{
  "index": "1947-B-1",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "7. Let \\( f(x) \\) be a function such that \\( f(1)=1 \\) and for \\( x \\geq 1 \\)\n\\[\nf^{\\prime}(x)=\\frac{1}{x^{2}+f^{2}(x)}\n\\]\n\nProve that\n\\[\n\\lim _{x \\rightarrow \\infty} f(x)\n\\]\nexists and is less than \\( 1+\\pi / 4 \\)",
  "solution": "Solution. Since \\( f^{\\prime} \\) is everywhere positive, \\( f \\) is strictly increasing and therefore\n\\[\nf(t)>f(1)=1 \\quad \\text { for } t>1\n\\]\n\nTherefore\n\\[\nf^{\\prime}(t)=\\frac{1}{t^{2}+f^{2}(t)}<\\frac{1}{t^{2}+1} \\quad \\text { for } t>1\n\\]\n\nSo\n\\[\n\\begin{array}{c}\nf(x)=1+\\int_{1}^{x} f^{\\prime}(t) d t \\\\\n<1+\\int_{1}^{x} \\frac{1}{1+t^{2}} d t<1+\\int_{1}^{\\infty} \\frac{d t}{1+t^{2}}=1+\\pi / 4\n\\end{array}\n\\]\n\nSince \\( f \\) is increasing and bounded, \\( \\lim _{x \\rightarrow \\infty} f(x) \\) exists and is at most \\( 1+\\pi / 4 \\). Strict inequality also follows from (1) because\n\\[\n\\lim _{x \\rightarrow \\infty} f(x)=1+\\int_{1}^{\\infty} f^{\\prime}(t) d t<1+\\int_{1}^{\\infty} \\frac{1}{1+t^{2}} d t=1+\\pi / 4\n\\]",
  "vars": [
    "f",
    "x",
    "t"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "growth",
        "x": "abscissa",
        "t": "parameter"
      },
      "question": "7. Let \\( growth(abscissa) \\) be a function such that \\( growth(1)=1 \\) and for \\( abscissa \\geq 1 \\)\n\\[\ngrowth^{\\prime}(abscissa)=\\frac{1}{abscissa^{2}+growth^{2}(abscissa)}\n\\]\n\nProve that\n\\[\n\\lim _{abscissa \\rightarrow \\infty} growth(abscissa)\n\\]\nexists and is less than \\( 1+\\pi / 4 \\)",
      "solution": "Solution. Since \\( growth^{\\prime} \\) is everywhere positive, \\( growth \\) is strictly increasing and therefore\n\\[\ngrowth(parameter)>growth(1)=1 \\quad \\text { for } parameter>1\n\\]\n\nTherefore\n\\[\ngrowth^{\\prime}(parameter)=\\frac{1}{parameter^{2}+growth^{2}(parameter)}<\\frac{1}{parameter^{2}+1} \\quad \\text { for } parameter>1\n\\]\n\nSo\n\\[\n\\begin{array}{c}\ngrowth(abscissa)=1+\\int_{1}^{abscissa} growth^{\\prime}(parameter) \\, d\\,parameter \\\\\n<1+\\int_{1}^{abscissa} \\frac{1}{1+parameter^{2}} \\, d\\,parameter<1+\\int_{1}^{\\infty} \\frac{d\\,parameter}{1+parameter^{2}}=1+\\pi / 4\n\\end{array}\n\\]\n\nSince \\( growth \\) is increasing and bounded, \\( \\lim _{abscissa \\rightarrow \\infty} growth(abscissa) \\) exists and is at most \\( 1+\\pi / 4 \\). Strict inequality also follows from (1) because\n\\[\n\\lim _{abscissa \\rightarrow \\infty} growth(abscissa)=1+\\int_{1}^{\\infty} growth^{\\prime}(parameter) \\, d\\,parameter<1+\\int_{1}^{\\infty} \\frac{1}{1+parameter^{2}} \\, d\\,parameter=1+\\pi / 4\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "sandstone",
        "x": "cloverleaf",
        "t": "driftwood"
      },
      "question": "7. Let \\( sandstone(cloverleaf) \\) be a function such that \\( sandstone(1)=1 \\) and for \\( cloverleaf \\geq 1 \\)\n\\[\nsandstone^{\\prime}(cloverleaf)=\\frac{1}{cloverleaf^{2}+sandstone^{2}(cloverleaf)}\n\\]\n\nProve that\n\\[\n\\lim _{cloverleaf \\rightarrow \\infty} sandstone(cloverleaf)\n\\]\nexists and is less than \\( 1+\\pi / 4 \\)",
