{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0001","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[0]；F0是常數項固定零的canonical polynomial primitive。求 x^1 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"0/1","explanation_zh":"原函數指定係數=0/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095001,"parameters":{"f":[0],"k":1,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[0,1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0002","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[2]；F0是常數項固定零的canonical polynomial primitive。求 x^4 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"0/1","explanation_zh":"原函數指定係數=0/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095002,"parameters":{"f":[2],"k":4,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[0,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0003","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-10, 3, -4, 5]；F0是常數項固定零的canonical polynomial primitive。求 x^3 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-4/3","explanation_zh":"原函數指定係數=-4/3。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095003,"parameters":{"f":[-10,3,-4,5],"k":3,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-4,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0004","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-9, 4, -5, -6]；F0是常數項固定零的canonical polynomial primitive。求 x^4 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-3/2","explanation_zh":"原函數指定係數=-3/2。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095004,"parameters":{"f":[-9,4,-5,-6],"k":4,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-6,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0005","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-8, 5, -6, 7]；F0是常數項固定零的canonical polynomial primitive。求 x^1 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-8/1","explanation_zh":"原函數指定係數=-8/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095005,"parameters":{"f":[-8,5,-6,7],"k":1,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-8,1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0006","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-7, 6, -7, -8]；F0是常數項固定零的canonical polynomial primitive。求 x^2 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"3/1","explanation_zh":"原函數指定係數=3/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095006,"parameters":{"f":[-7,6,-7,-8],"k":2,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[6,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0007","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-6, 7, -8, 9]；F0是常數項固定零的canonical polynomial primitive。求 x^3 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-8/3","explanation_zh":"原函數指定係數=-8/3。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095007,"parameters":{"f":[-6,7,-8,9],"k":3,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-8,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0008","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-5, 8, -9, -10]；F0是常數項固定零的canonical polynomial primitive。求 x^4 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-5/2","explanation_zh":"原函數指定係數=-5/2。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095008,"parameters":{"f":[-5,8,-9,-10],"k":4,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-10,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0009","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-4, 9, -10, 11]；F0是常數項固定零的canonical polynomial primitive。求 x^1 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-4/1","explanation_zh":"原函數指定係數=-4/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095009,"parameters":{"f":[-4,9,-10,11],"k":1,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-4,1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0010","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-3, 10, -11, -12]；F0是常數項固定零的canonical polynomial primitive。求 x^2 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"5/1","explanation_zh":"原函數指定係數=5/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095010,"parameters":{"f":[-3,10,-11,-12],"k":2,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[10,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0011","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-2, 11, -12, 13]；F0是常數項固定零的canonical polynomial primitive。求 x^3 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-4/1","explanation_zh":"原函數指定係數=-4/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095011,"parameters":{"f":[-2,11,-12,13],"k":3,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-12,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0012","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-1, 12, -13, -14]；F0是常數項固定零的canonical polynomial primitive。求 x^4 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-7/2","explanation_zh":"原函數指定係數=-7/2。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095012,"parameters":{"f":[-1,12,-13,-14],"k":4,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-14,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0013","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[1, 14, -15, -16]；F0是常數項固定零的canonical polynomial primitive。求 x^2 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"7/1","explanation_zh":"原函數指定係數=7/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095013,"parameters":{"f":[1,14,-15,-16],"k":2,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[14,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0014","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[2, 15, -16, 17]；F0是常數項固定零的canonical polynomial primitive。