{"schema_version":"ocme-question-v0.1","id":"q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0001","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[0]。令 G(t)=∫_-1^t f(u)du，求 G′(-2)，採最簡正分母分數。","answer":{"kind":"expression","value":"0/1","explanation_zh":"累積導數=0/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091001,"parameters":{"f":[0],"a":-1,"x":-2,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[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-r01-fundamental-theorem-calculus-0002","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[3]。令 G(t)=∫_0^t f(u)du，求 G′(-1)，採最簡正分母分數。","answer":{"kind":"expression","value":"3/1","explanation_zh":"累積導數=3/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091002,"parameters":{"f":[3],"a":0,"x":-1,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[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-r01-fundamental-theorem-calculus-0003","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-10, 3, 1, 3]。令 G(t)=∫_1^t f(u)du，求 G′(0)，採最簡正分母分數。","answer":{"kind":"expression","value":"-10/1","explanation_zh":"累積導數=-10/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091003,"parameters":{"f":[-10,3,1,3],"a":1,"x":0,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[-10,3,1,3],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-r01-fundamental-theorem-calculus-0004","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-9, 4, -1, 4]。令 G(t)=∫_-1^t f(u)du，求 G′(1)，採最簡正分母分數。","answer":{"kind":"expression","value":"-2/1","explanation_zh":"累積導數=-2/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091004,"parameters":{"f":[-9,4,-1,4],"a":-1,"x":1,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[-9,4,-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-r01-fundamental-theorem-calculus-0005","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-8, 5, 0, 1]。令 G(t)=∫_0^t f(u)du，求 G′(2)，採最簡正分母分數。","answer":{"kind":"expression","value":"10/1","explanation_zh":"累積導數=10/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091005,"parameters":{"f":[-8,5,0,1],"a":0,"x":2,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[-8,5,0,1],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-r01-fundamental-theorem-calculus-0006","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-7, 6, 1, 2]。令 G(t)=∫_1^t f(u)du，求 G′(-2)，採最簡正分母分數。","answer":{"kind":"expression","value":"-31/1","explanation_zh":"累積導數=-31/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091006,"parameters":{"f":[-7,6,1,2],"a":1,"x":-2,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[-7,6,1,2],-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-r01-fundamental-theorem-calculus-0007","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-6, 7, -1, 3]。令 G(t)=∫_-1^t f(u)du，求 G′(-1)，採最簡正分母分數。","answer":{"kind":"expression","value":"-17/1","explanation_zh":"累積導數=-17/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091007,"parameters":{"f":[-6,7,-1,3],"a":-1,"x":-1,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[-6,7,-1,3],-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-r01-fundamental-theorem-calculus-0008","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-5, 8, 0, 4]。令 G(t)=∫_0^t f(u)du，求 G′(0)，採最簡正分母分數。","answer":{"kind":"expression","value":"-5/1","explanation_zh":"累積導數=-5/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091008,"parameters":{"f":[-5,8,0,4],"a":0,"x":0,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[-5,8,0,4],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-r01-fundamental-theorem-calculus-0009","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-4, 9, 1, 1]。令 G(t)=∫_1^t f(u)du，求 G′(1)，採最簡正分母分數。","answer":{"kind":"expression","value":"7/1","explanation_zh":"累積導數=7/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091009,"parameters":{"f":[-4,9,1,1],"a":1,"x":1,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[-4,9,1,1],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-r01-fundamental-theorem-calculus-0010","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-3, 10, -1, 2]。令 G(t)=∫_-1^t f(u)du，求 G′(2)，採最簡正分母分數。","answer":{"kind":"expression","value":"29/1","explanation_zh":"累積導數=29/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091010,"parameters":{"f":[-3,10,-1,2],"a":-1,"x":2,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[-3,10,-1,2],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-r01-fundamental-theorem-calculus-0011","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-2, 11, 0, 3]。令 G(t)=∫_0^t f(u)du，求 G′(-2)，採最簡正分母分數。","answer":{"kind":"expression","value":"-48/1","explanation_zh":"累積導數=-48/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091011,"parameters":{"f":[-2,11,0,3],"a":0,"x":-2,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[-2,11,0,3],-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-r01-fundamental-theorem-calculus-0012","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-1, 12, 1, 4]。令 G(t)=∫_1^t f(u)du，求 G′(-1)，採最簡正分母分數。","answer":{"kind":"expression","value":"-16/1","explanation_zh":"累積導數=-16/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091012,"parameters":{"f":[-1,12,1,4],"a":1,"x":-1,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[-1,12,1,4],-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-r01-fundamental-theorem-calculus-0013","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[0, 13, -1, 1]。令 G(t)=∫_-1^t f(u)du，求 G′(0)，採最簡正分母分數。","answer":{"kind":"expression","value":"0/1","explanation_zh":"累積導數=0/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091013,"parameters":{"f":[0,13,-1,1],"a":-1,"x":0,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[0,13,-1,1],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-r01-fundamental-theorem-calculus-0014","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[1, 14, 0, 2]。令 G(t)=∫_0^t f(u)du，求 G′(1)，採最簡正分母分數。","answer":{"kind":"expression","value":"17/1","explanation_zh":"累積導數=17/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091014,"parameters":{"f":[1,14,0,2],"a":0,"x":1,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[1,14,0,2],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-r01-fundamental-theorem-calculus-0015","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[2, 15, 1, 3]。