{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0001","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dee4b9f2ee00515acc31eef698b67cd7fd13552d011d9a71bd242e81debe8e88","family_id":"qf-mod47-ivt-hypothesis","seed":202611162001,"semantic_template":"qf-mod47-ivt-hypothesis/{\"a\":[-13,1],\"b\":[-11,1],\"coefficients\":[[12,1],[1,1]],\"target\":[0,1]}"},"fields":{"title":"Polynomial Continuity and Inclusive Endpoint Bracketing Are Separate Obligations","stem":"Coefficients=[12, 1] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Use a=-13=0, -1, 1, 0, Yes, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (discontinuous_jump, f(x)=-1_for_x<0_and_1_for_x>=0, -1, 1, 0, Yes, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0077","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b22542519310892ef699c528fad2b92116136a0c38f3a15cb7555587ce9b832d","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162077,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonbracket\",\"scale\":[2,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonbracket and positive exact scale=2. Give (mode, function_description, a, b, target, inclusive_bracket, root_exists).","answer":{"kind":"expression","value":"(nonbracket, f(x)=2*x, 0, 1, 4, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonbracket, f(x)=2*x, 0, 1, 4, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0078","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"887dae98a5c4432af10eb99c7c9c03b28a7fcf47287abe138d6c723013231e27","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162078,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonunique\",\"scale\":[3,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonunique and positive exact scale=3. Give (mode, function_description, a, b, target, inclusive_bracket, multiplicity_statement, uniqueness_claim).","answer":{"kind":"expression","value":"(nonunique, f(x)=x^2, -3, 3, 9, Yes, two_roots_possible, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonunique, f(x)=x^2, -3, 3, 9, Yes, two_roots_possible, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0079","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce5f8238a06a41aefd9176458ef71d0cfbbfe49476b0ba00ae0ec2b5c1e4b441","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162079,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"discontinuous_jump\",\"scale\":[4,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=discontinuous_jump and positive exact scale=4. Give (mode, function_description, a, b, target, inclusive_bracket, continuous, root_exists).","answer":{"kind":"expression","value":"(discontinuous_jump, f(x)=-4_for_x<0_and_4_for_x>=0, -1, 1, 0, Yes, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (discontinuous_jump, f(x)=-4_for_x<0_and_4_for_x>=0, -1, 1, 0, Yes, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0080","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"315b8cbc7aa7602ec85d376a6b920c82aa735f82a4bbbf19eb3e8ee0bfc90af6","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162080,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonbracket\",\"scale\":[5,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonbracket and positive exact scale=5. Give (mode, function_description, a, b, target, inclusive_bracket, root_exists).","answer":{"kind":"expression","value":"(nonbracket, f(x)=5*x, 0, 1, 10, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonbracket, f(x)=5*x, 0, 1, 10, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0081","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de80c5dbdd584675dcf109e9d7aaa3f02b6879d0ad52573c8b9f991f3d633509","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162081,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonunique\",\"scale\":[6,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonunique and positive exact scale=6. Give (mode, function_description, a, b, target, inclusive_bracket, multiplicity_statement, uniqueness_claim).","answer":{"kind":"expression","value":"(nonunique, f(x)=x^2, -6, 6, 36, Yes, two_roots_possible, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonunique, f(x)=x^2, -6, 6, 36, Yes, two_roots_possible, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0082","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f94d38dfc181ec24cba0ad6384cb015223760e1cabe612ca6e18408523fc410d","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162082,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"discontinuous_jump\",\"scale\":[7,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=discontinuous_jump and positive exact scale=7. Give (mode, function_description, a, b, target, inclusive_bracket, continuous, root_exists).","answer":{"kind":"expression","value":"(discontinuous_jump, f(x)=-7_for_x<0_and_7_for_x>=0, -1, 1, 0, Yes, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (discontinuous_jump, f(x)=-7_for_x<0_and_7_for_x>=0, -1, 1, 0, Yes, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0083","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4a94b70d67e65df3bbc84c5cb08e9412349a54954691a7764293e1d0e074ced3","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162083,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonbracket\",\"scale\":[8,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonbracket and positive exact scale=8. Give (mode, function_description, a, b, target, inclusive_bracket, root_exists).","answer":{"kind":"expression","value":"(nonbracket, f(x)=8*x, 0, 1, 16, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonbracket, f(x)=8*x, 0, 1, 16, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0084","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"003062f9dd3144f11fba8798deaf901211b102f91c818617a1f3549deb6f4c56","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162084,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonunique\",\"scale\":[9,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonunique and positive exact scale=9. Give (mode, function_description, a, b, target, inclusive_bracket, multiplicity_statement, uniqueness_claim).","answer":{"kind":"expression","value":"(nonunique, f(x)=x^2, -9, 