求 x^3 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-16/3","explanation_zh":"原函數指定係數=-16/3。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095014,"parameters":{"f":[2,15,-16,17],"k":3,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-16,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0015","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[3, 16, -17, -18]；F0是常數項固定零的canonical polynomial primitive。求 x^4 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-9/2","explanation_zh":"原函數指定係數=-9/2。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095015,"parameters":{"f":[3,16,-17,-18],"k":4,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-18,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0016","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[4, 17, -18, 19]；F0是常數項固定零的canonical polynomial primitive。求 x^1 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"4/1","explanation_zh":"原函數指定係數=4/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095016,"parameters":{"f":[4,17,-18,19],"k":1,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[4,1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0017","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[5, 18, -19, -20]；F0是常數項固定零的canonical polynomial primitive。求 x^2 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"9/1","explanation_zh":"原函數指定係數=9/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095017,"parameters":{"f":[5,18,-19,-20],"k":2,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[18,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0018","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[6, 19, -20, 21]；F0是常數項固定零的canonical polynomial primitive。求 x^3 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-20/3","explanation_zh":"原函數指定係數=-20/3。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095018,"parameters":{"f":[6,19,-20,21],"k":3,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-20,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0019","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[7, 20, -21, -22]；F0是常數項固定零的canonical polynomial primitive。求 x^4 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-11/2","explanation_zh":"原函數指定係數=-11/2。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095019,"parameters":{"f":[7,20,-21,-22],"k":4,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-22,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0020","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[8, 21, -22, 23]；F0是常數項固定零的canonical polynomial primitive。求 x^1 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"8/1","explanation_zh":"原函數指定係數=8/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095020,"parameters":{"f":[8,21,-22,23],"k":1,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[8,1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0021","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[9, 22, -23, -24]；F0是常數項固定零的canonical polynomial primitive。求 x^2 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"11/1","explanation_zh":"原函數指定係數=11/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095021,"parameters":{"f":[9,22,-23,-24],"k":2,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[22,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0022","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[10, 23, -24, 25]；F0是常數項固定零的canonical polynomial primitive。求 x^3 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-8/1","explanation_zh":"原函數指定係數=-8/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095022,"parameters":{"f":[10,23,-24,25],"k":3,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-24,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0023","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[11, 24, -25, -26]；F0是常數項固定零的canonical polynomial primitive。求 x^4 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"-13/2","explanation_zh":"原函數指定係數=-13/2。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095023,"parameters":{"f":[11,24,-25,-26],"k":4,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[-26,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0024","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[12, 25, -26, 27]；F0是常數項固定零的canonical polynomial primitive。求 x^1 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"12/1","explanation_zh":"原函數指定係數=12/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095024,"parameters":{"f":[12,25,-26,27],"k":1,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[12,1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0025","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"零常數原函數的指定係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[13, 26, -27, -28]；F0是常數項固定零的canonical polynomial primitive。求 x^2 的係數，不存在的項按零，採最簡分數。","answer":{"kind":"expression","value":"13/1","explanation_zh":"原函數指定係數=13/1。積分係數除以次方加一；指定不存在高項為零，不截掉本來存在的項。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["零常數原函數的指定係數"]},"variant":{"family_id":"qf-density-primitive-coefficient","seed":202610095025,"parameters":{"f":[13,26,-27,-28],"k":2,"operation":"density_primitive_coefficient","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-coefficient/[26,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0026","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-12、C2=0。