令 G(t)=∫_1^t f(u)du，求 G′(2)，採最簡正分母分數。","answer":{"kind":"expression","value":"60/1","explanation_zh":"累積導數=60/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091015,"parameters":{"f":[2,15,1,3],"a":1,"x":2,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[2,15,1,3],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-r01-fundamental-theorem-calculus-0016","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[3, 16, -1, 4]。令 G(t)=∫_-1^t f(u)du，求 G′(-2)，採最簡正分母分數。","answer":{"kind":"expression","value":"-65/1","explanation_zh":"累積導數=-65/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091016,"parameters":{"f":[3,16,-1,4],"a":-1,"x":-2,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[3,16,-1,4],-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-r01-fundamental-theorem-calculus-0017","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[4, 17, 0, 1]。令 G(t)=∫_0^t f(u)du，求 G′(-1)，採最簡正分母分數。","answer":{"kind":"expression","value":"-14/1","explanation_zh":"累積導數=-14/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091017,"parameters":{"f":[4,17,0,1],"a":0,"x":-1,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[4,17,0,1],-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-r01-fundamental-theorem-calculus-0018","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[5, 18, 1, 2]。令 G(t)=∫_1^t f(u)du，求 G′(0)，採最簡正分母分數。","answer":{"kind":"expression","value":"5/1","explanation_zh":"累積導數=5/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091018,"parameters":{"f":[5,18,1,2],"a":1,"x":0,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[5,18,1,2],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-r01-fundamental-theorem-calculus-0019","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[6, 19, -1, 3]。令 G(t)=∫_-1^t f(u)du，求 G′(1)，採最簡正分母分數。","answer":{"kind":"expression","value":"27/1","explanation_zh":"累積導數=27/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091019,"parameters":{"f":[6,19,-1,3],"a":-1,"x":1,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[6,19,-1,3],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-r01-fundamental-theorem-calculus-0020","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[7, 20, 0, 4]。令 G(t)=∫_0^t f(u)du，求 G′(2)，採最簡正分母分數。","answer":{"kind":"expression","value":"79/1","explanation_zh":"累積導數=79/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091020,"parameters":{"f":[7,20,0,4],"a":0,"x":2,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[7,20,0,4],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-r01-fundamental-theorem-calculus-0021","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[8, 21, 1, 1]。令 G(t)=∫_1^t f(u)du，求 G′(-2)，採最簡正分母分數。","answer":{"kind":"expression","value":"-38/1","explanation_zh":"累積導數=-38/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091021,"parameters":{"f":[8,21,1,1],"a":1,"x":-2,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[8,21,1,1],-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-r01-fundamental-theorem-calculus-0022","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[9, 22, -1, 2]。令 G(t)=∫_-1^t f(u)du，求 G′(-1)，採最簡正分母分數。","answer":{"kind":"expression","value":"-16/1","explanation_zh":"累積導數=-16/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091022,"parameters":{"f":[9,22,-1,2],"a":-1,"x":-1,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[9,22,-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-r01-fundamental-theorem-calculus-0023","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[10, 23, 0, 3]。令 G(t)=∫_0^t f(u)du，求 G′(0)，採最簡正分母分數。","answer":{"kind":"expression","value":"10/1","explanation_zh":"累積導數=10/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091023,"parameters":{"f":[10,23,0,3],"a":0,"x":0,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[10,23,0,3],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-r01-fundamental-theorem-calculus-0024","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[11, 24, 1, 4]。令 G(t)=∫_1^t f(u)du，求 G′(1)，採最簡正分母分數。","answer":{"kind":"expression","value":"40/1","explanation_zh":"累積導數=40/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091024,"parameters":{"f":[11,24,1,4],"a":1,"x":1,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[11,24,1,4],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-r01-fundamental-theorem-calculus-0025","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積積分的點導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[12, 25, -1, 1]。令 G(t)=∫_-1^t f(u)du，求 G′(2)，採最簡正分母分數。","answer":{"kind":"expression","value":"66/1","explanation_zh":"累積導數=66/1。使用連續前提下的累積導數關係，在指定點求 f；積分起點不改此導數，未新增一般FTC形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積積分的點導數"]},"variant":{"family_id":"qf-ftc-accumulation-derivative","seed":202610091025,"parameters":{"f":[12,25,-1,1],"a":-1,"x":2,"operation":"ftc_accumulation_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-accumulation-derivative/[[12,25,-1,1],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-r01-fundamental-theorem-calculus-0026","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[1]，l(t)=1t+(0)、u(t)=2t+(0)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(0)。","answer":{"kind":"expression","value":"1/1","explanation_zh":"移動端點導數=1/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091026,"parameters":{"f":[1],"lower_m":1,"lower_b":0,"upper_m":2,"upper_b":0,"x":0,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[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-r01-fundamental-theorem-calculus-0027","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[0]，l(t)=1t+(0)、u(t)=3t+(0)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(1)。","answer":{"kind":"expression","value":"0/1","explanation_zh":"移動端點導數=0/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091027,"parameters":{"f":[0],"lower_m":1,"lower_b":0,"upper_m":3,"upper_b":0,"x":1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[\"zero_integrand\"]"}},"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-r01-fundamental-theorem-calculus-0028","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[3, 