{
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  "question_id": "q-20261007-advance-20261009-r06-definite-integral-0100",
  "pilot_number": "R0006-03850",
  "number": 3850,
  "family_id": "qf-density-integral-endpoint-recovery",
  "title": {
    "zh-Hant": "第 3850 題：非零常數積分的整數終點反解",
    "en": "Question 3850: Integer Endpoint Recovered from a Nonzero Constant Integral"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261007-advance-20261009-r06-definite-integral-0100",
    "batch_id": "batch-20261007-advance-20261009-r06-definite-integral-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-definite-integral"
    ],
    "title_zh": "非零常數積分的整數終點反解",
    "stem_zh": "全實數論域；升冪係數的受限多項式模型。非零常數f(x)=25，升冪係數=[25]。起點a=12，I(a,b)=-625/1，本輪保證b為整數。求有向終點b，仍保留 /1。",
    "answer": {
      "kind": "expression",
      "value": "-13/1",
      "explanation_zh": "有向終點=-13/1。用a加目標積分除以非零c求終點，不先排序或取絕對值；目標零時b=a，保持既有整數端點切片。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 3,
      "skills": [
        "非零常數積分的整數終點反解"
      ]
    },
    "variant": {
      "family_id": "qf-density-integral-endpoint-recovery",
      "seed": 202610096100,
      "parameters": {
        "c": 25,
        "a": 12,
        "target_n": -625,
        "target_d": 1,
        "operation": "density_integral_endpoint_recovery",
        "domain": "reals",
        "semantic_template": "qf-density-integral-endpoint-recovery/[25,12,\"-625/1\"]"
      }
    },
    "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"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261007-advance-20261009-r06-definite-integral-0100",
    "batch_id": "batch-20261007-advance-20261009-r06-definite-integral-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "70487747c78c996e72927866d1646c4fc257af7848d43c9c157fec3ea88d98fd",
      "family_id": "qf-density-integral-endpoint-recovery",
      "seed": 202610096100,
      "semantic_template": "qf-density-integral-endpoint-recovery/[25,12,\"-625/1\"]"
    },
    "fields": {
      "title": "Integer Endpoint Recovered from a Nonzero Constant Integral",
      "stem": "On all R, the constant integrand is c=25≠0. Starting at a=12, its oriented integral has target -625/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.",
      "answer": {
        "kind": "expression",
        "value": "-13/1",
        "explanation": "Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."
      },
      "skills": [
        "Integer Endpoint Recovered from a Nonzero Constant Integral"
      ],
      "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。非零常數f(x)=25，升冪係數=[25]。起點a=12，I(a,b)=-625/1，本輪保證b為整數。求有向終點b，仍保留 /1。",
        "en": "On all R, the constant integrand is c=25≠0. Starting at a=12, its oriented integral has target -625/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "explicit real polynomial moment or constant integrand model",
      "after": "oriented I(a,b)=H(b)−H(a);do not sort endpoints",
      "relation": "implies",
      "rule_id": "polynomial_moment_scope",
      "conditions": [],
      "explanation": {
        "zh-Hant": "受限多項式在實數上連續可積；用逐項原函數端點差求精確有向積分，不把絕對面積或有限和當積分。",
        "en": "Real polynomials are continuous and integrable;use termwise primitive endpoint differences for exact oriented integrals,not absolute area or a finite sum."
      },
      "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": "oriented interval displacement := div(-625, 25)",
      "after": "oriented interval displacement = -25",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "由c(b−a)=目標值除以非零c，不取絕對值。",
        "en": "Divide c(b−a)=target by nonzero c,without absolute values."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 oriented interval displacement，不以小數近似替代。",
          "en": "Recompute oriented interval displacement with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "-625",
          "25"
        ],
        "output": "-25",
        "label": "oriented interval displacement"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "recovered endpoint := sum(12, -25)",
      "after": "recovered endpoint = -13",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "把有向位移加回起點a；本題切片要求所得b仍為整數。",
        "en": "Add the oriented displacement to a;this frozen exercise slice requires an integer endpoint."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 recovered endpoint，不以小數近似替代。",
          "en": "Recompute recovered endpoint with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "12",
          "-25"
        ],
        "output": "-13",
        "label": "recovered endpoint"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "endpoint reverse substitution := mul(25, -25)",
      "after": "endpoint reverse substitution = -625",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "代回c(b−a)驗算原目標；目標0時b=a。",
        "en": "Substitute into c(b−a) to audit the target;target0 gives b=a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 endpoint reverse substitution，不以小數近似替代。",
          "en": "Recompute endpoint reverse substitution with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "25",
          "-25"
        ],
        "output": "-625",
        "label": "endpoint reverse substitution"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "c=25≠0;b=-13",
      "after": "I(a,b)=-625=target",
      "relation": "implies",
      "rule_id": "unique_constant_endpoint",
      "conditions": [],
      "explanation": {
        "zh-Hant": "非零常數使方程對b唯一；負c或負目標並不允許任意排序端點。",
        "en": "A nonzero constant makes the endpoint unique;negative c or target does not permit arbitrary endpoint reordering."
      },
      "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": {
    "result": "-13",
    "target": "-625"
  },
  "conclusion": {
    "canonical_value": "-13/1",
    "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."
    }
  },
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    "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."
  },
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      "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."
    },
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    "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.",
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