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"-12/1","explanation_zh":"共享原函數常數差=-12/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095026,"parameters":{"C1":-12,"C2":0,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"-12\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0027","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-11、C2=-1。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"-10/1","explanation_zh":"共享原函數常數差=-10/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095027,"parameters":{"C1":-11,"C2":-1,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"-10\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0028","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-10、C2=-2。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"-8/1","explanation_zh":"共享原函數常數差=-8/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095028,"parameters":{"C1":-10,"C2":-2,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"-8\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0029","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-9、C2=-3。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"-6/1","explanation_zh":"共享原函數常數差=-6/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095029,"parameters":{"C1":-9,"C2":-3,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"-6\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0030","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-8、C2=-4。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"-4/1","explanation_zh":"共享原函數常數差=-4/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095030,"parameters":{"C1":-8,"C2":-4,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"-4\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0031","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-7、C2=-5。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"-2/1","explanation_zh":"共享原函數常數差=-2/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095031,"parameters":{"C1":-7,"C2":-5,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"-2\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0032","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-6、C2=-6。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"0/1","explanation_zh":"共享原函數常數差=0/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095032,"parameters":{"C1":-6,"C2":-6,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"0\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0033","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-5、C2=-7。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"2/1","explanation_zh":"共享原函數常數差=2/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095033,"parameters":{"C1":-5,"C2":-7,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"2\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0034","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-4、C2=-8。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"4/1","explanation_zh":"共享原函數常數差=4/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095034,"parameters":{"C1":-4,"C2":-8,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"4\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0035","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-3、C2=-9。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"6/1","explanation_zh":"共享原函數常數差=6/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095035,"parameters":{"C1":-3,"C2":-9,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"6\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0036","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-2、C2=-10。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"8/1","explanation_zh":"共享原函數常數差=8/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095036,"parameters":{"C1":-2,"C2":-10,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"8\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0037","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-1、C2=-11。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"10/1","explanation_zh":"共享原函數常數差=10/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095037,"parameters":{"C1":-1,"C2":-11,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"10\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0038","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=0、C2=-12。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"12/1","explanation_zh":"共享原函數常數差=12/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095038,"parameters":{"C1":0,"C2":-12,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"12\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0039","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=1、C2=-13。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"14/1","explanation_zh":"共享原函數常數差=14/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095039,"parameters":{"C1":1,"C2":-13,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"14\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0040","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=2、C2=-14。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"16/1","explanation_zh":"共享原函數常數差=16/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095040,"parameters":{"C1":2,"C2":-14,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"16\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0041","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=3、C2=-15。