1]，l(t)=2t+(1)、u(t)=2t+(1)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-1)。","answer":{"kind":"expression","value":"0/1","explanation_zh":"移動端點導數=0/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091028,"parameters":{"f":[3,1],"lower_m":2,"lower_b":1,"upper_m":2,"upper_b":1,"x":-1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[\"identical_bounds\"]"}},"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-r01-fundamental-theorem-calculus-0029","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-9, 4, -1, 4]，l(t)=-1t+(-3)、u(t)=0t+(-9)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(1)。","answer":{"kind":"expression","value":"-297/1","explanation_zh":"移動端點導數=-297/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091029,"parameters":{"f":[-9,4,-1,4],"lower_m":-1,"lower_b":-3,"upper_m":0,"upper_b":-9,"x":1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[-9,4,-1,4],-1,-3,0,-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-r01-fundamental-theorem-calculus-0030","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-8, 5, 0, 1]，l(t)=2t+(0)、u(t)=4t+(-8)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-2)。","answer":{"kind":"expression","value":"-16552/1","explanation_zh":"移動端點導數=-16552/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091030,"parameters":{"f":[-8,5,0,1],"lower_m":2,"lower_b":0,"upper_m":4,"upper_b":-8,"x":-2,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[-8,5,0,1],2,0,4,-8,-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-r01-fundamental-theorem-calculus-0031","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-7, 6, 1, 2]，l(t)=-3t+(-1)、u(t)=0t+(-7)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-1)。","answer":{"kind":"expression","value":"75/1","explanation_zh":"移動端點導數=75/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091031,"parameters":{"f":[-7,6,1,2],"lower_m":-3,"lower_b":-1,"upper_m":0,"upper_b":-7,"x":-1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[-7,6,1,2],-3,-1,0,-7,-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-r01-fundamental-theorem-calculus-0032","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-6, 7, -1, 3]，l(t)=1t+(-2)、u(t)=2t+(-6)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(0)。","answer":{"kind":"expression","value":"-1416/1","explanation_zh":"移動端點導數=-1416/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091032,"parameters":{"f":[-6,7,-1,3],"lower_m":1,"lower_b":-2,"upper_m":2,"upper_b":-6,"x":0,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[-6,7,-1,3],1,-2,2,-6,0]"}},"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-r01-fundamental-theorem-calculus-0033","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-5, 8, 0, 4]，l(t)=-2t+(-3)、u(t)=0t+(-5)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(1)。","answer":{"kind":"expression","value":"-1090/1","explanation_zh":"移動端點導數=-1090/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091033,"parameters":{"f":[-5,8,0,4],"lower_m":-2,"lower_b":-3,"upper_m":0,"upper_b":-5,"x":1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[-5,8,0,4],-2,-3,0,-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-r01-fundamental-theorem-calculus-0034","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-4, 9, 1, 1]，l(t)=3t+(0)、u(t)=6t+(-4)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-2)。","answer":{"kind":"expression","value":"-23214/1","explanation_zh":"移動端點導數=-23214/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091034,"parameters":{"f":[-4,9,1,1],"lower_m":3,"lower_b":0,"upper_m":6,"upper_b":-4,"x":-2,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[-4,9,1,1],3,0,6,-4,-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-r01-fundamental-theorem-calculus-0035","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-3, 10, -1, 2]，l(t)=-1t+(-1)、u(t)=0t+(-3)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-1)。","answer":{"kind":"expression","value":"-3/1","explanation_zh":"移動端點導數=-3/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091035,"parameters":{"f":[-3,10,-1,2],"lower_m":-1,"lower_b":-1,"upper_m":0,"upper_b":-3,"x":-1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[-3,10,-1,2],-1,-1,0,-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-r01-fundamental-theorem-calculus-0036","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-2, 11, 0, 3]，l(t)=2t+(-2)、u(t)=4t+(-2)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(0)。","answer":{"kind":"expression","value":"-96/1","explanation_zh":"移動端點導數=-96/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091036,"parameters":{"f":[-2,11,0,3],"lower_m":2,"lower_b":-2,"upper_m":4,"upper_b":-2,"x":0,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[-2,11,0,3],2,-2,4,-2,0]"}},"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-r01-fundamental-theorem-calculus-0037","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-1, 12, 1, 4]，l(t)=-3t+(-3)、u(t)=0t+(-1)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(1)。","answer":{"kind":"expression","value":"-2703/1","explanation_zh":"移動端點導數=-2703/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091037,"parameters":{"f":[-1,12,1,4],"lower_m":-3,"lower_b":-3,"upper_m":0,"upper_b":-1,"x":1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[-1,12,1,4],-3,-3,0,-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-r01-fundamental-theorem-calculus-0038","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[0, 13, -1, 1]，l(t)=1t+(0)、u(t)=2t+(0)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-2)。","answer":{"kind":"expression","value":"-226/1","explanation_zh":"移動端點導數=-226/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091038,"parameters":{"f":[0,13,-1,1],"lower_m":1,"lower_b":0,"upper_m":2,"upper_b":0,"x":-2,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[0,13,-1,1],1,0,2,0,-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-r01-fundamental-theorem-calculus-0039","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[1, 14, 0, 