{
  "index": "2009-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Functions $f,g,h$ are differentiable on some open interval around $0$\nand satisfy the equations and initial conditions\n\\begin{gather*}\nf' = 2f^2gh+\\frac{1}{gh},\\quad f(0)=1, \\\\\ng'=fg^2h+\\frac{4}{fh}, \\quad g(0)=1, \\\\\nh'=3fgh^2+\\frac{1}{fg}, \\quad h(0)=1.\n\\end{gather*}\nFind an explicit formula for $f(x)$, valid in some open interval around $0$.",
  "solution": "Multiplying the first differential equation by $gh$, the second by $fh$,\nand the third by $fg$, and summing gives\n\\[\n(fgh)' = 6(fgh)^2+6.\n\\]\nWrite $k(x) = f(x)g(x)h(x)$; then $k' = 6k^2+6$ and $k(0) = 1$. One\nsolution for this differential equation with this initial condition is\n$k(x) = \\tan(6x+\\pi/4)$; by standard uniqueness, this must necessarily\nhold for $x$ in some open interval around $0$. Now the first given\nequation becomes\n\\begin{align*}\nf'/f &= 2k(x)+1/k(x) \\\\\n&= 2\\tan(6x+\\pi/4)+\\cot(6x+\\pi/4);\n\\end{align*}\nintegrating both sides gives\n\\[\n\\ln(f(x)) = \\frac{-2\\ln\\cos(6x+\\pi/4) + \\ln\\sin(6x+\\pi/4)}{6}+c,\n\\]\nwhence $f(x) = e^c\n\\left(\\frac{\\sin(6x+\\pi/4)}{\\cos^2(6x+\\pi/4)}\\right)^{1/6}$.\nSubstituting $f(0)=1$ gives $e^c = 2^{-1/12}$ and thus $f(x) = 2^{-1/12}\n\\left(\\frac{\\sin(6x+\\pi/4)}{\\cos^2(6x+\\pi/4)}\\right)^{1/6}$.\n\n\\textbf{Remark.} The answer can be put in alternate forms using\ntrigonometric identities. One particularly simple one is\n\\[\n f(x) = (\\sec 12x)^{1/12} (\\sec 12x + \\tan 12x)^{1/4}.\n\\]",
  "vars": [
    "f",
    "g",
    "h",
    "k",
    "x"
  ],
  "params": [
    "c"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "funcone",
        "g": "functwo",
        "h": "functhr",
        "k": "prodcomp",
        "x": "varaxis",
        "c": "constan"
      },
      "question": "Functions $funcone,functwo,functhr$ are differentiable on some open interval around $0$\nand satisfy the equations and initial conditions\n\\begin{gather*}\nfuncone' = 2funcone^2functwofuncthr+\\frac{1}{functwofuncthr},\\quad funcone(0)=1, \\\\\nfunctwo'=funconefunctwo^2functhr+\\frac{4}{funconefuncthr}, \\quad functwo(0)=1, \\\\\nfuncthr'=3funconefunctwofuncthr^2+\\frac{1}{funconefunctwo}, \\quad functhr(0)=1.\n\\end{gather*}\nFind an explicit formula for $funcone(varaxis)$, valid in some open interval around $0$.",
      "solution": "Multiplying the first differential equation by $functwofuncthr$, the second by $funconefuncthr$,\nand the third by $funconefunctwo$, and summing gives\n\\[\n(funconefunctwofuncthr)' = 6(funconefunctwofuncthr)^2+6.