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"18/1","explanation_zh":"共享原函數常數差=18/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095041,"parameters":{"C1":3,"C2":-15,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"18\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0042","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=4、C2=-16。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"20/1","explanation_zh":"共享原函數常數差=20/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095042,"parameters":{"C1":4,"C2":-16,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"20\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0043","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=5、C2=-17。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"22/1","explanation_zh":"共享原函數常數差=22/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095043,"parameters":{"C1":5,"C2":-17,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"22\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0044","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=6、C2=-18。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"24/1","explanation_zh":"共享原函數常數差=24/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095044,"parameters":{"C1":6,"C2":-18,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"24\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0045","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=7、C2=-19。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"26/1","explanation_zh":"共享原函數常數差=26/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095045,"parameters":{"C1":7,"C2":-19,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"26\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0046","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=8、C2=-20。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"28/1","explanation_zh":"共享原函數常數差=28/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095046,"parameters":{"C1":8,"C2":-20,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"28\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0047","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=9、C2=-21。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"30/1","explanation_zh":"共享原函數常數差=30/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095047,"parameters":{"C1":9,"C2":-21,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"30\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0048","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=10、C2=-22。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"32/1","explanation_zh":"共享原函數常數差=32/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095048,"parameters":{"C1":10,"C2":-22,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"32\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0049","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=11、C2=-23。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"34/1","explanation_zh":"共享原函數常數差=34/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095049,"parameters":{"C1":11,"C2":-23,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"34\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0050","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"共享原函數的兩常數差","stem_zh":"全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=12、C2=-24。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。","answer":{"kind":"expression","value":"36/1","explanation_zh":"共享原函數常數差=36/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["共享原函數的兩常數差"]},"variant":{"family_id":"qf-density-primitive-constant-difference","seed":202610095050,"parameters":{"C1":12,"C2":-24,"operation":"density_primitive_constant_difference","domain":"reals","semantic_template":"qf-density-primitive-constant-difference/[\"36\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0051","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[1, 1]，F0常數項固定零。令 F=F0+C 且 F(0)=1/2，求C。","answer":{"kind":"expression","value":"1/2","explanation_zh":"錨點所需常數=1/2。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095051,"parameters":{"f":[1,1],"a":0,"value_n":1,"value_d":2,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[\"zero_primitive_at_anchor\",\"1/2\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0052","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[0]，F0常數項固定零。令 F=F0+C 且 F(3)=-2/1，求C。","answer":{"kind":"expression","value":"-2/1","explanation_zh":"錨點所需常數=-2/1。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095052,"parameters":{"f":[0],"a":3,"value_n":-2,"value_d":1,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[\"zero_primitive_at_anchor\",\"-2/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0053","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-10, 3, 1, 3]，F0常數項固定零。令 F=F0+C 且 F(-3)=-10/3，求C。","answer":{"kind":"expression","value":"-1183/12","explanation_zh":"錨點所需常數=-1183/12。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095053,"parameters":{"f":[-10,3,1,3],"a":-3,"value_n":-10,"value_d":3,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[-10,3,1,3],-3,\"-10/3\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0054","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-9, 4, -1, 4]，F0常數項固定零。令 F=F0+C 且 F(-4)=-9/1，求C。","answer":{"kind":"expression","value":"-1063/3","explanation_zh":"錨點所需常數=-1063/3。