2]，l(t)=-2t+(-1)、u(t)=0t+(1)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-1)。","answer":{"kind":"expression","value":"34/1","explanation_zh":"移動端點導數=34/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091039,"parameters":{"f":[1,14,0,2],"lower_m":-2,"lower_b":-1,"upper_m":0,"upper_b":1,"x":-1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[1,14,0,2],-2,-1,0,1,-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-r01-fundamental-theorem-calculus-0040","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[2, 15, 1, 3]，l(t)=3t+(-2)、u(t)=6t+(2)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(0)。","answer":{"kind":"expression","value":"504/1","explanation_zh":"移動端點導數=504/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091040,"parameters":{"f":[2,15,1,3],"lower_m":3,"lower_b":-2,"upper_m":6,"upper_b":2,"x":0,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[2,15,1,3],3,-2,6,2,0]"}},"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-r01-fundamental-theorem-calculus-0041","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[3, 16, -1, 4]，l(t)=-1t+(-3)、u(t)=0t+(3)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(1)。","answer":{"kind":"expression","value":"-333/1","explanation_zh":"移動端點導數=-333/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091041,"parameters":{"f":[3,16,-1,4],"lower_m":-1,"lower_b":-3,"upper_m":0,"upper_b":3,"x":1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[3,16,-1,4],-1,-3,0,3,1]"}},"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-r01-fundamental-theorem-calculus-0042","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[4, 17, 0, 1]，l(t)=2t+(0)、u(t)=4t+(4)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-2)。","answer":{"kind":"expression","value":"-256/1","explanation_zh":"移動端點導數=-256/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091042,"parameters":{"f":[4,17,0,1],"lower_m":2,"lower_b":0,"upper_m":4,"upper_b":4,"x":-2,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[4,17,0,1],2,0,4,4,-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-r01-fundamental-theorem-calculus-0043","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[5, 18, 1, 2]，l(t)=-3t+(-1)、u(t)=0t+(5)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-1)。","answer":{"kind":"expression","value":"183/1","explanation_zh":"移動端點導數=183/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091043,"parameters":{"f":[5,18,1,2],"lower_m":-3,"lower_b":-1,"upper_m":0,"upper_b":5,"x":-1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[5,18,1,2],-3,-1,0,5,-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-r01-fundamental-theorem-calculus-0044","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[6, 19, -1, 3]，l(t)=1t+(-2)、u(t)=2t+(6)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(0)。","answer":{"kind":"expression","value":"1524/1","explanation_zh":"移動端點導數=1524/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091044,"parameters":{"f":[6,19,-1,3],"lower_m":1,"lower_b":-2,"upper_m":2,"upper_b":6,"x":0,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[6,19,-1,3],1,-2,2,6,0]"}},"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-r01-fundamental-theorem-calculus-0045","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[7, 20, 0, 4]，l(t)=-2t+(-3)、u(t)=0t+(7)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(1)。","answer":{"kind":"expression","value":"-1186/1","explanation_zh":"移動端點導數=-1186/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091045,"parameters":{"f":[7,20,0,4],"lower_m":-2,"lower_b":-3,"upper_m":0,"upper_b":7,"x":1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[7,20,0,4],-2,-3,0,7,1]"}},"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-r01-fundamental-theorem-calculus-0046","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[8, 21, 1, 1]，l(t)=3t+(0)、u(t)=6t+(8)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-2)。","answer":{"kind":"expression","value":"150/1","explanation_zh":"移動端點導數=150/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091046,"parameters":{"f":[8,21,1,1],"lower_m":3,"lower_b":0,"upper_m":6,"upper_b":8,"x":-2,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[8,21,1,1],3,0,6,8,-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-r01-fundamental-theorem-calculus-0047","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[9, 22, -1, 2]，l(t)=-1t+(-1)、u(t)=0t+(9)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-1)。","answer":{"kind":"expression","value":"9/1","explanation_zh":"移動端點導數=9/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091047,"parameters":{"f":[9,22,-1,2],"lower_m":-1,"lower_b":-1,"upper_m":0,"upper_b":9,"x":-1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[9,22,-1,2],-1,-1,0,9,-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-r01-fundamental-theorem-calculus-0048","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[10, 23, 0, 3]，l(t)=2t+(-2)、u(t)=4t+(10)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(0)。","answer":{"kind":"expression","value":"13080/1","explanation_zh":"移動端點導數=13080/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091048,"parameters":{"f":[10,23,0,3],"lower_m":2,"lower_b":-2,"upper_m":4,"upper_b":10,"x":0,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[10,23,0,3],2,-2,4,10,0]"}},"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-r01-fundamental-theorem-calculus-0049","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[11, 24, 1, 4]，l(t)=-3t+(-3)、u(t)=0t+(11)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(1)。","answer":{"kind":"expression","value":"-2883/1","explanation_zh":"移動端點導數=-2883/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091049,"parameters":{"f":[11,24,1,4],"lower_m":-3,"lower_b":-3,"upper_m":0,"upper_b":11,"x":1,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[11,24,1,4],-3,-3,0,11,1]"}},"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-r01-fundamental-theorem-calculus-0050","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"移動上下限的有向積分導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[12, 25, -1, 1]，l(t)=1t+(0)、u(t)=2t+(12)。H(t)=∫_l(t)^u(t) f(s)ds，求 H′(-2)。","answer":{"kind":"expression","value":"1370/1","explanation_zh":"移動端點導數=1370/1。