\n\\]\nWrite $prodcomp(varaxis) = funcone(varaxis)functwo(varaxis)functhr(varaxis)$; then $prodcomp' = 6prodcomp^2+6$ and $prodcomp(0) = 1$. One\nsolution for this differential equation with this initial condition is\n$prodcomp(varaxis) = \\tan(6varaxis+\\pi/4)$; by standard uniqueness, this must necessarily\nhold for $varaxis$ in some open interval around $0$. Now the first given\nequation becomes\n\\begin{align*}\nfuncone'/funcone &= 2prodcomp(varaxis)+1/prodcomp(varaxis) \\\\\n&= 2\\tan(6varaxis+\\pi/4)+\\cot(6varaxis+\\pi/4);\n\\end{align*}\nintegrating both sides gives\n\\[\n\\ln(funcone(varaxis)) = \\frac{-2\\ln\\cos(6varaxis+\\pi/4) + \\ln\\sin(6varaxis+\\pi/4)}{6}+constan,\n\\]\nwhence $funcone(varaxis) = e^{constan}\n\\left(\\frac{\\sin(6varaxis+\\pi/4)}{\\cos^2(6varaxis+\\pi/4)}\\right)^{1/6}$.\nSubstituting $funcone(0)=1$ gives $e^{constan} = 2^{-1/12}$ and thus $funcone(varaxis) = 2^{-1/12}\n\\left(\\frac{\\sin(6varaxis+\\pi/4)}{\\cos^2(6varaxis+\\pi/4)}\\right)^{1/6}$.\n\n\\textbf{Remark.} The answer can be put in alternate forms using\ntrigonometric identities. One particularly simple one is\n\\[\n funcone(varaxis) = (\\sec 12varaxis)^{1/12} (\\sec 12varaxis + \\tan 12varaxis)^{1/4}.\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "sunflower",
        "g": "teacupholder",
        "h": "bookshelf",
        "k": "watermelon",
        "x": "skylight",
        "c": "muffinslice"
      },
      "question": "Functions $sunflower,teacupholder,bookshelf$ are differentiable on some open interval around $0$\nand satisfy the equations and initial conditions\n\\begin{gather*}\nsunflower' = 2sunflower^2teacupholderbookshelf+\\frac{1}{teacupholderbookshelf},\\quad sunflower(0)=1, \\\\\nteacupholder'=sunflowerteacupholder^2bookshelf+\\frac{4}{sunflowerbookshelf}, \\quad teacupholder(0)=1, \\\\\nbookshelf'=3sunflowerteacupholderbookshelf^2+\\frac{1}{sunflowerteacupholder}, \\quad bookshelf(0)=1.\n\\end{gather*}\nFind an explicit formula for $sunflower(skylight)$, valid in some open interval around $0$.",
      "solution": "Multiplying the first differential equation by $teacupholderbookshelf$, the second by $sunflowerbookshelf$,\nand the third by $sunflowerteacupholder$, and summing gives\n\\[\n(sunflowerteacupholderbookshelf)' = 6(sunflowerteacupholderbookshelf)^2+6.\n\\]\nWrite $watermelon(skylight) = sunflower(skylight)teacupholder(skylight)bookshelf(skylight)$; then $watermelon' = 6watermelon^2+6$ and $watermelon(0) = 1$. One\nsolution for this differential equation with this initial condition is\n$watermelon(skylight) = \\tan(6skylight+\\pi/4)$; by standard uniqueness, this must necessarily\nhold for $skylight$ in some open interval around $0$. Now the first given\nequation becomes\n\\begin{align*}\nsunflower'/sunflower &= 2watermelon(skylight)+1/watermelon(skylight) \\\\\n&= 2\\tan(6skylight+\\pi/4)+\\cot(6skylight+\\pi/4);\n\\end{align*}\nintegrating both sides gives\n\\[\n\\ln(sunflower(skylight)) = \\frac{-2\\ln\\cos(6skylight+\\pi/4) + \\ln\\sin(6skylight+\\pi/4)}{6}+muffinslice,\n\\]\nwhence $sunflower(skylight) = e^{muffinslice}\n\\left(\\frac{\\sin(6skylight+\\pi/4)}{\\cos^2(6skylight+\\pi/4)}\\right)^{1/6}$.