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095054,"parameters":{"f":[-9,4,-1,4],"a":-4,"value_n":-9,"value_d":1,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[-9,4,-1,4],-4,\"-9/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0055","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-8, 5, 0, 1]，F0常數項固定零。令 F=F0+C 且 F(-1)=-8/2，求C。","answer":{"kind":"expression","value":"-59/4","explanation_zh":"錨點所需常數=-59/4。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095055,"parameters":{"f":[-8,5,0,1],"a":-1,"value_n":-8,"value_d":2,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[-8,5,0,1],-1,\"-4/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0056","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-7, 6, 1, 2]，F0常數項固定零。令 F=F0+C 且 F(-2)=-7/3，求C。","answer":{"kind":"expression","value":"-101/3","explanation_zh":"錨點所需常數=-101/3。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095056,"parameters":{"f":[-7,6,1,2],"a":-2,"value_n":-7,"value_d":3,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[-7,6,1,2],-2,\"-7/3\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0057","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-6, 7, -1, 3]，F0常數項固定零。令 F=F0+C 且 F(-3)=-6/1，求C。","answer":{"kind":"expression","value":"-501/4","explanation_zh":"錨點所需常數=-501/4。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095057,"parameters":{"f":[-6,7,-1,3],"a":-3,"value_n":-6,"value_d":1,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[-6,7,-1,3],-3,\"-6/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0058","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-5, 8, 0, 4]，F0常數項固定零。令 F=F0+C 且 F(-4)=-5/2，求C。","answer":{"kind":"expression","value":"-685/2","explanation_zh":"錨點所需常數=-685/2。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095058,"parameters":{"f":[-5,8,0,4],"a":-4,"value_n":-5,"value_d":2,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[-5,8,0,4],-4,\"-5/2\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0059","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-4, 9, 1, 1]，F0常數項固定零。令 F=F0+C 且 F(-1)=-4/3，求C。","answer":{"kind":"expression","value":"-39/4","explanation_zh":"錨點所需常數=-39/4。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095059,"parameters":{"f":[-4,9,1,1],"a":-1,"value_n":-4,"value_d":3,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[-4,9,1,1],-1,\"-4/3\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0060","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-3, 10, -1, 2]，F0常數項固定零。令 F=F0+C 且 F(-2)=-3/1，求C。","answer":{"kind":"expression","value":"-119/3","explanation_zh":"錨點所需常數=-119/3。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095060,"parameters":{"f":[-3,10,-1,2],"a":-2,"value_n":-3,"value_d":1,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[-3,10,-1,2],-2,\"-3/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0061","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-2, 11, 0, 3]，F0常數項固定零。令 F=F0+C 且 F(-3)=-2/2，求C。","answer":{"kind":"expression","value":"-469/4","explanation_zh":"錨點所需常數=-469/4。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095061,"parameters":{"f":[-2,11,0,3],"a":-3,"value_n":-2,"value_d":2,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[-2,11,0,3],-3,\"-1/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0062","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[-1, 12, 1, 4]，F0常數項固定零。令 F=F0+C 且 F(-4)=-1/3，求C。","answer":{"kind":"expression","value":"-335/1","explanation_zh":"錨點所需常數=-335/1。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095062,"parameters":{"f":[-1,12,1,4],"a":-4,"value_n":-1,"value_d":3,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[-1,12,1,4],-4,\"-1/3\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0063","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[0, 13, -1, 1]，F0常數項固定零。令 F=F0+C 且 F(-1)=0/1，求C。","answer":{"kind":"expression","value":"-85/12","explanation_zh":"錨點所需常數=-85/12。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095063,"parameters":{"f":[0,13,-1,1],"a":-1,"value_n":0,"value_d":1,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[0,13,-1,1],-1,\"0/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0064","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[1, 14, 0, 2]，F0常數項固定零。令 F=F0+C 且 F(-2)=1/2，求C。","answer":{"kind":"expression","value":"-67/2","explanation_zh":"錨點所需常數=-67/2。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095064,"parameters":{"f":[1,14,0,2],"a":-2,"value_n":1,"value_d":2,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[1,14,0,2],-2,\"1/2\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0065","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[2, 15, 1, 3]，F0常數項固定零。令 F=F0+C 且 F(-3)=2/3，求C。","answer":{"kind":"expression","value":"-1351/12","explanation_zh":"錨點所需常數=-1351/12。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095065,"parameters":{"f":[2,15,1,3],"a":-3,"value_n":2,"value_d":3,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[2,15,1,3],-3,\"2/3\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0066","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[3, 16, -1, 4]，F0常數項固定零。令 F=F0+C 且 F(-4)=3/1，求C。","answer":{"kind":"expression","value":"-1171/3","explanation_zh":"錨點所需常數=-1171/3。