{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261007-advance-20261008-r06-antiderivative-0040",
  "pilot_number": "R0001-01340",
  "number": 1340,
  "family_id": "qf-antiderivative-polynomial",
  "title": {
    "zh-Hant": "第 1340 題：多項式的零常數原函數",
    "en": "Question 1340: Zero-Constant Antiderivative of a Polynomial"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261007-advance-20261008-r06-antiderivative-0040",
    "batch_id": "batch-20261007-advance-20261008-r06-antiderivative-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-antiderivative"
    ],
    "title_zh": "多項式的零常數原函數",
    "stem_zh": "全實數論域，以下函數／點均以明示實數模型解讀。升冪係數 f=[2, 15, 1, 3]。求常數項固定零的原函數升冪有理係數，採最簡正分母。",
    "answer": {
      "kind": "expression",
      "value": "[0/1, 2/1, 15/2, 1/3, 3/4]",
      "explanation_zh": "零常數原函數係數=[0/1, 2/1, 15/2, 1/3, 3/4]。逐係數積成指定的 canonical polynomial primitive，不聲稱一般反導函數唯一性已被有限資料證明。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 2,
      "skills": [
        "多項式的零常數原函數"
      ]
    },
    "variant": {
      "family_id": "qf-antiderivative-polynomial",
      "seed": 202610086040,
      "parameters": {
        "f": [
          2,
          15,
          1,
          3
        ],
        "operation": "polynomial_primitive",
        "domain": "reals",
        "representation": "ascending_powers",
        "semantic_template": "qf-antiderivative-polynomial/{\"inputs\":[\"2\",\"15\",\"1\",\"3\"],\"operation\":\"polynomial_primitive\"}"
      }
    },
    "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-20261008-r06-antiderivative-0040",
    "batch_id": "batch-20261007-advance-20261008-r06-antiderivative-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "52e10283ac1336bd71a9bf9e1afe6bbe04dc286a391d229a3e257a49e7aa136e",
      "family_id": "qf-antiderivative-polynomial",
      "seed": 202610086040,
      "semantic_template": "qf-antiderivative-polynomial/{\"inputs\":[\"2\",\"15\",\"1\",\"3\"],\"operation\":\"polynomial_primitive\"}"
    },
    "fields": {
      "title": "Zero-Constant Antiderivative of a Polynomial",
      "stem": "On all R, f has ascending coefficients [2, 15, 1, 3], starting with its constant term. Give ascending reduced positive-denominator rational coefficients of the zero-constant primitive, removing trailing zero terms.",
      "answer": {
        "kind": "expression",
        "value": "[0/1, 2/1, 15/2, 1/3, 3/4]",
        "explanation": "Place a_k/(k+1) at power k+1, with constant zero, obtaining [0/1, 2/1, 15/2, 1/3, 3/4]. The actual antiderivative family also includes adding a constant."
      },
      "skills": [
        "Zero-Constant Antiderivative of a Polynomial"
      ],
      "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，以下函數／點均以明示實數模型解讀。升冪係數 f=[2, 15, 1, 3]。求常數項固定零的原函數升冪有理係數，採最簡正分母。",
        "en": "On all R, f has ascending coefficients [2, 15, 1, 3], starting with its constant term. Give ascending reduced positive-denominator rational coefficients of the zero-constant primitive, removing trailing zero terms."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "domain=R; coefficient lists are in ascending powers",
      "after": "constant term first; rational pairs=[numerator,denominator]",
      "relation": "implies",
      "rule_id": "polynomial_representation",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先固定係數次序，不能把升冪清單當降冪，也不能把分子分母交換。",
        "en": "Fix coefficient order;do not read ascending coefficients as descending or swap numerator and denominator."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "f coefficients ascending=[2, 15, 1, 3]",
      "after": "seek F₀ with constant coefficient0 and F₀′=f",
      "relation": "implies",
      "rule_id": "canonical_primitive_domain",
      "conditions": [],
      "explanation": {
        "zh-Hant": "原函數並不一般唯一；本題額外固定常數項為0。",
        "en": "Antiderivatives are not generally unique; this task additionally fixes the constant coefficient at0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "primitive coefficient at degree1 := div(2, 1)",
      "after": "primitive coefficient at degree1 = 2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "由(x^(k+1))′=(k+1)x^k，係數須除以非零正整數k+1。",
        "en": "Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 primitive coefficient at degree1，不以小數近似替代。",
          "en": "Recompute primitive coefficient at degree1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "2",
          "1"
        ],
        "output": "2",
        "label": "primitive coefficient at degree1"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "differentiate back coefficient0 := mul(2, 1)",
      "after": "differentiate back coefficient0 = 2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "微分乘回k+1，應恢復原多項式對應係數。",
        "en": "Differentiation multiplies back by k+1 and must recover the original coefficient."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 differentiate back coefficient0，不以小數近似替代。",
          "en": "Recompute differentiate back coefficient0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "2",
          "1"
        ],
        "output": "2",
        "label": "differentiate back coefficient0"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "primitive coefficient at degree2 := div(15, 2)",
      "after": "primitive coefficient at degree2 = 15/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "由(x^(k+1))′=(k+1)x^k，係數須除以非零正整數k+1。",
        "en": "Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 primitive coefficient at degree2，不以小數近似替代。",
          "en": "Recompute primitive coefficient at degree2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "15",
          "2"
        ],
        "output": "15/2",
        "label": "primitive coefficient at degree2"
      }
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "differentiate back coefficient1 := mul(15/2, 2)",
      "after": "differentiate back coefficient1 = 15",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "微分乘回k+1，應恢復原多項式對應係數。",
