{
  "index": "1958-A-1",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "1. If \\( a_{0}, a_{1}, \\ldots, a_{n} \\) are real numbers satisfying\n\\[\n\\frac{a_{0}}{1}+\\frac{a_{1}}{2}+\\cdots+\\frac{a_{n}}{n+1}=0\n\\]\nshow that the equation \\( a_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n}=0 \\) has at least one real root.",
  "solution": "Solution. If \\( f(x)=a_{0}+a_{1} x+\\cdots+a_{n} x^{n} \\), then\n\\[\n\\int_{0}^{1} f(x) d x=\\frac{a_{0}}{1}+\\frac{a_{1}}{2}+\\cdots+\\frac{a_{n}}{n+1}=0 .\n\\]\n\nHence, by the mean value theorem for integrals, there exists a number \\( \\boldsymbol{\\xi} \\) between 0 and 1 such that\n\\[\nf(\\xi)=\\int_{0}^{1} f(x) d x=0\n\\]\n\nRemark. This problem appears in G. H. Hardy, A Course in Pure Mathematics, 7th ed., Cambridge University Press, 1938, page 243. It is stated there that the problem appeared in the Cambridge Mathematical Tripos for 1929.",
  "vars": [
    "x",
    "f",
    "\\\\xi"
  ],
  "params": [
    "a_0",
    "a_1",
    "a_2",
    "a_n",
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "varrealx",
        "f": "polynfun",
        "\\xi": "meanptxi",
        "a_0": "coeffzero",
        "a_1": "coeffone",
        "a_2": "coefftwo",
        "a_n": "coeffenn",
        "n": "degree"
      },
      "question": "1. If \\( coeffzero, coeffone, \\ldots, coeffenn \\) are real numbers satisfying\n\\[\n\\frac{coeffzero}{1}+\\frac{coeffone}{2}+\\cdots+\\frac{coeffenn}{degree+1}=0\n\\]\nshow that the equation \\( coeffzero+coeffone varrealx+coefftwo varrealx^{2}+\\cdots+coeffenn varrealx^{degree}=0 \\) has at least one real root.",
      "solution": "Solution. If \\( polynfun(varrealx)=coeffzero+coeffone varrealx+\\cdots+coeffenn varrealx^{degree} \\), then\n\\[\n\\int_{0}^{1} polynfun(varrealx) d varrealx=\\frac{coeffzero}{1}+\\frac{coeffone}{2}+\\cdots+\\frac{coeffenn}{degree+1}=0 .\n\\]\n\nHence, by the mean value theorem for integrals, there exists a number \\( \\boldsymbol{meanptxi} \\) between 0 and 1 such that\n\\[\npolynfun(meanptxi)=\\int_{0}^{1} polynfun(varrealx) d varrealx=0\n\\]\n\nRemark. This problem appears in G. H. Hardy, A Course in Pure Mathematics, 7th ed., Cambridge University Press, 1938, page 243. It is stated there that the problem appeared in the Cambridge Mathematical Tripos for 1929."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marigold",
        "f": "trombone",
        "\\xi": "pendulum",
        "a_0": "sapphire",
        "a_1": "graphite",
        "a_2": "lemonade",
        "a_n": "kangaroo",
        "n": "accordion"
      },
      "question": "Problem:\n<<<\n1. If \\( sapphire, graphite, \\ldots, kangaroo \\) are real numbers satisfying\n\\[\n\\frac{sapphire}{1}+\\frac{graphite}{2}+\\cdots+\\frac{kangaroo}{accordion+1}=0\n\\]\nshow that the equation \\( sapphire+graphite marigold+lemonade marigold^{2}+\\cdots+kangaroo marigold^{accordion}=0 \\) has at least one real root.\n>>>\n",
      "solution": "Solution:\n<<<\nSolution. If \\( trombone(marigold)=sapphire+graphite marigold+\\cdots+kangaroo marigold^{accordion} \\), then\n\\[\n\\int_{0}^{1} trombone(marigold) d marigold=\\frac{sapphire}{1}+\\frac{graphite}{2}+\\cdots+\\frac{kangaroo}{accordion+1}=0 .\n\\]\n\nHence, by the mean value theorem for integrals, there exists a number \\( \\boldsymbol{pendulum} \\) between 0 and 1 such that\n\\[\ntrombone(pendulum)=\\int_{0}^{1} trombone(marigold) d marigold=0\n\\]\n\nRemark. This problem appears in G. H. Hardy, A Course in Pure Mathematics, 7th ed., Cambridge University Press, 1938, page 243. It is stated there that the problem appeared in the Cambridge Mathematical Tripos for 1929.\n>>>\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knownpoint",
        "f": "unvarying",
        "\\\\xi": "exterior",
        "a_0": "unrelated",
        "a_1": "detached",
        "a_2": "separate",
        "a_n": "foreignco",
        "n": "boundless"
      },
      "question": "1. If \\( unrelated, detached, \\ldots, foreignco \\) are real numbers satisfying\n\\[\n\\frac{unrelated}{1}+\\frac{detached}{2}+\\cdots+\\frac{foreignco}{boundless+1}=0\n\\]\nshow that the equation \\( unrelated+detached\\, knownpoint+separate\\, knownpoint^{2}+\\cdots+foreignco\\, knownpoint^{boundless}=0 \\) has at least one real root.",
      "solution": "Solution. If \\( unvarying(knownpoint)=unrelated+detached\\, knownpoint+\\cdots+foreignco\\, knownpoint^{boundless} \\), then\n\\[\n\\int_{0}^{1} unvarying(knownpoint)\\, d\\, knownpoint=\\frac{unrelated}{1}+\\frac{detached}{2}+\\cdots+\\frac{foreignco}{boundless+1}=0 .\n\\]\n\nHence, by the mean value theorem for integrals, there exists a number \\( \\boldsymbol{exterior} \\) between 0 and 1 such that\n\\[\nunvarying(exterior)=\\int_{0}^{1} unvarying(knownpoint)\\, d\\, knownpoint=0\n\\]\n\nRemark. This problem appears in G. H. Hardy, A Course in Pure Mathematics, 7th ed., Cambridge University Press, 1938, page 243. It is stated there that the problem appeared in the Cambridge Mathematical Tripos for 1929."