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095066,"parameters":{"f":[3,16,-1,4],"a":-4,"value_n":3,"value_d":1,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[3,16,-1,4],-4,\"3/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0067","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[4, 17, 0, 1]，F0常數項固定零。令 F=F0+C 且 F(-1)=4/2，求C。","answer":{"kind":"expression","value":"-11/4","explanation_zh":"錨點所需常數=-11/4。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095067,"parameters":{"f":[4,17,0,1],"a":-1,"value_n":4,"value_d":2,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[4,17,0,1],-1,\"2/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0068","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[5, 18, 1, 2]，F0常數項固定零。令 F=F0+C 且 F(-2)=5/3，求C。","answer":{"kind":"expression","value":"-89/3","explanation_zh":"錨點所需常數=-89/3。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095068,"parameters":{"f":[5,18,1,2],"a":-2,"value_n":5,"value_d":3,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[5,18,1,2],-2,\"5/3\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0069","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[6, 19, -1, 3]，F0常數項固定零。令 F=F0+C 且 F(-3)=6/1，求C。","answer":{"kind":"expression","value":"-525/4","explanation_zh":"錨點所需常數=-525/4。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095069,"parameters":{"f":[6,19,-1,3],"a":-3,"value_n":6,"value_d":1,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[6,19,-1,3],-3,\"6/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0070","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[7, 20, 0, 4]，F0常數項固定零。令 F=F0+C 且 F(-4)=7/2，求C。","answer":{"kind":"expression","value":"-769/2","explanation_zh":"錨點所需常數=-769/2。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095070,"parameters":{"f":[7,20,0,4],"a":-4,"value_n":7,"value_d":2,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[7,20,0,4],-4,\"7/2\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0071","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[8, 21, 1, 1]，F0常數項固定零。令 F=F0+C 且 F(-1)=8/3，求C。","answer":{"kind":"expression","value":"1/4","explanation_zh":"錨點所需常數=1/4。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095071,"parameters":{"f":[8,21,1,1],"a":-1,"value_n":8,"value_d":3,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[8,21,1,1],-1,\"8/3\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0072","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[9, 22, -1, 2]，F0常數項固定零。令 F=F0+C 且 F(-2)=9/1，求C。","answer":{"kind":"expression","value":"-83/3","explanation_zh":"錨點所需常數=-83/3。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095072,"parameters":{"f":[9,22,-1,2],"a":-2,"value_n":9,"value_d":1,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[9,22,-1,2],-2,\"9/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0073","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[10, 23, 0, 3]，F0常數項固定零。令 F=F0+C 且 F(-3)=10/2，求C。","answer":{"kind":"expression","value":"-517/4","explanation_zh":"錨點所需常數=-517/4。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095073,"parameters":{"f":[10,23,0,3],"a":-3,"value_n":10,"value_d":2,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[10,23,0,3],-3,\"5/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0074","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[11, 24, 1, 4]，F0常數項固定零。令 F=F0+C 且 F(-4)=11/3，求C。","answer":{"kind":"expression","value":"-379/1","explanation_zh":"錨點所需常數=-379/1。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095074,"parameters":{"f":[11,24,1,4],"a":-4,"value_n":11,"value_d":3,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[11,24,1,4],-4,\"11/3\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0075","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"由明示錨點解原函數常數","stem_zh":"全實數論域；升冪係數的受限多項式模型。升冪係數 f=[12, 25, -1, 1]，F0常數項固定零。令 F=F0+C 且 F(-1)=12/1，求C。","answer":{"kind":"expression","value":"131/12","explanation_zh":"錨點所需常數=131/12。求指定錨點的精確F0值再相減；保留實際點與目標值，C不由求導資格單獨決定。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["由明示錨點解原函數常數"]},"variant":{"family_id":"qf-density-primitive-anchor-constant","seed":202610095075,"parameters":{"f":[12,25,-1,1],"a":-1,"value_n":12,"value_d":1,"operation":"density_primitive_anchor_constant","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-anchor-constant/[[12,25,-1,1],-1,\"12/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0076","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[7,1]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[0/1]","explanation_zh":"候選導數升冪係數=[0/1]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095076,"parameters":{"F":[[7,1]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0077","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[3,1],[2,3]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[2/3]","explanation_zh":"候選導數升冪係數=[2/3]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095077,"parameters":{"F":[[3,1],[2,3]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"2/3\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0078","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[-10,1],[3,2],[-1,3],[1,1],[2,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[3/2, -2/3, 3/1, 8/5]","explanation_zh":"候選導數升冪係數=[3/2, -2/3, 3/1, 8/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095078,"parameters":{"F":[[-10,1],[3,2],[-1,3],[1,1],[2,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"3/2\",\"-1/3\",\"1/1\",\"2/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0079","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[-9,1],[2,1],[-2,3],[5,4],[-3,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[2/1, -4/3, 