\nSubstituting $sunflower(0)=1$ gives $e^{muffinslice} = 2^{-1/12}$ and thus $sunflower(skylight) = 2^{-1/12}\n\\left(\\frac{\\sin(6skylight+\\pi/4)}{\\cos^2(6skylight+\\pi/4)}\\right)^{1/6}$.\n\n\\textbf{Remark.} The answer can be put in alternate forms using\ntrigonometric identities. One particularly simple one is\n\\[\n sunflower(skylight) = (\\sec 12skylight)^{1/12} (\\sec 12skylight + \\tan 12skylight)^{1/4}.\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "fixedvalue",
        "g": "steadyparm",
        "h": "unchanging",
        "k": "quotient",
        "x": "constant",
        "c": "variable"
      },
      "question": "Functions $fixedvalue,steadyparm,unchanging$ are differentiable on some open interval around $0$\nand satisfy the equations and initial conditions\n\\begin{gather*}\nfixedvalue' = 2fixedvalue^2steadyparm unchanging+\\frac{1}{steadyparm unchanging},\\quad fixedvalue(0)=1, \\\\\nsteadyparm'=fixedvalue steadyparm^2 unchanging+\\frac{4}{fixedvalue unchanging}, \\quad steadyparm(0)=1, \\\\\nunchanging'=3fixedvalue steadyparm unchanging^2+\\frac{1}{fixedvalue steadyparm}, \\quad unchanging(0)=1.\n\\end{gather*}\nFind an explicit formula for $fixedvalue(constant)$, valid in some open interval around $0$.",
      "solution": "Multiplying the first differential equation by $steadyparm unchanging$, the second by $fixedvalue unchanging$,\nand the third by $fixedvalue steadyparm$, and summing gives\n\\[\n(fixedvalue steadyparm unchanging)' = 6(fixedvalue steadyparm unchanging)^2+6.\n\\]\nWrite $quotient(constant) = fixedvalue(constant) steadyparm(constant) unchanging(constant)$; then $quotient' = 6quotient^2+6$ and $quotient(0) = 1$. One\nsolution for this differential equation with this initial condition is\n$quotient(constant) = \\tan(6constant+\\pi/4)$; by standard uniqueness, this must necessarily\nhold for $constant$ in some open interval around $0$. Now the first given\nequation becomes\n\\begin{align*}\nfixedvalue'/fixedvalue &= 2quotient(constant)+1/quotient(constant) \\\\\n&= 2\\tan(6constant+\\pi/4)+\\cot(6constant+\\pi/4);\n\\end{align*}\nintegrating both sides gives\n\\[\n\\ln(fixedvalue(constant)) = \\frac{-2\\ln\\cos(6constant+\\pi/4) + \\ln\\sin(6constant+\\pi/4)}{6}+variable,\n\\]\nwhence $fixedvalue(constant) = e^{variable}\n\\left(\\frac{\\sin(6constant+\\pi/4)}{\\cos^2(6constant+\\pi/4)}\\right)^{1/6}$.\nSubstituting $fixedvalue(0)=1$ gives $e^{variable} = 2^{-1/12}$ and thus $fixedvalue(constant) = 2^{-1/12}\n\\left(\\frac{\\sin(6constant+\\pi/4)}{\\cos^2(6constant+\\pi/4)}\\right)^{1/6}$.\n\n\\textbf{Remark.} The answer can be put in alternate forms using\ntrigonometric identities. One particularly simple one is\n\\[\n fixedvalue(constant) = (\\sec 12constant)^{1/12} (\\sec 12constant + \\tan 12constant)^{1/4}.\n\\]"
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "g": "hjgrksla",
