{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261007-advance-20261009-r05-antiderivative-0031", "pilot_number": "R0059-30006", "number": 30006, "family_id": "qf-density-primitive-constant-difference", "title": { "zh-Hant": "第 30006 題：共享原函數的兩常數差", "en": "Question 30006: Difference of Constants for a Shared Primitive" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261007-advance-20261009-r05-antiderivative-0031", "batch_id": "batch-20261007-advance-20261009-r05-antiderivative-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-antiderivative" ], "title_zh": "共享原函數的兩常數差", "stem_zh": "全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-7、C2=-5。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。", "answer": { "kind": "expression", "value": "-2/1", "explanation_zh": "共享原函數常數差=-2/1。共同原函數消去，只剩實際C1−C2；沒有加入無關多項式或求值點充量。" }, "difficulty": { "band": "undergraduate_core", "level": 1, "skills": [ "共享原函數的兩常數差" ] }, "variant": { "family_id": "qf-density-primitive-constant-difference", "seed": 202610095031, "parameters": { "C1": -7, "C2": -5, "operation": "density_primitive_constant_difference", "domain": "reals", "semantic_template": "qf-density-primitive-constant-difference/[\"-2\"]" } }, "verification": { "status": "candidate", "checks": [ "producer_exact_integer_rational_arithmetic", "producer_finite_enumeration", "producer_source_snapshot_check", "producer_substantive_template_deduplication", "producer_json_schema" ], "warning_zh": "僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。" }, "provenance": { "generator_role": "question_producer", "generator_model": "model_unknown", "generated_at": "2026-10-07", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20261007-advance-20261009-r05-antiderivative-0031", "batch_id": "batch-20261007-advance-20261009-r05-antiderivative-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "f432b9ed54c6d76bb42547f62b541284ca34d696f2d2ba53cba63c437e26e76c", "family_id": "qf-density-primitive-constant-difference", "seed": 202610095031, "semantic_template": "qf-density-primitive-constant-difference/[\"-2\"]" }, "fields": { "title": "Difference of Constants for a Shared Primitive", "stem": "Two full-real polynomial primitives share F0: F1=F0+(-7) and F2=F0+(-5). Find the constant F1−F2. Use reduced positive-denominator fractions, retaining /1 for integers.", "answer": { "kind": "expression", "value": "-2/1", "explanation": "The shared primitive cancels, leaving C1−C2=-2/1. No irrelevant polynomial or point is introduced, and this scoped algebraic identity does not establish a theorem on disconnected domains." }, "skills": [ "Difference of Constants for a Shared Primitive" ], "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review." } }, "givens": [ { "id": "G001", "text": { "zh-Hant": "全實數論域；升冪係數的受限多項式模型。兩函數為同一全R原函數F0加上 C1=-7、C2=-5。求 (F0+C1)−(F0+C2) 的常數值；升冪係數只在共享部分，採 p/q。", "en": "Two full-real polynomial primitives share F0: F1=F0+(-7) and F2=F0+(-5). Find the constant F1−F2. Use reduced positive-denominator fractions, retaining /1 for integers." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "source roles = [qf-density-primitive-constant-difference, density_primitive_constant_difference, reals]", "after": "derived = [[shared_primitive_constant_difference_fraction]]", "relation": "implies", "rule_id": "R59_EXACT_family_domain_FULLprefix_operation_and_answer_roles", "conditions": [], "explanation": { "zh-Hant": "保原bank的完整operation前綴、domain及輸出欄序；frame只用orthogonal_equal_unit，角度unit只用degrees。分數整數保/1；不增不存在的參數，不改原中英來源。", "en": "Retain the ORIGINAL bank full operation prefix,domain and ordered answer roles.Use orthogonal_equal_unit only where a frame is declared,and degrees where an angle unit is declared.Integer fractions retain/1.Do not invent parameters or alter original bilingual sources." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以原始資料、索引與前提反向代回本步，不聲稱可由結論唯一還原輸入。有限精確核對、獨立審查、形式證明及發布資格分開。", "en": "Reverse-check this step against the ORIGINAL data,indices and premises,without claiming the conclusion uniquely recovers the inputs.Finite exact checking,independent review,formal proof and publication clearance remain separate." } }, "arithmetic": null }, { "id": "S002", "uses": [ "G001", "S001" ], "before": "Primitive difference ORIGINAL C1 minus C2 := sub(-7, -5)", "after": "Primitive difference ORIGINAL C1 minus C2 = -2", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按原角色精確代值，不作浮點近似；SAFE只驗raw輸入，衍生整數／有理數不截斷，不套其他銀行的component cap。", "en": "Substitute exact ORIGINAL roles,without floating approximation.SAFE bounds raw inputs only;derived integers/rationals are unbounded and do not inherit another bank component cap." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Primitive difference ORIGINAL C1 minus C2，不以小數近似替代。", "en": "Recompute Primitive difference ORIGINAL C1 minus C2 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sub", "inputs": [ "-7", "-5" ], "output": "-2", "label": "Primitive difference ORIGINAL C1 minus C2" } }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "source roles = [-7, -5]", "after": "derived = [-2/1, (F0+C1)-(F0+C2)=C1-C2]", "relation": "implies", "rule_id": "Primitive_difference_SHARED_F0_cancels_not_general_disconnected_theorem", "conditions": [], "explanation": { "zh-Hant": "題設兩式共用同一F0，展開減法後F0與-F0消去，只剩C1-C2。這不要求新poly或anchor，也不推任意不連通論域原函數差皆一個constant。", "en": "The TWO GIVEN expressions share the SAME F0;expanding subtraction cancels F0 and -F0,leaving C1-C2.No extra polynomial or anchor is needed,and no theorem on arbitrary disconnected-domain primitives is inferred." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以原始資料、索引與前提反向代回本步，不聲稱可由結論唯一還原輸入。有限精確核對、獨立審查、形式證明及發布資格分開。", "en": "Reverse-check this step against the ORIGINAL data,indices and premises,without claiming the conclusion uniquely recovers the inputs.Finite exact checking,independent review,formal proof and publication clearance remain separate." } }, "arithmetic": null } ], "math_model": { "result": "-2/1" }, "conclusion": { "canonical_value": "-2/1", "explanation": { "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。", "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged." } }, "definition_note": { "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。", "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate." }, "source": { "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db", "question_binding": { "batch_id": "batch-20261007-advance-20261009-r05-antiderivative-foundations", "batch_path": "public/data/questions/batches/batch-20261007-advance-20261009-r05-antiderivative-foundations/questions.jsonl", "batch_sha256": "085ade26180bd6f2bb25667f7334717d6fd8127fddfdf06e74e67bdbe897b44e", "original_line": 31, "original_line_sha256": "3a1e5de61553529fd3c6e1b3dcfbad3b63848b365fd358599f17a2ff50e8d8fb", "question_compact_sha256": "f432b9ed54c6d76bb42547f62b541284ca34d696f2d2ba53cba63c437e26e76c", "english_row_compact_sha256": "6a3f6d0d7d0e2f1216593ae742642b13a9fcdf9e571c1782068131bbbcd3e0e4", "english_manifest_path": "public/data/i18n/editions/batch-20261007-advance-20261009-r05-antiderivative-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20261007-advance-20261009-r05-antiderivative-foundations.en.jsonl" }, "mko_bindings": { "mko-antiderivative": { "sha256": "5f2ecff6866e471d1f8da7fea1fa5e121364ae2d24b2fea405d2300fe9ff1cae", "english_sha256": "f983dd8b898bc9eac9d5237c7895e5f7dc182ac46efb5f9b58b925fc6c2ecb9d", "original_path": "public/data/mko/antiderivative.json", "english_path": "public/data/i18n/editions/mko-antiderivative.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "17d593a551933064f195bd9277d5085fe2b6c98e11cbda0c8aa27d11b7b07949", "original_candidate_sha256": "046fc15e9f375cb2ee8c5e8534bdaa3cb55c3c5e36c4d2d896bcba5c1c1ceb7a", "original_candidate_line_sha256": "046fc15e9f375cb2ee8c5e8534bdaa3cb55c3c5e36c4d2d896bcba5c1c1ceb7a" }, "historical_source_state": { "review_state": { "derivation": "candidate_authored_and_locally_checked", "source_question_state": "candidate", "independent_AI_or_human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN" }, "scope_note": { "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。", "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged." }, "label": "Frozen producer metadata; current per-question review is separate below." }, "revision": { "id": "review-20261026-011-p60", "review_date": "2026-10-26", "owner_authorized_reader_publication": true, "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.", "review_receipts": { "semantic_review_sha256": "4c93dea3956f58509e20b3b8bbbf716dbc340b424a2a5177bc04aba9e8ea9ec5", "question_first_witness_sha256": "a5e56e0d64d622d7202840cef7019839c5a99244c602694190779155207dafdf" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 3, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "af896503020a80b885137fef541f7a6c3be427e9d1f2403e78cc3d0e44a064f8", "arithmetic_steps": 1 }, "uses_answer_as_premise": false, "human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN", "Canonical_truth_promotion": false, "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged." } }