15/4, -12/5]","explanation_zh":"候選導數升冪係數=[2/1, -4/3, 15/4, -12/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095079,"parameters":{"F":[[-9,1],[2,1],[-2,3],[5,4],[-3,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"2/1\",\"-2/3\",\"5/4\",\"-3/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0080","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[-8,1],[5,2],[-1,1],[3,2],[4,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[5/2, -2/1, 9/2, 16/5]","explanation_zh":"候選導數升冪係數=[5/2, -2/1, 9/2, 16/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095080,"parameters":{"F":[[-8,1],[5,2],[-1,1],[3,2],[4,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"5/2\",\"-1/1\",\"3/2\",\"4/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0081","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[-7,1],[3,1],[-4,3],[7,4],[-1,1]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[3/1, -8/3, 21/4, -4/1]","explanation_zh":"候選導數升冪係數=[3/1, -8/3, 21/4, -4/1]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095081,"parameters":{"F":[[-7,1],[3,1],[-4,3],[7,4],[-1,1]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"3/1\",\"-4/3\",\"7/4\",\"-1/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0082","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[-6,1],[7,2],[-5,3],[2,1],[6,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[7/2, -10/3, 6/1, 24/5]","explanation_zh":"候選導數升冪係數=[7/2, -10/3, 6/1, 24/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095082,"parameters":{"F":[[-6,1],[7,2],[-5,3],[2,1],[6,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"7/2\",\"-5/3\",\"2/1\",\"6/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0083","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[-5,1],[4,1],[-2,1],[9,4],[-7,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[4/1, -4/1, 27/4, -28/5]","explanation_zh":"候選導數升冪係數=[4/1, -4/1, 27/4, -28/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095083,"parameters":{"F":[[-5,1],[4,1],[-2,1],[9,4],[-7,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"4/1\",\"-2/1\",\"9/4\",\"-7/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0084","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[-4,1],[9,2],[-7,3],[5,2],[8,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[9/2, -14/3, 15/2, 32/5]","explanation_zh":"候選導數升冪係數=[9/2, -14/3, 15/2, 32/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095084,"parameters":{"F":[[-4,1],[9,2],[-7,3],[5,2],[8,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"9/2\",\"-7/3\",\"5/2\",\"8/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0085","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[-3,1],[5,1],[-8,3],[11,4],[-9,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[5/1, -16/3, 33/4, -36/5]","explanation_zh":"候選導數升冪係數=[5/1, -16/3, 33/4, -36/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095085,"parameters":{"F":[[-3,1],[5,1],[-8,3],[11,4],[-9,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"5/1\",\"-8/3\",\"11/4\",\"-9/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0086","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[-2,1],[11,2],[-3,1],[3,1],[2,1]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[11/2, -6/1, 9/1, 8/1]","explanation_zh":"候選導數升冪係數=[11/2, -6/1, 9/1, 8/1]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095086,"parameters":{"F":[[-2,1],[11,2],[-3,1],[3,1],[2,1]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"11/2\",\"-3/1\",\"3/1\",\"2/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0087","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[-1,1],[6,1],[-10,3],[13,4],[-11,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[6/1, -20/3, 39/4, -44/5]","explanation_zh":"候選導數升冪係數=[6/1, -20/3, 39/4, -44/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095087,"parameters":{"F":[[-1,1],[6,1],[-10,3],[13,4],[-11,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"6/1\",\"-10/3\",\"13/4\",\"-11/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0088","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[0,1],[13,2],[-11,3],[7,2],[12,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[13/2, -22/3, 21/2, 48/5]","explanation_zh":"候選導數升冪係數=[13/2, -22/3, 21/2, 48/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095088,"parameters":{"F":[[0,1],[13,2],[-11,3],[7,2],[12,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"13/2\",\"-11/3\",\"7/2\",\"12/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0089","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[1,1],[7,1],[-4,1],[15,4],[-13,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[7/1, -8/1, 45/4, -52/5]","explanation_zh":"候選導數升冪係數=[7/1, -8/1, 45/4, -52/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095089,"parameters":{"F":[[1,1],[7,1],[-4,1],[15,4],[-13,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"7/1\",\"-4/1\",\"15/4\",\"-13/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0090","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[2,1],[15,2],[-13,3],[4,1],[14,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[15/2, -26/3, 12/1, 56/5]","explanation_zh":"候選導數升冪係數=[15/2, -26/3, 12/1, 