分別以 f(u)u′ 與 f(l)l′ 計算再相減；某點上下限相等不保證導數零，這是連續模型的條件計算。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["移動上下限的有向積分導數"]},"variant":{"family_id":"qf-ftc-moving-bounds-derivative","seed":202610091050,"parameters":{"f":[12,25,-1,1],"lower_m":1,"lower_b":0,"upper_m":2,"upper_b":12,"x":-2,"operation":"ftc_moving_bounds_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-moving-bounds-derivative/[[12,25,-1,1],1,0,2,12,-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-r01-fundamental-theorem-calculus-0051","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[0]，a=-1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 0/1)","explanation_zh":"起點值與導數=(0/1, 0/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091051,"parameters":{"f":[0],"a":-1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[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-r01-fundamental-theorem-calculus-0052","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[3]，a=0，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 3/1)","explanation_zh":"起點值與導數=(0/1, 3/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091052,"parameters":{"f":[3],"a":0,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[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-r01-fundamental-theorem-calculus-0053","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-10, 3, 1, 3]，a=1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, -3/1)","explanation_zh":"起點值與導數=(0/1, -3/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091053,"parameters":{"f":[-10,3,1,3],"a":1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[-10,3,1,3],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-r01-fundamental-theorem-calculus-0054","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-9, 4, -1, 4]，a=2，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 27/1)","explanation_zh":"起點值與導數=(0/1, 27/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091054,"parameters":{"f":[-9,4,-1,4],"a":2,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[-9,4,-1,4],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-r01-fundamental-theorem-calculus-0055","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-8, 5, 0, 1]，a=-1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, -14/1)","explanation_zh":"起點值與導數=(0/1, -14/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091055,"parameters":{"f":[-8,5,0,1],"a":-1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[-8,5,0,1],-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-r01-fundamental-theorem-calculus-0056","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-7, 6, 1, 2]，a=0，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, -7/1)","explanation_zh":"起點值與導數=(0/1, -7/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091056,"parameters":{"f":[-7,6,1,2],"a":0,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[-7,6,1,2],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-r01-fundamental-theorem-calculus-0057","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-6, 7, -1, 3]，a=1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 3/1)","explanation_zh":"起點值與導數=(0/1, 3/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091057,"parameters":{"f":[-6,7,-1,3],"a":1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[-6,7,-1,3],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-r01-fundamental-theorem-calculus-0058","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-5, 8, 0, 4]，a=2，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 43/1)","explanation_zh":"起點值與導數=(0/1, 43/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091058,"parameters":{"f":[-5,8,0,4],"a":2,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[-5,8,0,4],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-r01-fundamental-theorem-calculus-0059","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-4, 9, 1, 1]，a=-1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, -13/1)","explanation_zh":"起點值與導數=(0/1, -13/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091059,"parameters":{"f":[-4,9,1,1],"a":-1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[-4,9,1,1],-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-r01-fundamental-theorem-calculus-0060","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-3, 10, -1, 2]，a=0，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, -3/1)","explanation_zh":"起點值與導數=(0/1, -3/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091060,"parameters":{"f":[-3,10,-1,2],"a":0,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[-3,10,-1,2],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-r01-fundamental-theorem-calculus-0061","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-2, 11, 0, 3]，a=1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 12/1)","explanation_zh":"起點值與導數=(0/1, 12/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091061,"parameters":{"f":[-2,11,0,3],"a":1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[-2,11,0,3],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-r01-fundamental-theorem-calculus-0062","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-1, 12, 1, 4]，a=2，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 59/1)","explanation_zh":"起點值與導數=(0/1, 59/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091062,"parameters":{"f":[-1,12,1,4],"a":2,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[-1,12,1,4],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-r01-fundamental-theorem-calculus-0063","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[0, 13, -1, 1]，a=-1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, -15/1)","explanation_zh":"起點值與導數=(0/1, -15/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091063,"parameters":{"f":[0,13,-1,1],"a":-1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[0,13,-1,1],-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-r01-fundamental-theorem-calculus-0064","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[1, 14, 0, 2]，a=0，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 1/1)","explanation_zh":"起點值與導數=(0/1, 