      "solution": "Solution. Since \\( sandstone^{\\prime} \\) is everywhere positive, \\( sandstone \\) is strictly increasing and therefore\n\\[\nsandstone(driftwood)>sandstone(1)=1 \\quad \\text { for } driftwood>1\n\\]\n\nTherefore\n\\[\nsandstone^{\\prime}(driftwood)=\\frac{1}{driftwood^{2}+sandstone^{2}(driftwood)}<\\frac{1}{driftwood^{2}+1} \\quad \\text { for } driftwood>1\n\\]\n\nSo\n\\[\n\\begin{array}{c}\nsandstone(cloverleaf)=1+\\int_{1}^{cloverleaf} sandstone^{\\prime}(driftwood) d driftwood \\\\\n<1+\\int_{1}^{cloverleaf} \\frac{1}{1+driftwood^{2}} d driftwood<1+\\int_{1}^{\\infty} \\frac{d driftwood}{1+driftwood^{2}}=1+\\pi / 4\n\\end{array}\n\\]\n\nSince \\( sandstone \\) is increasing and bounded, \\( \\lim _{cloverleaf \\rightarrow \\infty} sandstone(cloverleaf) \\) exists and is at most \\( 1+\\pi / 4 \\). Strict inequality also follows from (1) because\n\\[\n\\lim _{cloverleaf \\rightarrow \\infty} sandstone(cloverleaf)=1+\\int_{1}^{\\infty} sandstone^{\\prime}(driftwood) d driftwood<1+\\int_{1}^{\\infty} \\frac{1}{1+driftwood^{2}} d driftwood=1+\\pi / 4\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "constantmap",
        "x": "fixedpoint",
        "t": "distance"
      },
      "question": "7. Let \\( constantmap(fixedpoint) \\) be a function such that \\( constantmap(1)=1 \\) and for \\( fixedpoint \\geq 1 \\)\n\\[\nconstantmap^{\\prime}(fixedpoint)=\\frac{1}{fixedpoint^{2}+constantmap^{2}(fixedpoint)}\n\\]\n\nProve that\n\\[\n\\lim _{fixedpoint \\rightarrow \\infty} constantmap(fixedpoint)\n\\]\nexists and is less than \\( 1+\\pi / 4 \\)",
      "solution": "Solution. Since \\( constantmap^{\\prime} \\) is everywhere positive, \\( constantmap \\) is strictly increasing and therefore\n\\[\nconstantmap(distance)>constantmap(1)=1 \\quad \\text { for } distance>1\n\\]\n\nTherefore\n\\[\nconstantmap^{\\prime}(distance)=\\frac{1}{distance^{2}+constantmap^{2}(distance)}<\\frac{1}{distance^{2}+1} \\quad \\text { for } distance>1\n\\]\n\nSo\n\\[\n\\begin{array}{c}\nconstantmap(fixedpoint)=1+\\int_{1}^{fixedpoint} constantmap^{\\prime}(distance) d distance \\\\\n<1+\\int_{1}^{fixedpoint} \\frac{1}{1+distance^{2}} d distance<1+\\int_{1}^{\\infty} \\frac{d distance}{1+distance^{2}}=1+\\pi / 4\n\\end{array}\n\\]\n\nSince \\( constantmap \\) is increasing and bounded, \\( \\lim _{fixedpoint \\rightarrow \\infty} constantmap(fixedpoint) \\) exists and is at most \\( 1+\\pi / 4 \\). Strict inequality also follows from (1) because\n\\[\n\\lim _{fixedpoint \\rightarrow \\infty} constantmap(fixedpoint)=1+\\int_{1}^{\\infty} constantmap^{\\prime}(distance) d distance<1+\\int_{1}^{\\infty} \\frac{1}{1+distance^{2}} d distance=1+\\pi / 4\n\\]"
    },
    "garbled_string": {
      "map": {
        "f": "mqsdlkpa",
        "x": "vprnlogu",
        "t": "chsweorf"
      },
      "question": "7. Let \\( mqsdlkpa(vprnlogu) \\) be a function such that \\( mqsdlkpa(1)=1 \\) and for \\( vprnlogu \\geq 1 \\)\n\\[\nmqsdlkpa^{\\prime}(vprnlogu)=\\frac{1}{vprnlogu^{2}+mqsdlkpa^{2}(vprnlogu)}\n\\]\n\nProve that\n\\[\n\\lim _{vprnlogu \\rightarrow \\infty} mqsdlkpa(vprnlogu)\n\\]\nexists and is less than \\( 1+\\pi / 4 \\)",