56/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095090,"parameters":{"F":[[2,1],[15,2],[-13,3],[4,1],[14,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"15/2\",\"-13/3\",\"4/1\",\"14/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0091","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[3,1],[8,1],[-14,3],[17,4],[-3,1]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[8/1, -28/3, 51/4, -12/1]","explanation_zh":"候選導數升冪係數=[8/1, -28/3, 51/4, -12/1]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095091,"parameters":{"F":[[3,1],[8,1],[-14,3],[17,4],[-3,1]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"8/1\",\"-14/3\",\"17/4\",\"-3/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0092","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[4,1],[17,2],[-5,1],[9,2],[16,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[17/2, -10/1, 27/2, 64/5]","explanation_zh":"候選導數升冪係數=[17/2, -10/1, 27/2, 64/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095092,"parameters":{"F":[[4,1],[17,2],[-5,1],[9,2],[16,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"17/2\",\"-5/1\",\"9/2\",\"16/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0093","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[5,1],[9,1],[-16,3],[19,4],[-17,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[9/1, -32/3, 57/4, -68/5]","explanation_zh":"候選導數升冪係數=[9/1, -32/3, 57/4, -68/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095093,"parameters":{"F":[[5,1],[9,1],[-16,3],[19,4],[-17,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"9/1\",\"-16/3\",\"19/4\",\"-17/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0094","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[6,1],[19,2],[-17,3],[5,1],[18,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[19/2, -34/3, 15/1, 72/5]","explanation_zh":"候選導數升冪係數=[19/2, -34/3, 15/1, 72/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095094,"parameters":{"F":[[6,1],[19,2],[-17,3],[5,1],[18,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"19/2\",\"-17/3\",\"5/1\",\"18/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0095","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[7,1],[10,1],[-6,1],[21,4],[-19,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[10/1, -12/1, 63/4, -76/5]","explanation_zh":"候選導數升冪係數=[10/1, -12/1, 63/4, -76/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095095,"parameters":{"F":[[7,1],[10,1],[-6,1],[21,4],[-19,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"10/1\",\"-6/1\",\"21/4\",\"-19/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0096","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[8,1],[21,2],[-19,3],[11,2],[4,1]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[21/2, -38/3, 33/2, 16/1]","explanation_zh":"候選導數升冪係數=[21/2, -38/3, 33/2, 16/1]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095096,"parameters":{"F":[[8,1],[21,2],[-19,3],[11,2],[4,1]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"21/2\",\"-19/3\",\"11/2\",\"4/1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0097","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[9,1],[11,1],[-20,3],[23,4],[-21,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[11/1, -40/3, 69/4, -84/5]","explanation_zh":"候選導數升冪係數=[11/1, -40/3, 69/4, -84/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095097,"parameters":{"F":[[9,1],[11,1],[-20,3],[23,4],[-21,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"11/1\",\"-20/3\",\"23/4\",\"-21/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0098","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[10,1],[23,2],[-7,1],[6,1],[22,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[23/2, -14/1, 18/1, 88/5]","explanation_zh":"候選導數升冪係數=[23/2, -14/1, 18/1, 88/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095098,"parameters":{"F":[[10,1],[23,2],[-7,1],[6,1],[22,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"23/2\",\"-7/1\",\"6/1\",\"22/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0099","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[11,1],[12,1],[-22,3],[25,4],[-23,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[12/1, -44/3, 75/4, -92/5]","explanation_zh":"候選導數升冪係數=[12/1, -44/3, 75/4, -92/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095099,"parameters":{"F":[[11,1],[12,1],[-22,3],[25,4],[-23,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"12/1\",\"-22/3\",\"25/4\",\"-23/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r05-antiderivative-0100","batch_id":"batch-20261007-advance-20261009-r05-antiderivative-foundations","language":"zh-Hant","source_mko_ids":["mko-antiderivative"],"title_zh":"有理候選的導數係數","stem_zh":"全實數論域；升冪係數的受限多項式模型。候選F的升冪有理係數為 [[12,1],[25,2],[-23,3],[13,2],[24,5]]，每對為[分子,正分母]。列F′的完整升冪最簡有理係數，零式列[0/1]。","answer":{"kind":"expression","value":"[25/2, -46/3, 39/2, 96/5]","explanation_zh":"候選導數升冪係數=[25/2, -46/3, 39/2, 96/5]。以係數乘次方求導並去除末端零；候選常數項不影響導數，也不以代入抽樣作一般證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有理候選的導數係數"]},"variant":{"family_id":"qf-density-primitive-candidate-derivative","seed":202610095100,"parameters":{"F":[[12,1],[25,2],[-23,3],[13,2],[24,5]],"operation":"density_primitive_candidate_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-density-primitive-candidate-derivative/[\"25/2\",\"-23/3\",\"13/2\",\"24/5\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-07","contract_version":"ocme-question-factory-v0.1"}}