        "h": "bcyrmteu",
        "k": "vnqpslrz",
        "x": "mwetrzql",
        "c": "lfdsuaqo"
      },
      "question": "Problem:\n<<<\nFunctions $qzxwvtnp,hjgrksla,bcyrmteu$ are differentiable on some open interval around $0$\nand satisfy the equations and initial conditions\n\\begin{gather*}\nqzxwvtnp' = 2 qzxwvtnp^2 hjgrksla bcyrmteu+\\frac{1}{ hjgrksla bcyrmteu},\\quad qzxwvtnp(0)=1, \\\\\nhjgrksla' = qzxwvtnp hjgrksla^2 bcyrmteu+\\frac{4}{ qzxwvtnp bcyrmteu}, \\quad hjgrksla(0)=1, \\\\\nbcyrmteu' = 3 qzxwvtnp hjgrksla bcyrmteu^2+\\frac{1}{ qzxwvtnp hjgrksla}, \\quad bcyrmteu(0)=1.\n\\end{gather*}\nFind an explicit formula for $qzxwvtnp(mwetrzql)$, valid in some open interval around $0$.\n>>>\n",
      "solution": "Solution:\n<<<\nMultiplying the first differential equation by $hjgrksla bcyrmteu$, the second by $qzxwvtnp bcyrmteu$,\nand the third by $qzxwvtnp hjgrksla$, and summing gives\n\\[\n(qzxwvtnp hjgrksla bcyrmteu)' = 6(qzxwvtnp hjgrksla bcyrmteu)^2+6.\n\\]\nWrite $vnqpslrz(mwetrzql) = qzxwvtnp(mwetrzql)hjgrksla(mwetrzql)bcyrmteu(mwetrzql)$; then $vnqpslrz' = 6 vnqpslrz^2+6$ and $vnqpslrz(0) = 1$. One\nsolution for this differential equation with this initial condition is\n$vnqpslrz(mwetrzql) = \\tan(6 mwetrzql+\\pi/4)$; by standard uniqueness, this must necessarily\nhold for $mwetrzql$ in some open interval around $0$. Now the first given\nequation becomes\n\\begin{align*}\nqzxwvtnp'/qzxwvtnp &= 2 vnqpslrz(mwetrzql)+1/vnqpslrz(mwetrzql) \\\\\n&= 2\\tan(6 mwetrzql+\\pi/4)+\\cot(6 mwetrzql+\\pi/4);\n\\end{align*}\nintegrating both sides gives\n\\[\n\\ln(qzxwvtnp(mwetrzql)) = \\frac{-2\\ln\\cos(6 mwetrzql+\\pi/4) + \\ln\\sin(6 mwetrzql+\\pi/4)}{6}+lfdsuaqo,\n\\]\nwhence $qzxwvtnp(mwetrzql) = e^{lfdsuaqo}\n\\left(\\frac{\\sin(6 mwetrzql+\\pi/4)}{\\cos^2(6 mwetrzql+\\pi/4)}\\right)^{1/6}$.\nSubstituting $qzxwvtnp(0)=1$ gives $e^{lfdsuaqo} = 2^{-1/12}$ and thus $qzxwvtnp(mwetrzql) = 2^{-1/12}\n\\left(\\frac{\\sin(6 mwetrzql+\\pi/4)}{\\cos^2(6 mwetrzql+\\pi/4)}\\right)^{1/6}$.\n\n\\textbf{Remark.} The answer can be put in alternate forms using\ntrigonometric identities. One particularly simple one is\n\\[\n qzxwvtnp(mwetrzql) = (\\sec 12 mwetrzql)^{1/12} (\\sec 12 mwetrzql + \\tan 12 mwetrzql)^{1/4}.\n\\]\n>>>\n"
    },
    "kernel_variant": {
      "question": "Let $f,g,h,p,q$ be differentiable on some open interval containing $0$ and suppose they satisfy the nonlinear system  \n\\[\n\\begin{aligned}\nf'(x)&=11\\,f(x)^{2}g(x)h(x)p(x)q(x)\\;+\\;\\dfrac{3}{g(x)h(x)p(x)q(x)},\\qquad &&f(0)=1,\\\\[4pt]\ng'(x)&= 7\\,f(x)g(x)^{2}h(x)p(x)q(x)\\;+\\;\\dfrac{8}{f(x)h(x)p(x)q(x)},\\qquad &&g(0)=1,\\\\[4pt]\nh'(x)&= 2\\,f(x)g(x)h(x)^{2}p(x)q(x)\\;+\\;\\dfrac{5}{f(x)g(x)p(x)q(x)},\\qquad &&h(0)=1,\\\\[4pt]\np'(x)&= 3\\,f(x)g(x)h(x)p(x)^{2}q(x)\\;+\\;\\dfrac{4}{f(x)g(x)h(x)q(x)},\\qquad &&p(0)=1,\\\\[4pt]\nq'(x)&= 2\\,f(x)g(x)h(x)p(x)q(x)^{2}\\;+\\;\\dfrac{5}{f(x)g(x)h(x)p(x)},\\qquad &&q(0)=1 .\n\\end{aligned}\n\\]\n\nDefine  \n\\[\nK(x)=f(x)g(x)h(x)p(x)q(x).\n\\]\n\n(a) Show that $K$ satisfies the differential equation  \n\\[\nK'(x)=25\\bigl(K(x)^{2}+1\\bigr),\\qquad K(0)=1 .\n\\]\n\n(b) Obtain an explicit closed formula for $f(x)$.\n\n(c) Determine an open interval about $x=0$ on which your formula for $f(x)$ is real-valued.\n\n\\vspace{0.3cm}",