      "solution": "Solution. Since \\( mqsdlkpa^{\\prime} \\) is everywhere positive, \\( mqsdlkpa \\) is strictly increasing and therefore\n\\[\nmqsdlkpa(chsweorf)>mqsdlkpa(1)=1 \\quad \\text { for } chsweorf>1\n\\]\n\nTherefore\n\\[\nmqsdlkpa^{\\prime}(chsweorf)=\\frac{1}{chsweorf^{2}+mqsdlkpa^{2}(chsweorf)}<\\frac{1}{chsweorf^{2}+1} \\quad \\text { for } chsweorf>1\n\\]\n\nSo\n\\[\n\\begin{array}{c}\nmqsdlkpa(vprnlogu)=1+\\int_{1}^{vprnlogu} mqsdlkpa^{\\prime}(chsweorf) d chsweorf \\\\\n<1+\\int_{1}^{vprnlogu} \\frac{1}{1+chsweorf^{2}} d chsweorf<1+\\int_{1}^{\\infty} \\frac{d chsweorf}{1+chsweorf^{2}}=1+\\pi / 4\n\\end{array}\n\\]\n\nSince \\( mqsdlkpa \\) is increasing and bounded, \\( \\lim _{vprnlogu \\rightarrow \\infty} mqsdlkpa(vprnlogu) \\) exists and is at most \\( 1+\\pi / 4 \\). Strict inequality also follows from (1) because\n\\[\n\\lim _{vprnlogu \\rightarrow \\infty} mqsdlkpa(vprnlogu)=1+\\int_{1}^{\\infty} mqsdlkpa^{\\prime}(chsweorf) d chsweorf<1+\\int_{1}^{\\infty} \\frac{1}{1+chsweorf^{2}} d chsweorf=1+\\pi / 4\n\\]"
    },
    "kernel_variant": {
      "question": "Let\\(\\;f:[2,\\infty)\\to\\mathbb R\\) satisfy\n\\[\n f(2)=\\tfrac12\\quad\\text{and}\\quad f'(x)=\\frac{2}{x^{3}+f^{2}(x)}\\qquad(x\\ge 2).\n\\]\nProve that the limit\n\\[\\displaystyle \\lim_{x\\to\\infty}f(x)\\]\nexists and that\n\\[\\displaystyle \\lim_{x\\to\\infty}f(x)<\\tfrac34.\\]",
      "solution": "Let f:[2,∞)→ℝ satisfy f(2)=½ and f′(x)=2/(x³+f²(x)) for x≥2.\n\n1. Since the numerator 2>0 and the denominator x³+f²(x)>0, we have f′(x)>0. Hence f is strictly increasing, so for x>2,\n   f(x)>f(2)=½.\n\n2. Thus for t≥2 we get f²(t)>¼ and\n   0<f′(t)=2/(t³+f²(t))<2/(t³+¼).\n\n3. Integrating from 2 to x≥2 gives\n   f(x)=f(2)+∫₂ˣ f′(t)dt<½+∫₂ˣ2/(t³+¼)dt<½+∫₂ˣ2/t³dt<½+∫₂^∞2/t³dt=½+¼=¾.\n   Hence f(x)<¾ for all x, so f is bounded above by ¾.\n\n4. Because f is increasing and bounded above, the limit L=limₓ→∞f(x) exists.\n\n5. Applying the same integrand estimate over [2,∞) yields\n   L=½+∫₂^∞f′(t)dt<½+∫₂^∞2/t³dt=¾.\n\nTherefore, limₓ→∞f(x) exists and is strictly less than ¾. ∎",
      "_meta": {
        "core_steps": [
          "Observe f′(x)>0 ⇒ f is strictly increasing",
          "Since f(t)>f(1), bound derivative: f′(t)=1/(t²+f²(t)) < 1/(t²+f(1)²)",
          "Integrate this inequality from the base point to x to obtain an explicit upper bound for f(x)",
          "Show the improper integral ∫ 1/(t²+f(1)²) dt converges, so f is bounded above",
          "Increasing + bounded ⇒ limit exists; strict inequality carried through the ‘<’ in the integral comparison"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "chosen base point where the initial value of f is given",
            "original": "1"
          },
          "slot2": {
            "description": "initial value f(slot1) that supplies a positive constant to compare with f(t)",
            "original": "1"
          },
          "slot3": {
            "description": "power on the independent variable in the denominator of f′ (any exponent >1 still gives an integrable comparison function)",
            "original": "2"
          },
          "slot4": {
            "description": "positive constant numerator in f′; any fixed positive number keeps f′ positive and merely rescales all bounds",
            "original": "1"
          },
          "slot5": {
            "description": "numerical value of the arctangent-based upper bound (comes from integrating 1/(t²+slot2²) from slot1 to ∞)",
            "original": "π/4"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}