1/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091064,"parameters":{"f":[1,14,0,2],"a":0,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[1,14,0,2],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-r01-fundamental-theorem-calculus-0065","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[2, 15, 1, 3]，a=1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 21/1)","explanation_zh":"起點值與導數=(0/1, 21/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091065,"parameters":{"f":[2,15,1,3],"a":1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[2,15,1,3],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-r01-fundamental-theorem-calculus-0066","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[3, 16, -1, 4]，a=2，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 63/1)","explanation_zh":"起點值與導數=(0/1, 63/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091066,"parameters":{"f":[3,16,-1,4],"a":2,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[3,16,-1,4],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-r01-fundamental-theorem-calculus-0067","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[4, 17, 0, 1]，a=-1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, -14/1)","explanation_zh":"起點值與導數=(0/1, -14/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091067,"parameters":{"f":[4,17,0,1],"a":-1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[4,17,0,1],-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-r01-fundamental-theorem-calculus-0068","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[5, 18, 1, 2]，a=0，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 5/1)","explanation_zh":"起點值與導數=(0/1, 5/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091068,"parameters":{"f":[5,18,1,2],"a":0,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[5,18,1,2],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-r01-fundamental-theorem-calculus-0069","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[6, 19, -1, 3]，a=1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 27/1)","explanation_zh":"起點值與導數=(0/1, 27/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091069,"parameters":{"f":[6,19,-1,3],"a":1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[6,19,-1,3],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-r01-fundamental-theorem-calculus-0070","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[7, 20, 0, 4]，a=2，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 79/1)","explanation_zh":"起點值與導數=(0/1, 79/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091070,"parameters":{"f":[7,20,0,4],"a":2,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[7,20,0,4],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-r01-fundamental-theorem-calculus-0071","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[8, 21, 1, 1]，a=-1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, -13/1)","explanation_zh":"起點值與導數=(0/1, -13/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091071,"parameters":{"f":[8,21,1,1],"a":-1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[8,21,1,1],-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-r01-fundamental-theorem-calculus-0072","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[9, 22, -1, 2]，a=0，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 9/1)","explanation_zh":"起點值與導數=(0/1, 9/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091072,"parameters":{"f":[9,22,-1,2],"a":0,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[9,22,-1,2],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-r01-fundamental-theorem-calculus-0073","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[10, 23, 0, 3]，a=1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 36/1)","explanation_zh":"起點值與導數=(0/1, 36/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091073,"parameters":{"f":[10,23,0,3],"a":1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[10,23,0,3],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-r01-fundamental-theorem-calculus-0074","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[11, 24, 1, 4]，a=2，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, 95/1)","explanation_zh":"起點值與導數=(0/1, 95/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091074,"parameters":{"f":[11,24,1,4],"a":2,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[11,24,1,4],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-r01-fundamental-theorem-calculus-0075","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"累積函數在起點的值與導數","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[12, 25, -1, 1]，a=-1，G(t)=∫_a^t f(s)ds。依順序求 (G(a), G′(a))，每項以 p/q 表示。","answer":{"kind":"expression","value":"(0/1, -15/1)","explanation_zh":"起點值與導數=(0/1, -15/1)。起點值為零，但導數是 f(a)；兩種數值不能混同，仍保留全R連續前提。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["累積函數在起點的值與導數"]},"variant":{"family_id":"qf-ftc-anchor-value-derivative","seed":202610091075,"parameters":{"f":[12,25,-1,1],"a":-1,"operation":"ftc_anchor_value_derivative","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-anchor-value-derivative/[[12,25,-1,1],-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-r01-fundamental-theorem-calculus-0076","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[0]；候選 F 的升冪有理係數為 [[5,1]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[0/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091076,"parameters":{"f":[0],"F":[[5,1]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[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-r01-fundamental-theorem-calculus-0077","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[1]；候選 F 的升冪有理係數為 [[0,1],[1,1],[0,1],[0,1],[1,1]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"否","explanation_zh":"候選導數係數=[1/1, 0/1, 0/1, 4/1]；FTC前提=否。