    },
    "garbled_string": {
      "map": {
        "x": "hjgrksla",
        "f": "bvcmrtye",
        "\\\\xi": "qzxwvtnp",
        "a_0": "pmcfriad",
        "a_1": "knzghqtm",
        "a_2": "ybrsxedl",
        "a_n": "udqkplaz",
        "n": "wjosifbl"
      },
      "question": "1. If \\( pmcfriad, knzghqtm, \\ldots, udqkplaz \\) are real numbers satisfying\n\\[\n\\frac{pmcfriad}{1}+\\frac{knzghqtm}{2}+\\cdots+\\frac{udqkplaz}{wjosifbl+1}=0\n\\]\nshow that the equation \\( pmcfriad+knzghqtm hjgrksla+ybrsxedl hjgrksla^{2}+\\cdots+udqkplaz hjgrksla^{wjosifbl}=0 \\) has at least one real root.",
      "solution": "Solution. If \\( bvcmrtye(hjgrksla)=pmcfriad+knzghqtm hjgrksla+\\cdots+udqkplaz hjgrksla^{wjosifbl} \\), then\n\\[\n\\int_{0}^{1} bvcmrtye(hjgrksla) d hjgrksla=\\frac{pmcfriad}{1}+\\frac{knzghqtm}{2}+\\cdots+\\frac{udqkplaz}{wjosifbl+1}=0 .\n\\]\n\nHence, by the mean value theorem for integrals, there exists a number \\( qzxwvtnp \\) between 0 and 1 such that\n\\[\nbvcmrtye(qzxwvtnp)=\\int_{0}^{1} bvcmrtye(hjgrksla) d hjgrksla=0\n\\]\n\nRemark. This problem appears in G. H. Hardy, A Course in Pure Mathematics, 7th ed., Cambridge University Press, 1938, page 243. It is stated there that the problem appeared in the Cambridge Mathematical Tripos for 1929."
    },
    "kernel_variant": {
      "question": "Let n≥1 and let real numbers a0,a1,…,an satisfy the two simultaneous moment conditions  \n  ∑_{k=0}^{n} a_k (3^{k+1}−2^{k+1})/(k+1)=0,   ∑_{k=0}^{n} a_k (3^{k+2}−2^{k+2})/(k+2)=0.  \nShow that the polynomial  \n  P(x)=a0+a1x+⋯+anx^n  \npossesses a real zero ξ and its derivative P′ possesses a (possibly different) real zero ζ, both lying in the open interval (2,3).",
      "solution": "Let f(x)=P(x).  Because ∫_{2}^{3}x^{k}dx=(3^{k+1}−2^{k+1})/(k+1), the first condition gives ∫_{2}^{3}f(x)dx=0.  Set F(x)=∫_{2}^{x}f(t)dt; note that F is differentiable on [2,3] and F(2)=F(3)=0.  \n\nBy Rolle’s theorem there exists ξ∈(2,3) with F′(ξ)=f(ξ)=0, producing one root of P.  The second moment condition yields ∫_{2}^{3}t f(t)dt=0.  Define G(x)=∫_{2}^{x}t f(t)dt; again G(2)=G(3)=0.  \n\nRolle applied to G furnishes η∈(2,3) with G′(η)=η f(η)=0, hence f(η)=0 with η≠ξ.  Finally, Rolle applied to f on [ξ,η] produces ζ∈(2,3) with f′(ζ)=P′(ζ)=0.  Thus P and P′ each vanish inside (2,3), as required.",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.024544",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}