9, 81, Yes, two_roots_possible, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonunique, f(x)=x^2, -9, 9, 81, Yes, two_roots_possible, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0085","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"031a910af5eab42adc7687ed12ac525252d1d89f301796b4f4070aec417f8b60","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162085,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"discontinuous_jump\",\"scale\":[10,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=discontinuous_jump and positive exact scale=10. Give (mode, function_description, a, b, target, inclusive_bracket, continuous, root_exists).","answer":{"kind":"expression","value":"(discontinuous_jump, f(x)=-10_for_x<0_and_10_for_x>=0, -1, 1, 0, Yes, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (discontinuous_jump, f(x)=-10_for_x<0_and_10_for_x>=0, -1, 1, 0, Yes, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0086","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f4f1db26cfcfc4bef0d0c75d66f0fd72619c92abc22b3ef75a484250595c887","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162086,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonbracket\",\"scale\":[11,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonbracket and positive exact scale=11. Give (mode, function_description, a, b, target, inclusive_bracket, root_exists).","answer":{"kind":"expression","value":"(nonbracket, f(x)=11*x, 0, 1, 22, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonbracket, f(x)=11*x, 0, 1, 22, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0087","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e3a9b891c8c290585df841bed812c00f36215db6d1b5d700ba9994d9603d508","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162087,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonunique\",\"scale\":[12,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonunique and positive exact scale=12. Give (mode, function_description, a, b, target, inclusive_bracket, multiplicity_statement, uniqueness_claim).","answer":{"kind":"expression","value":"(nonunique, f(x)=x^2, -12, 12, 144, Yes, two_roots_possible, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonunique, f(x)=x^2, -12, 12, 144, Yes, two_roots_possible, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0088","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9784fc40401af8dccd183084335a804f1c2af1de485c211ed13851cbbfef4b1c","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162088,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"discontinuous_jump\",\"scale\":[13,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=discontinuous_jump and positive exact scale=13. Give (mode, function_description, a, b, target, inclusive_bracket, continuous, root_exists).","answer":{"kind":"expression","value":"(discontinuous_jump, f(x)=-13_for_x<0_and_13_for_x>=0, -1, 1, 0, Yes, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (discontinuous_jump, f(x)=-13_for_x<0_and_13_for_x>=0, -1, 1, 0, Yes, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0089","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff654dd679d7dd3448629a38e78601d0c3294f96736a6ff0f40ba4eeec0394b0","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162089,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonbracket\",\"scale\":[14,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonbracket and positive exact scale=14. Give (mode, function_description, a, b, target, inclusive_bracket, root_exists).","answer":{"kind":"expression","value":"(nonbracket, f(x)=14*x, 0, 1, 28, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonbracket, f(x)=14*x, 0, 1, 28, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0090","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"580c38fe473c38d0b3edd1c30131e7604eecfd5219a9b521bf182fb73e41c106","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162090,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonunique\",\"scale\":[15,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonunique and positive exact scale=15. Give (mode, function_description, a, b, target, inclusive_bracket, multiplicity_statement, uniqueness_claim).","answer":{"kind":"expression","value":"(nonunique, f(x)=x^2, -15, 15, 225, Yes, two_roots_possible, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonunique, f(x)=x^2, -15, 15, 225, Yes, two_roots_possible, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0091","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6eb2e4aedd2c2f31d2cfaabf56db458ab790eadb0302f477525b379540c1b4a8","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162091,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"discontinuous_jump\",\"scale\":[16,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=discontinuous_jump and positive exact scale=16. Give (mode, function_description, a, b, target, inclusive_bracket, continuous, root_exists).","answer":{"kind":"expression","value":"(discontinuous_jump, f(x)=-16_for_x<0_and_16_for_x>=0, -1, 1, 0, Yes, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (discontinuous_jump, f(x)=-16_for_x<0_and_16_for_x>=0, -1, 1, 0, Yes, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0092","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed43d5b1cedcec6796ee8d322b90f8e1f6b59c4510b043f55aac5675f553bf82","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162092,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonbracket\",\"scale\":[17,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonbracket and positive exact scale=17. Give (mode, function_description, a, b, target, inclusive_bracket, root_exists).","answer":{"kind":"expression","value":"(nonbracket, f(x)=17*x, 0, 1, 34, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonbracket, f(x)=17*x, 0, 1, 34, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0093","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"53b4989a16cb9d7c1e20baa85432688986f17dbb697ce4e51e1bd5da5fe55e47","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162093,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonunique\",\"scale\":[18,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonunique and positive exact scale=18. Give (mode, function_description, a, b, target, inclusive_bracket, multiplicity_statement, uniqueness_claim).","answer":{"kind":"expression","value":"(nonunique, f(x)=x^2, -18, 18, 324, Yes, two_roots_possible, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonunique, f(x)=x^2, -18, 18, 324, Yes, two_roots_possible, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0094","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90cdc7a7189bc5a262e709b082b10ddfd788a6664e00b157332196b9f7260d80","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162094,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"discontinuous_jump\",\"scale\":[19,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=discontinuous_jump and positive exact scale=19. Give (mode, function_description, a, b, target, inclusive_bracket, continuous, root_exists).","answer":{"kind":"expression","value":"(discontinuous_jump, f(x)=-19_for_x<0_and_19_for_x>=0, -1, 1, 0, Yes, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (discontinuous_jump, f(x)=-19_for_x<0_and_19_for_x>=0, -1, 1, 0, Yes, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0095","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"279fd7adbe4bad2cfa319d89a6a031cc41321417d5dfb4545cff8b477915b9b0","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162095,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonbracket\",\"scale\":[20,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonbracket and positive exact scale=20. Give (mode, function_description, a, b, target, inclusive_bracket, root_exists).","answer":{"kind":"expression","value":"(nonbracket, f(x)=20*x, 0, 1, 40, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonbracket, f(x)=20*x, 0, 1, 40, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0096","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4e82a57e60ebc57f1e6e05314e7198a0ab3c9fa48ea7f3a2ed36303e34119443","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162096,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonunique\",\"scale\":[21,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonunique and positive exact scale=21. Give (mode, function_description, a, b, target, inclusive_bracket, multiplicity_statement, uniqueness_claim).","answer":{"kind":"expression","value":"(nonunique, f(x)=x^2, -21, 21, 441, Yes, two_roots_possible, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonunique, f(x)=x^2, -21, 21, 441, Yes, two_roots_possible, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0097","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d832b4a183b9631a2f76b65aaeef205878a48c1ea8509177e046197a0f85aa2","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162097,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"discontinuous_jump\",\"scale\":[22,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=discontinuous_jump and positive exact scale=22. Give (mode, function_description, a, b, target, inclusive_bracket, continuous, root_exists).","answer":{"kind":"expression","value":"(discontinuous_jump, f(x)=-22_for_x<0_and_22_for_x>=0, -1, 1, 0, Yes, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (discontinuous_jump, f(x)=-22_for_x<0_and_22_for_x>=0, -1, 1, 0, Yes, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0098","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c468fce5ce71fa9303d907271ce7a22d3d68e294b6904a7c7a58a3061c1048c1","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162098,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonbracket\",\"scale\":[23,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonbracket and positive exact scale=23. Give (mode, function_description, a, b, target, inclusive_bracket, root_exists).","answer":{"kind":"expression","value":"(nonbracket, f(x)=23*x, 0, 1, 46, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonbracket, f(x)=23*x, 0, 1, 46, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0099","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f1c206263c915dd9fdcc926db6a5f3095c97aae76608266fd8bd806b626b8c65","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162099,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"nonunique\",\"scale\":[24,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=nonunique and positive exact scale=24. Give (mode, function_description, a, b, target, inclusive_bracket, multiplicity_statement, uniqueness_claim).","answer":{"kind":"expression","value":"(nonunique, f(x)=x^2, -24, 24, 576, Yes, two_roots_possible, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (nonunique, f(x)=x^2, -24, 24, 576, Yes, two_roots_possible, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r02-intermediate-value-theorem-0100","batch_id":"batch-20261015-advance-20261116-r02-intermediate-value-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d255b35e0485c6cc742931708839f534f42ef0c796638d0f044f3a727dc0ba52","family_id":"qf-mod47-ivt-discontinuity-boundary","seed":202611162100,"semantic_template":"qf-mod47-ivt-discontinuity-boundary/{\"mode\":\"discontinuous_jump\",\"scale\":[25,1]}"},"fields":{"title":"A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries","stem":"Use boundary mode=discontinuous_jump and positive exact scale=25. Give (mode, function_description, a, b, target, inclusive_bracket, continuous, root_exists).","answer":{"kind":"expression","value":"(discontinuous_jump, f(x)=-25_for_x<0_and_25_for_x>=0, -1, 1, 0, Yes, No, No)","explanation":"The discontinuous jump takes only ±scale and misses0 despite bracketing it on [−1,1]. A scaled identity on [0,1] misses target2*scale, so the bracket fails. For x^2 on [−scale,scale], target scale^2 has both endpoint roots: existence is not uniqueness. Preserve each branch field meaning and do not infer continuity from samples. The exact result, retaining the requested order, is (discontinuous_jump, f(x)=-25_for_x<0_and_25_for_x>=0, -1, 1, 0, Yes, No, No)."},"skills":["A Discontinuous Sign Jump and Failure of Uniqueness Have Different Boundaries"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}