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091077,"parameters":{"f":[1],"F":[[0,1],[1,1],[0,1],[0,1],[1,1]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[1],[\"1/1\",\"0/1\",\"0/1\",\"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-r01-fundamental-theorem-calculus-0078","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-10, 3, 1, 3]；候選 F 的升冪有理係數為 [[-10,1],[-10,1],[3,2],[1,3],[3,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[-10/1, 3/1, 1/1, 3/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091078,"parameters":{"f":[-10,3,1,3],"F":[[-10,1],[-10,1],[3,2],[1,3],[3,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[-10,3,1,3],[\"-10/1\",\"3/2\",\"1/3\",\"3/4\"]]"}},"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-r01-fundamental-theorem-calculus-0079","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-9, 4, -1, 4]；候選 F 的升冪有理係數為 [[-9,1],[-9,1],[2,1],[-1,3],[1,1]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[-9/1, 4/1, -1/1, 4/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091079,"parameters":{"f":[-9,4,-1,4],"F":[[-9,1],[-9,1],[2,1],[-1,3],[1,1]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[-9,4,-1,4],[\"-9/1\",\"2/1\",\"-1/3\",\"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-r01-fundamental-theorem-calculus-0080","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-8, 5, 0, 1]；候選 F 的升冪有理係數為 [[-8,1],[-8,1],[5,2],[0,1],[5,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"否","explanation_zh":"候選導數係數=[-8/1, 5/1, 0/1, 5/1]；FTC前提=否。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091080,"parameters":{"f":[-8,5,0,1],"F":[[-8,1],[-8,1],[5,2],[0,1],[5,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[-8,5,0,1],[\"-8/1\",\"5/2\",\"0/1\",\"5/4\"]]"}},"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-r01-fundamental-theorem-calculus-0081","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-7, 6, 1, 2]；候選 F 的升冪有理係數為 [[-7,1],[-7,1],[3,1],[1,3],[1,2]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[-7/1, 6/1, 1/1, 2/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091081,"parameters":{"f":[-7,6,1,2],"F":[[-7,1],[-7,1],[3,1],[1,3],[1,2]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[-7,6,1,2],[\"-7/1\",\"3/1\",\"1/3\",\"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-r01-fundamental-theorem-calculus-0082","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-6, 7, -1, 3]；候選 F 的升冪有理係數為 [[-6,1],[-6,1],[7,2],[-1,3],[3,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[-6/1, 7/1, -1/1, 3/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091082,"parameters":{"f":[-6,7,-1,3],"F":[[-6,1],[-6,1],[7,2],[-1,3],[3,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[-6,7,-1,3],[\"-6/1\",\"7/2\",\"-1/3\",\"3/4\"]]"}},"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-r01-fundamental-theorem-calculus-0083","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-5, 8, 0, 4]；候選 F 的升冪有理係數為 [[-5,1],[-5,1],[4,1],[0,1],[2,1]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"否","explanation_zh":"候選導數係數=[-5/1, 8/1, 0/1, 8/1]；FTC前提=否。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091083,"parameters":{"f":[-5,8,0,4],"F":[[-5,1],[-5,1],[4,1],[0,1],[2,1]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[-5,8,0,4],[\"-5/1\",\"4/1\",\"0/1\",\"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-r01-fundamental-theorem-calculus-0084","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-4, 9, 1, 1]；候選 F 的升冪有理係數為 [[-4,1],[-4,1],[9,2],[1,3],[1,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[-4/1, 9/1, 1/1, 1/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091084,"parameters":{"f":[-4,9,1,1],"F":[[-4,1],[-4,1],[9,2],[1,3],[1,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[-4,9,1,1],[\"-4/1\",\"9/2\",\"1/3\",\"1/4\"]]"}},"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-r01-fundamental-theorem-calculus-0085","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-3, 10, -1, 2]；候選 F 的升冪有理係數為 [[-3,1],[-3,1],[5,1],[-1,3],[1,2]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[-3/1, 10/1, -1/1, 2/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091085,"parameters":{"f":[-3,10,-1,2],"F":[[-3,1],[-3,1],[5,1],[-1,3],[1,2]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[-3,10,-1,2],[\"-3/1\",\"5/1\",\"-1/3\",\"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-r01-fundamental-theorem-calculus-0086","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-2, 11, 0, 3]；候選 F 的升冪有理係數為 [[-2,1],[-2,1],[11,2],[0,1],[7,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"否","explanation_zh":"候選導數係數=[-2/1, 11/1, 0/1, 7/1]；FTC前提=否。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091086,"parameters":{"f":[-2,11,0,3],"F":[[-2,1],[-2,1],[11,2],[0,1],[7,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[-2,11,0,3],[\"-2/1\",\"11/2\",\"0/1\",\"7/4\"]]"}},"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-r01-fundamental-theorem-calculus-0087","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[-1, 12, 1, 4]；候選 F 的升冪有理係數為 [[-1,1],[-1,1],[6,1],[1,3],[1,1]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[-1/1, 12/1, 1/1, 4/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091087,"parameters":{"f":[-1,12,1,4],"F":[[-1,1],[-1,1],[6,1],[1,3],[1,1]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[-1,12,1,4],[\"-1/1\",\"6/1\",\"1/3\",\"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-r01-fundamental-theorem-calculus-0088","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[0, 13, -1, 1]；候選 F 的升冪有理係數為 [[0,1],[0,1],[13,2],[-1,3],[1,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[0/1, 13/1, -1/1, 1/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091088,"parameters":{"f":[0,13,-1,1],"F":[[0,1],[0,1],[13,2],[-1,3],[1,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[0,13,-1,1],[\"0/1\",\"13/2\",\"-1/3\",\"1/4\"]]"}},"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-r01-fundamental-theorem-calculus-0089","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[1, 14, 0, 2]；候選 F 的升冪有理係數為 [[1,1],[1,1],[7,1],[0,1],[3,2]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"否","explanation_zh":"候選導數係數=[1/1, 14/1, 0/1, 6/1]；FTC前提=否。