      "solution": "(a) Differentiating $K$ and using the product rule,\n\\[\n\\begin{aligned}\nK'&=(f'g h p q)+(f g' h p q)+(f g h' p q)+(f g h p' q)+(f g h p q')\\\\\n  &=\\frac{K}{f}f'+\\frac{K}{g}g'+\\frac{K}{h}h'+\\frac{K}{p}p'+\\frac{K}{q}q'.\n\\end{aligned}\n\\]\nInsert the five given differential equations, multiply each by the indicated factor $K/f_i$, and collect the coefficients:\n\\[\n\\begin{aligned}\nK'&=\\bigl(11+7+2+3+2\\bigr)K^{2}+\\bigl(3+8+5+4+5\\bigr)\\\\\n   &=25K^{2}+25\\\\\n   &=25\\bigl(K^{2}+1\\bigr).\n\\end{aligned}\n\\]\nWith the initial data $K(0)=1$ the statement of part (a) is proved.\n\n(b) The ODE for $K$ is separable:\n\\[\n\\frac{dK}{K^{2}+1}=25\\,dx\\quad\\Longrightarrow\\quad \\arctan K=25x+C.\n\\]\nBecause $\\arctan K(0)=\\pi/4$, we have $C=\\pi/4$, hence\n\\[\nK(x)=\\tan\\bigl(25x+\\pi/4\\bigr).\n\\]\n\nNow use the first differential equation again, writing it in logarithmic form:\n\\[\n\\frac{f'}{f}=11K+\\frac{3}{K}.\n\\]\nSet \n\\[\n\\theta(x)=25x+\\frac{\\pi}{4},\\qquad\\text{so that }K=\\tan\\theta,\\; d\\theta=25\\,dx.\n\\]\nThen\n\\[\n\\int\\frac{f'}{f}\\,dx\n     =\\frac{1}{25}\\int\\!\\Bigl(11\\tan\\theta+3\\cot\\theta\\Bigr)\\,d\\theta\n     =-\\frac{11}{25}\\ln\\lvert\\cos\\theta\\rvert+\\frac{3}{25}\\ln\\lvert\\sin\\theta\\rvert+C_{0}.\n\\]\nExponentiating gives\n\\[\nf(x)=C\\,\\bigl[\\sin\\theta(x)\\bigr]^{3/25}\\bigl[\\cos\\theta(x)\\bigr]^{-11/25}.\n\\]\n\nTo find $C$, use $f(0)=1$ and $\\theta(0)=\\pi/4$:\n\\[\n1=C\\bigl(\\tfrac{\\sqrt{2}}{2}\\bigr)^{3/25}\\bigl(\\tfrac{\\sqrt{2}}{2}\\bigr)^{-11/25}\n  =C\\bigl(\\tfrac{\\sqrt{2}}{2}\\bigr)^{-8/25},\n\\]\nhence\n\\[\nC=\\Bigl(\\tfrac{\\sqrt{2}}{2}\\Bigr)^{8/25}.\n\\]\nConsequently\n\\[\n\\boxed{\\,f(x)=\\Bigl(\\tfrac{\\sqrt{2}}{2}\\Bigr)^{8/25}\n             \\bigl[\\sin(25x+\\pi/4)\\bigr]^{3/25}\n             \\bigl[\\cos(25x+\\pi/4)\\bigr]^{-11/25}\\,}.\n\\]\n\n(c) Domain of real-valuedness.  \nBecause fractional powers are involved, we insist on\n\\[\n\\sin\\theta(x)>0,\\qquad\\cos\\theta(x)\\neq0,\n\\]\nand, to stay within the principal real branch, we further require $\\cos\\theta(x)>0$.  \nThe simultaneous conditions $\\sin\\theta>0$ and $\\cos\\theta>0$ are met exactly when\n\\[\n0<\\theta(x)<\\frac{\\pi}{2}.\n\\]\nSubstituting $\\theta(x)=25x+\\pi/4$,\n\\[\n0<25x+\\frac{\\pi}{4}<\\frac{\\pi}{2}\\quad\\Longrightarrow\\quad -\\frac{\\pi}{100}<x<\\frac{\\pi}{100}.