degenerate_interval, 3, 3, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0080","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6cfebaa2a21698b79e95fac0165359f351997c9ef6d335c2949554d9c402292","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102080,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-5,\"b\":6,\"mode\":\"jump_at_zero\",\"offset\":4,\"scale\":5}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=jump_at_zero, a=-5, b=6, scale=5, offset=4. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(jump_at_zero, -5, 6, 5, 4, No, No, No, endpoint_interval_continuity_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (jump_at_zero, -5, 6, 5, 4, No, No, No, endpoint_interval_continuity_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0081","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eef3407c4cfa2df4be1ea0e15542610d96cb28379bb7dc7a6654f7ea0e078059","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102081,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-6,\"b\":7,\"mode\":\"cusp_at_zero\",\"offset\":5,\"scale\":6}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=cusp_at_zero, a=-6, b=7, scale=6, offset=5. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(cusp_at_zero, -6, 7, 6, 5, Yes, No, No, open_interval_differentiability_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (cusp_at_zero, -6, 7, 6, 5, Yes, No, No, open_interval_differentiability_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0082","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9d665b1d58af68c0020efb0e4ebfc8699fd37803eefccd6675039386646dddb","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102082,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":6,\"b\":6,\"mode\":\"degenerate_interval\",\"offset\":-6,\"scale\":7}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=degenerate_interval, a=6, b=6, scale=7, offset=-6. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, ordered_endpoints, secant_slope_or_null, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(degenerate_interval, 6, 6, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (degenerate_interval, 6, 6, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0083","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c96ed88082fa385e57807bc963bb95895c6d0074f53cf753b494c53541be24e","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102083,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-8,\"b\":9,\"mode\":\"jump_at_zero\",\"offset\":7,\"scale\":8}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=jump_at_zero, a=-8, b=9, scale=8, offset=7. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(jump_at_zero, -8, 9, 8, 7, No, No, No, endpoint_interval_continuity_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (jump_at_zero, -8, 9, 8, 7, No, No, No, endpoint_interval_continuity_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0084","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14f57b87e476dc32d93c7fc7820eb1b4dad6c35c7f950639b3ba5d0f76fb2912","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102084,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-9,\"b\":10,\"mode\":\"cusp_at_zero\",\"offset\":8,\"scale\":9}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=cusp_at_zero, a=-9, b=10, scale=9, offset=8. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(cusp_at_zero, -9, 10, 9, 8, Yes, No, No, open_interval_differentiability_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (cusp_at_zero, -9, 10, 9, 8, Yes, No, No, open_interval_differentiability_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0085","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"521642a62fde12c9234c009aec5fa1a1c423b97486b59b69a066a29a41146c9b","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102085,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":9,\"b\":9,\"mode\":\"degenerate_interval\",\"offset\":-9,\"scale\":10}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=degenerate_interval, a=9, b=9, scale=10, offset=-9. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, ordered_endpoints, secant_slope_or_null, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(degenerate_interval, 9, 9, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (degenerate_interval, 9, 9, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0086","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d39137dd0a6bf4f716f8d808426c80a6e576f80d1d437f1167b2d857368591d1","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102086,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-11,\"b\":12,\"mode\":\"jump_at_zero\",\"offset\":10,\"scale\":11}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=jump_at_zero, a=-11, b=12, scale=11, offset=10. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(jump_at_zero, -11, 12, 11, 10, No, No, No, endpoint_interval_continuity_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (jump_at_zero, -11, 12, 11, 10, No, No, No, endpoint_interval_continuity_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0087","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a082171df8f143c91ec2a4971c62d2570db41d30fe8312c4a6a83ed678bc41f","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102087,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-12,\"b\":13,\"mode\":\"cusp_at_zero\",\"offset\":11,\"scale\":12}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=cusp_at_zero, a=-12, b=13, scale=12, offset=11. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(cusp_at_zero, -12, 13, 12, 11, Yes, No, No, open_interval_differentiability_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (cusp_at_zero, -12, 13, 12, 11, Yes, No, No, open_interval_differentiability_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0088","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f09755b84422c04f83d75d516627c3509cea8ba0ffba011b0858ac5c04446bc7","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102088,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":12,\"b\":12,\"mode\":\"degenerate_interval\",\"offset\":-12,\"scale\":13}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=degenerate_interval, a=12, b=12, scale=13, offset=-12. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, ordered_endpoints, secant_slope_or_null, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(degenerate_interval, 12, 12, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (degenerate_interval, 12, 12, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0089","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88a36239cb776dea2d43103fddddfb98f45a9ed64cdcc50aa3105f8e497c7206","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102089,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-14,\"b\":15,\"mode\":\"jump_at_zero\",\"offset\":13,\"scale\":14}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=jump_at_zero, a=-14, b=15, scale=14, offset=13. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(jump_at_zero, -14, 15, 14, 13, No, No, No, endpoint_interval_continuity_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (jump_at_zero, -14, 15, 14, 13, No, No, No, endpoint_interval_continuity_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0090","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc2e39affd4aa71145bd3b008aaa1a4d89e8c993af4d9bd2473c4082c4a4a56c","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102090,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-15,\"b\":16,\"mode\":\"cusp_at_zero\",\"offset\":14,\"scale\":15}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=cusp_at_zero, a=-15, b=16, scale=15, offset=14. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(cusp_at_zero, -15, 16, 15, 14, Yes, No, No, open_interval_differentiability_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (cusp_at_zero, -15, 16, 15, 14, Yes, No, No, open_interval_differentiability_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0091","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54cdeed0fbb85bdd64fa3378d2365617dfef774eeed7babc2dc14a76770565fb","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102091,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":15,\"b\":15,\"mode\":\"degenerate_interval\",\"offset\":-15,\"scale\":16}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=degenerate_interval, a=15, b=15, scale=16, offset=-15. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, ordered_endpoints, secant_slope_or_null, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(degenerate_interval, 15, 15, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (degenerate_interval, 15, 15, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0092","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cb5d914f6adc716f3afb0155ebb30f3a77924c50226d83cb092a7e91635ac3e","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102092,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-17,\"b\":18,\"mode\":\"jump_at_zero\",\"offset\":16,\"scale\":17}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=jump_at_zero, a=-17, b=18, scale=17, offset=16. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(jump_at_zero, -17, 18, 17, 16, No, No, No, endpoint_interval_continuity_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (jump_at_zero, -17, 18, 17, 16, No, No, No, endpoint_interval_continuity_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0093","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cafda0d2878dfa24330afe781d02a8fb9d6ad7a62755988486002dc2aed18607","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102093,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-18,\"b\":19,\"mode\":\"cusp_at_zero\",\"offset\":17,\"scale\":18}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=cusp_at_zero, a=-18, b=19, scale=18, offset=17. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(cusp_at_zero, -18, 19, 18, 17, Yes, No, No, open_interval_differentiability_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (cusp_at_zero, -18, 19, 18, 17, Yes, No, No, open_interval_differentiability_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0094","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"63b87170992b0aaa6028266c3f87a479d375e6a2644b0905d1c2707653a2a281","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102094,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":18,\"b\":18,\"mode\":\"degenerate_interval\",\"offset\":-18,\"scale\":19}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=degenerate_interval, a=18, b=18, scale=19, offset=-18. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, ordered_endpoints, secant_slope_or_null, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(degenerate_interval, 18, 18, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (degenerate_interval, 18, 18, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0095","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec1ec0b9f3d25ff59cadec7940835badcba3019b4a51114085d64d044f9d2355","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102095,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-20,\"b\":21,\"mode\":\"jump_at_zero\",\"offset\":19,\"scale\":20}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=jump_at_zero, a=-20, b=21, scale=20, offset=19. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(jump_at_zero, -20, 21, 20, 19, No, No, No, endpoint_interval_continuity_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (jump_at_zero, -20, 21, 20, 19, No, No, No, endpoint_interval_continuity_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0096","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"293a06dbce3afaa9fefc8a78f746f8e791e790e8b97e5f6b9e61709917d0e4c5","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102096,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-21,\"b\":22,\"mode\":\"cusp_at_zero\",\"offset\":20,\"scale\":21}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=cusp_at_zero, a=-21, b=22, scale=21, offset=20. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(cusp_at_zero, -21, 22, 21, 20, Yes, No, No, open_interval_differentiability_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (cusp_at_zero, -21, 22, 21, 20, Yes, No, No, open_interval_differentiability_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0097","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d54003d3cda57a2c789af77c2993e87c6b1dde1595f7580e51470b673a54366","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102097,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":21,\"b\":21,\"mode\":\"degenerate_interval\",\"offset\":-21,\"scale\":22}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=degenerate_interval, a=21, b=21, scale=22, offset=-21. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, ordered_endpoints, secant_slope_or_null, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(degenerate_interval, 21, 21, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (degenerate_interval, 21, 21, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0098","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b98fc894f9683dc3cb94070c9e12eb9536d955862303feca222a49f48eb395c5","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102098,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-23,\"b\":24,\"mode\":\"jump_at_zero\",\"offset\":22,\"scale\":23}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=jump_at_zero, a=-23, b=24, scale=23, offset=22. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(jump_at_zero, -23, 24, 23, 22, No, No, No, endpoint_interval_continuity_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (jump_at_zero, -23, 24, 23, 22, No, No, No, endpoint_interval_continuity_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0099","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f806447f94f0847715c9a7d2f8deafc58bc7634aa4f3bd5092708be92c665e1","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102099,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":-24,\"b\":25,\"mode\":\"cusp_at_zero\",\"offset\":23,\"scale\":24}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=cusp_at_zero, a=-24, b=25, scale=24, offset=23. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, scale, offset, closed_interval_continuity, open_interval_differentiability, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(cusp_at_zero, -24, 25, 24, 23, Yes, No, No, open_interval_differentiability_fails_at_zero)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (cusp_at_zero, -24, 25, 24, 23, Yes, No, No, open_interval_differentiability_fails_at_zero)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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-20261014-advance-20261110-r02-mean-value-theorem-0100","batch_id":"batch-20261014-advance-20261110-r02-mean-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc6633584223d5f7994092dfce5a7b0bca2ec1566831ad46fa586559144f101a","family_id":"qf-mod41-mvt-failure-boundary","seed":202611102100,"semantic_template":"qf-mod41-mvt-failure-boundary/{\"a\":24,\"b\":24,\"mode\":\"degenerate_interval\",\"offset\":-24,\"scale\":25}"},"fields":{"title":"Failed MVT Hypotheses Need Their Own Exact Output Layout","stem":"Use boundary mode=degenerate_interval, a=24, b=24, scale=25, offset=-24. degenerate_interval declares a≥b; jump_at_zero and cusp_at_zero declare a<0<b and nonzero scale, with the named interior defect. Give (mode, a, b, ordered_endpoints, secant_slope_or_null, theorem_applicability, boundary).","answer":{"kind":"expression","value":"(degenerate_interval, 24, 24, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)","explanation":"Use the original mode-selected fields, without padding. Degenerate/reversed endpoint mode withholds a secant value as null; even if a>b makes a numerical quotient possible, it fails the ordered-interval theorem scope. A jump fails continuity and differentiability at zero; a cusp is continuous but not differentiable there. No means this failed-hypothesis case cannot invoke MVT, not a general classification of every unrelated function. The exact result, retaining the requested order, is (degenerate_interval, 24, 24, No, null, No, a_less_than_b_and_nonzero_secant_denominator_required)."},"skills":["Failed MVT Hypotheses Need Their Own Exact Output Layout"],"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."}}