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091089,"parameters":{"f":[1,14,0,2],"F":[[1,1],[1,1],[7,1],[0,1],[3,2]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[1,14,0,2],[\"1/1\",\"7/1\",\"0/1\",\"3/2\"]]"}},"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-r01-fundamental-theorem-calculus-0090","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[2, 15, 1, 3]；候選 F 的升冪有理係數為 [[2,1],[2,1],[15,2],[1,3],[3,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[2/1, 15/1, 1/1, 3/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091090,"parameters":{"f":[2,15,1,3],"F":[[2,1],[2,1],[15,2],[1,3],[3,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[2,15,1,3],[\"2/1\",\"15/2\",\"1/3\",\"3/4\"]]"}},"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-r01-fundamental-theorem-calculus-0091","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[3, 16, -1, 4]；候選 F 的升冪有理係數為 [[3,1],[3,1],[8,1],[-1,3],[1,1]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[3/1, 16/1, -1/1, 4/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091091,"parameters":{"f":[3,16,-1,4],"F":[[3,1],[3,1],[8,1],[-1,3],[1,1]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[3,16,-1,4],[\"3/1\",\"8/1\",\"-1/3\",\"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-r01-fundamental-theorem-calculus-0092","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[4, 17, 0, 1]；候選 F 的升冪有理係數為 [[4,1],[4,1],[17,2],[0,1],[5,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"否","explanation_zh":"候選導數係數=[4/1, 17/1, 0/1, 5/1]；FTC前提=否。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091092,"parameters":{"f":[4,17,0,1],"F":[[4,1],[4,1],[17,2],[0,1],[5,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[4,17,0,1],[\"4/1\",\"17/2\",\"0/1\",\"5/4\"]]"}},"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-r01-fundamental-theorem-calculus-0093","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[5, 18, 1, 2]；候選 F 的升冪有理係數為 [[5,1],[5,1],[9,1],[1,3],[1,2]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[5/1, 18/1, 1/1, 2/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091093,"parameters":{"f":[5,18,1,2],"F":[[5,1],[5,1],[9,1],[1,3],[1,2]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[5,18,1,2],[\"5/1\",\"9/1\",\"1/3\",\"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-r01-fundamental-theorem-calculus-0094","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[6, 19, -1, 3]；候選 F 的升冪有理係數為 [[6,1],[6,1],[19,2],[-1,3],[3,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[6/1, 19/1, -1/1, 3/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091094,"parameters":{"f":[6,19,-1,3],"F":[[6,1],[6,1],[19,2],[-1,3],[3,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[6,19,-1,3],[\"6/1\",\"19/2\",\"-1/3\",\"3/4\"]]"}},"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-r01-fundamental-theorem-calculus-0095","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[7, 20, 0, 4]；候選 F 的升冪有理係數為 [[7,1],[7,1],[10,1],[0,1],[2,1]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"否","explanation_zh":"候選導數係數=[7/1, 20/1, 0/1, 8/1]；FTC前提=否。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091095,"parameters":{"f":[7,20,0,4],"F":[[7,1],[7,1],[10,1],[0,1],[2,1]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[7,20,0,4],[\"7/1\",\"10/1\",\"0/1\",\"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-r01-fundamental-theorem-calculus-0096","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[8, 21, 1, 1]；候選 F 的升冪有理係數為 [[8,1],[8,1],[21,2],[1,3],[1,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[8/1, 21/1, 1/1, 1/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091096,"parameters":{"f":[8,21,1,1],"F":[[8,1],[8,1],[21,2],[1,3],[1,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[8,21,1,1],[\"8/1\",\"21/2\",\"1/3\",\"1/4\"]]"}},"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-r01-fundamental-theorem-calculus-0097","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[9, 22, -1, 2]；候選 F 的升冪有理係數為 [[9,1],[9,1],[11,1],[-1,3],[1,2]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[9/1, 22/1, -1/1, 2/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091097,"parameters":{"f":[9,22,-1,2],"F":[[9,1],[9,1],[11,1],[-1,3],[1,2]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[9,22,-1,2],[\"9/1\",\"11/1\",\"-1/3\",\"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-r01-fundamental-theorem-calculus-0098","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[10, 23, 0, 3]；候選 F 的升冪有理係數為 [[10,1],[10,1],[23,2],[0,1],[7,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"否","explanation_zh":"候選導數係數=[10/1, 23/1, 0/1, 7/1]；FTC前提=否。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091098,"parameters":{"f":[10,23,0,3],"F":[[10,1],[10,1],[23,2],[0,1],[7,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[10,23,0,3],[\"10/1\",\"23/2\",\"0/1\",\"7/4\"]]"}},"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-r01-fundamental-theorem-calculus-0099","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[11, 24, 1, 4]；候選 F 的升冪有理係數為 [[11,1],[11,1],[12,1],[1,3],[1,1]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[11/1, 24/1, 1/1, 4/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091099,"parameters":{"f":[11,24,1,4],"F":[[11,1],[11,1],[12,1],[1,3],[1,1]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[11,24,1,4],[\"11/1\",\"12/1\",\"1/3\",\"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-r01-fundamental-theorem-calculus-0100","batch_id":"batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations","language":"zh-Hant","source_mko_ids":["mko-fundamental-theorem-calculus"],"title_zh":"端點公式的候選原函數前提","stem_zh":"全實數論域；連續多項式被積函數，FTC的明示充分前提。升冪係數 f=[12, 25, -1, 1]；候選 F 的升冪有理係數為 [[12,1],[12,1],[25,2],[-1,3],[1,4]]，每對為[分子,正分母]。是否滿足 F′=f 的係數恆等，以使此候選適用FTC端點差？","answer":{"kind":"short_text","value":"是","explanation_zh":"候選導數係數=[12/1, 25/1, -1/1, 1/1]；FTC前提=是。逐係數求導核對；候選失敗不否定其他原函數存在，加常數不影響此資格，有限核對不是一般定理的形式證明。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["端點公式的候選原函數前提"]},"variant":{"family_id":"qf-ftc-primitive-hypothesis","seed":202610091100,"parameters":{"f":[12,25,-1,1],"F":[[12,1],[12,1],[25,2],[-1,3],[1,4]],"operation":"ftc_primitive_hypothesis","domain":"reals","representation":"ascending_powers","semantic_template":"qf-ftc-primitive-hypothesis/[[12,25,-1,1],[\"12/1\",\"25/2\",\"-1/3\",\"1/4\"]]"}},"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"}}