\n\\]\nTherefore the formula for $f(x)$ is real-valued on the open interval\n\\[\n\\boxed{\\;(-\\pi/100,\\;\\pi/100)\\;}.\n\\]\n\n\\vspace{0.3cm}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.812062",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the original problem had 3 unknown functions; this variant has 5, increasing the algebraic and bookkeeping complexity.  \n• Additional parameters: unequal coefficient pairs (11,3), (7,8), … remove the visible symmetry of the original; detecting that their sums coincide is now essential.  \n• More intricate manipulation: five coupled nonlinear equations must be balanced simultaneously to extract the single Riccati equation.  \n• Heavier integration: the exponent pair (11/25, 3/25) forces fractional powers that are not obvious from pattern-matching the earlier solution.  \n• Domain justification: one must analyse both sine and cosine factors as well as the higher-frequency argument 25x+π/4 to state a correct interval.  \nCollectively these additions demand deeper insight, more algebraic endurance, and a clearer grasp of Riccati reductions than either the original or the previous kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $f,g,h,p,q$ be differentiable on some open interval containing $0$ and suppose they satisfy the nonlinear system  \n\\[\n\\begin{aligned}\nf'(x)&=11\\,f(x)^{2}g(x)h(x)p(x)q(x)\\;+\\;\\dfrac{3}{g(x)h(x)p(x)q(x)},\\qquad &&f(0)=1,\\\\[4pt]\ng'(x)&= 7\\,f(x)g(x)^{2}h(x)p(x)q(x)\\;+\\;\\dfrac{8}{f(x)h(x)p(x)q(x)},\\qquad &&g(0)=1,\\\\[4pt]\nh'(x)&= 2\\,f(x)g(x)h(x)^{2}p(x)q(x)\\;+\\;\\dfrac{5}{f(x)g(x)p(x)q(x)},\\qquad &&h(0)=1,\\\\[4pt]\np'(x)&= 3\\,f(x)g(x)h(x)p(x)^{2}q(x)\\;+\\;\\dfrac{4}{f(x)g(x)h(x)q(x)},\\qquad &&p(0)=1,\\\\[4pt]\nq'(x)&= 2\\,f(x)g(x)h(x)p(x)q(x)^{2}\\;+\\;\\dfrac{5}{f(x)g(x)h(x)p(x)},\\qquad &&q(0)=1 .\n\\end{aligned}\n\\]\n\nDefine  \n\\[\nK(x)=f(x)g(x)h(x)p(x)q(x).\n\\]\n\n(a) Show that $K$ satisfies the differential equation  \n\\[\nK'(x)=25\\bigl(K(x)^{2}+1\\bigr),\\qquad K(0)=1 .\n\\]\n\n(b) Obtain an explicit closed formula for $f(x)$.\n\n(c) Determine an open interval about $x=0$ on which your formula for $f(x)$ is real-valued.\n\n\\vspace{0.3cm}",
      "solution": "(a) Differentiating $K$ and using the product rule,\n\\[\n\\begin{aligned}\nK'&=(f'g h p q)+(f g' h p q)+(f g h' p q)+(f g h p' q)+(f g h p q')\\\\\n  &=\\frac{K}{f}f'+\\frac{K}{g}g'+\\frac{K}{h}h'+\\frac{K}{p}p'+\\frac{K}{q}q'.\n\\end{aligned}\n\\]\nInsert the five given differential equations, multiply each by the indicated factor $K/f_i$, and collect the coefficients:\n\\[\n\\begin{aligned}\nK'&=\\bigl(11+7+2+3+2\\bigr)K^{2}+\\bigl(3+8+5+4+5\\bigr)\\\\\n   &=25K^{2}+25\\\\\n   &=25\\bigl(K^{2}+1\\bigr).\n\\end{aligned}\n\\]\nWith the initial data $K(0)=1$ the statement of part (a) is proved.\n\n(b) The ODE for $K$ is separable:\n\\[\n\\frac{dK}{K^{2}+1}=25\\,dx\\quad\\Longrightarrow\\quad \\arctan K=25x+C.