        "en": "Differentiation multiplies back by k+1 and must recover the original coefficient."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 differentiate back coefficient1，不以小數近似替代。",
          "en": "Recompute differentiate back coefficient1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "15/2",
          "2"
        ],
        "output": "15",
        "label": "differentiate back coefficient1"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "primitive coefficient at degree3 := div(1, 3)",
      "after": "primitive coefficient at degree3 = 1/3",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "由(x^(k+1))′=(k+1)x^k，係數須除以非零正整數k+1。",
        "en": "Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 primitive coefficient at degree3，不以小數近似替代。",
          "en": "Recompute primitive coefficient at degree3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "1",
          "3"
        ],
        "output": "1/3",
        "label": "primitive coefficient at degree3"
      }
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "differentiate back coefficient2 := mul(1/3, 3)",
      "after": "differentiate back coefficient2 = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "微分乘回k+1，應恢復原多項式對應係數。",
        "en": "Differentiation multiplies back by k+1 and must recover the original coefficient."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 differentiate back coefficient2，不以小數近似替代。",
          "en": "Recompute differentiate back coefficient2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/3",
          "3"
        ],
        "output": "1",
        "label": "differentiate back coefficient2"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "primitive coefficient at degree4 := div(3, 4)",
      "after": "primitive coefficient at degree4 = 3/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "由(x^(k+1))′=(k+1)x^k，係數須除以非零正整數k+1。",
        "en": "Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 primitive coefficient at degree4，不以小數近似替代。",
          "en": "Recompute primitive coefficient at degree4 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "3",
          "4"
        ],
        "output": "3/4",
        "label": "primitive coefficient at degree4"
      }
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "differentiate back coefficient3 := mul(3/4, 4)",
      "after": "differentiate back coefficient3 = 3",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "微分乘回k+1，應恢復原多項式對應係數。",
        "en": "Differentiation multiplies back by k+1 and must recover the original coefficient."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 differentiate back coefficient3，不以小數近似替代。",
          "en": "Recompute differentiate back coefficient3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "3/4",
          "4"
        ],
        "output": "3",
        "label": "differentiate back coefficient3"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "[0, 2, 15/2, 1/3, 3/4]",
      "after": "all derivative coefficients agree with f",
      "relation": "implies",
      "rule_id": "global_polynomial_identity",
      "conditions": [],
      "explanation": {
        "zh-Hant": "係數逐項相同證明全實數上的多項式恆等，不是只在幾個點相等。",
        "en": "Coefficient agreement proves the polynomial identity on all reals,not merely equality at a few points."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "[0, 2, 15/2, 1/3, 3/4]",
      "after": "reduced coefficients;integer coefficients retain /1",
      "relation": "implies",
      "rule_id": "coefficient_output_protocol",
      "conditions": [],
      "explanation": {
        "zh-Hant": "輸出升冪係數清單；多餘最高次零係數去掉，但零多項式保留[0/1]。",
        "en": "Return ascending coefficients,trim redundant leading-in-degree zeros,and retain[0/1] for the zero polynomial."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "primitive": [
      "0",
      "2",
      "15/2",
      "1/3",
      "3/4"
    ]
  },
  "conclusion": {
    "canonical_value": "[0/1, 2/1, 15/2, 1/3, 3/4]",
    "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": {
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    "question_binding": {
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      "batch_path": "public/data/questions/batches/batch-20261007-advance-20261008-r06-antiderivative-foundations/questions.jsonl",
      "batch_sha256": "322101eaff25d0ec9ae01a8564baf7cf7de5678f3caeba49643ad8968169b98e",
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      "question_compact_sha256": "52e10283ac1336bd71a9bf9e1afe6bbe04dc286a391d229a3e257a49e7aa136e",
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      "english_rows_path": "public/data/i18n/questions/batch-20261007-advance-20261008-r06-antiderivative-foundations.en.jsonl"
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        "original_path": "public/data/mko/antiderivative.json",
        "english_path": "public/data/i18n/editions/mko-antiderivative.en.json"
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    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "b9fe70a7a264c90c20349be0449b2bd1a1b4d0e0a97092bc9fbcb7ca36ce0694",
    "original_candidate_sha256": "6fea96674ed0d52f427fecb772487d39cd640c088f625c92f925fcd169f8fd9b",
    "original_candidate_line_sha256": "6fea96674ed0d52f427fecb772487d39cd640c088f625c92f925fcd169f8fd9b"
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  "historical_source_state": {
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      "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-20261026-004-p2",
    "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": {
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      "question_first_witness_sha256": "3e91602adc7f81a2671a2a3ffaf6a447028b56abc5853e8271673d76f30cd7ad"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 12,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "72b6dde5baedf32e11f1bf0102cb9a24baef06353a8521e89b33a362c4316be0",
      "arithmetic_steps": 8
    },
    "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."
  }
}