\n\\]\nBecause $\\arctan K(0)=\\pi/4$, we have $C=\\pi/4$, hence\n\\[\nK(x)=\\tan\\bigl(25x+\\pi/4\\bigr).\n\\]\n\nNow use the first differential equation again, writing it in logarithmic form:\n\\[\n\\frac{f'}{f}=11K+\\frac{3}{K}.\n\\]\nSet \n\\[\n\\theta(x)=25x+\\frac{\\pi}{4},\\qquad\\text{so that }K=\\tan\\theta,\\; d\\theta=25\\,dx.\n\\]\nThen\n\\[\n\\int\\frac{f'}{f}\\,dx\n     =\\frac{1}{25}\\int\\!\\Bigl(11\\tan\\theta+3\\cot\\theta\\Bigr)\\,d\\theta\n     =-\\frac{11}{25}\\ln\\lvert\\cos\\theta\\rvert+\\frac{3}{25}\\ln\\lvert\\sin\\theta\\rvert+C_{0}.\n\\]\nExponentiating gives\n\\[\nf(x)=C\\,\\bigl[\\sin\\theta(x)\\bigr]^{3/25}\\bigl[\\cos\\theta(x)\\bigr]^{-11/25}.\n\\]\n\nTo find $C$, use $f(0)=1$ and $\\theta(0)=\\pi/4$:\n\\[\n1=C\\bigl(\\tfrac{\\sqrt{2}}{2}\\bigr)^{3/25}\\bigl(\\tfrac{\\sqrt{2}}{2}\\bigr)^{-11/25}\n  =C\\bigl(\\tfrac{\\sqrt{2}}{2}\\bigr)^{-8/25},\n\\]\nhence\n\\[\nC=\\Bigl(\\tfrac{\\sqrt{2}}{2}\\Bigr)^{8/25}.\n\\]\nConsequently\n\\[\n\\boxed{\\,f(x)=\\Bigl(\\tfrac{\\sqrt{2}}{2}\\Bigr)^{8/25}\n             \\bigl[\\sin(25x+\\pi/4)\\bigr]^{3/25}\n             \\bigl[\\cos(25x+\\pi/4)\\bigr]^{-11/25}\\,}.\n\\]\n\n(c) Domain of real-valuedness.  \nBecause fractional powers are involved, we insist on\n\\[\n\\sin\\theta(x)>0,\\qquad\\cos\\theta(x)\\neq0,\n\\]\nand, to stay within the principal real branch, we further require $\\cos\\theta(x)>0$.  \nThe simultaneous conditions $\\sin\\theta>0$ and $\\cos\\theta>0$ are met exactly when\n\\[\n0<\\theta(x)<\\frac{\\pi}{2}.\n\\]\nSubstituting $\\theta(x)=25x+\\pi/4$,\n\\[\n0<25x+\\frac{\\pi}{4}<\\frac{\\pi}{2}\\quad\\Longrightarrow\\quad -\\frac{\\pi}{100}<x<\\frac{\\pi}{100}.\n\\]\nTherefore the formula for $f(x)$ is real-valued on the open interval\n\\[\n\\boxed{\\;(-\\pi/100,\\;\\pi/100)\\;}.\n\\]\n\n\\vspace{0.3cm}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.621286",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the original problem had 3 unknown functions; this variant has 5, increasing the algebraic and bookkeeping complexity.  \n• Additional parameters: unequal coefficient pairs (11,3), (7,8), … remove the visible symmetry of the original; detecting that their sums coincide is now essential.  \n• More intricate manipulation: five coupled nonlinear equations must be balanced simultaneously to extract the single Riccati equation.  \n• Heavier integration: the exponent pair (11/25, 3/25) forces fractional powers that are not obvious from pattern-matching the earlier solution.  \n• Domain justification: one must analyse both sine and cosine factors as well as the higher-frequency argument 25x+π/4 to state a correct interval.  \nCollectively these additions demand deeper insight, more algebraic endurance, and a clearer grasp of Riccati reductions than either the original or the previous kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}