{
  "index": "2014-B-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Suppose that $f$ is a function on the interval $[1,3]$ such that $-1 \\leq f(x) \\leq 1$ for all $x$ and $\\int_1^3 f(x)\\,dx = 0$. How large can $\\int_1^3 \\frac{f(x)}{x}\\,dx$ be?\n\n\\,",
  "solution": "In all solutions, we assume that the function $f$ is integrable.\n\n\\textbf{First solution:}\nLet $g(x)$ be $1$ for $1\\leq x\\leq 2$ and $-1$ for $2<x\\leq 3$, and \ndefine $h(x)=g(x)-f(x)$. Then $\\int_1^3 h(x)\\,dx = 0$ and $h(x) \\geq 0$ \nfor $1\\leq x\\leq 2$, $h(x) \\leq 0$ for $2<x\\leq 3$. Now\n\\begin{align*}\n\\int_1^3 \\frac{h(x)}{x}\\,dx & = \\int_1^2 \\frac{|h(x)|}{x}\\,dx - \\int_2^3 \n\\frac{|h(x)|}{x}\\,dx\\\\\n&\\geq \\int_1^2 \\frac{|h(x)|}{2}\\,dx - \\int_2^3 \\frac{|h(x)|}{2}\\,dx = 0,\n\\end{align*}\nand thus $\\int_1^3 \\frac{f(x)}{x}\\,dx \\leq \\int_1^3 \\frac{g(x)}{x}\\,dx = \n2\\log 2-\\log 3 = \\log \\frac{4}{3}$. Since $g(x)$ achieves the upper bound, the answer is \n$\\log \\frac{4}{3}$.\n\n\\noindent\n\\textbf{Reformulation:}\n(by Karl Mahlburg and Karthik Adimurthi) \nSince $f$ is integrable, it can be expressed in terms of the Hadamard basis\n\\begin{align*}\nH_0(x) &= \\begin{cases} 1 & x \\in [1,2) \\\\\n-1 & x \\in [2,3] \\\\\n0 & x \\notin [1,3] \\end{cases} \\\\\nH_{n+1}(x) &= H_n( 2(x-1) + 1) + H_n( 2(x-2) + 1).\n\\end{align*}\nMore precisely, we have $f(x) = \\sum_n c_n H_n(x)$ for some $c_n$ \nwith $|c_0| + |c_1| + |\\cdots| \\leq 1$. Let $g_n = \\int_1^3 (H_n(x) / x)\\,dx$; \nit is easy to show that the $g_n$ are strictly decreasing in $n$. \nThus \\[\n\\int_1^3 (f(x) / x) dx = c_0 g_0 +\nc_1 g_1 + \\cdots \\leq 1 \\cdot g_0 = \\log \\frac{4}{3}.\n\\] \n\n\\noindent\n\\textbf{Second solution:}\n(Art of Problem Solving, user \\verb+libra_gold+)\nDefine the function $F(x) = \\int_1^x f(t)\\,dt$ for $1 \\leq x \\leq 3$; then $F(1) = F(3) = 0$ and $F(x) \\leq \\min\\{x-1,3-x\\}$. Using integration by parts, we obtain\n\\begin{align*}\n\\int_1^3 \\frac{f(x)}{x}\\,dx &= \\int_1^3 \\frac{F(x)}{x^2}\\,dx \\\\\n&\\leq \\int_1^2 \\frac{x-1}{x^2}\\,dx + \\int_2^3 \\frac{3-x}{x^2}\\,dx \\\\\n&= \\log \\frac{4}{3}.\n\\end{align*}\n(Some minor adjustment is needed to make this completely rigorous, e.g., approximating $f$ uniformly by continuous functions.)",
  "vars": [
    "x",
    "t",
    "n"
  ],
  "params": [
    "f",
    "g",
    "h",
    "H_0",
    "H_n",
    "c_n",
    "g_n",
    "F"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "t": "variablet",
        "n": "indexvar",
        "f": "boundedfunc",
        "g": "stepfunc",
        "h": "diffhelper",
        "H_0": "hadamardzero",
        "H_n": "hadamardgen",
        "c_n": "coeffindex",
        "g_n": "integrvalue",
        "F": "antideriv"
      },
      "question": "Suppose that $boundedfunc$ is a function on the interval $[1,3]$ such that $-1 \\leq boundedfunc(variablex) \\leq 1$ for all $variablex$ and $\\int_1^3 boundedfunc(variablex)\\,dvariablex = 0$. How large can $\\int_1^3 \\frac{boundedfunc(variablex)}{variablex}\\,dvariablex$ be?\n",
      "solution": "In all solutions, we assume that the function $boundedfunc$ is integrable.\n\n\\textbf{First solution:}\nLet $stepfunc(variablex)$ be $1$ for $1\\leq variablex\\leq 2$ and $-1$ for $2<variablex\\leq 3$, and define $diffhelper(variablex)=stepfunc(variablex)-boundedfunc(variablex)$. Then $\\int_1^3 diffhelper(variablex)\\,dvariablex = 0$ and $diffhelper(variablex) \\geq 0$ for $1\\leq variablex\\leq 2$, $diffhelper(variablex) \\leq 0$ for $2<variablex\\leq 3$. Now\n\\begin{align*}\n\\int_1^3 \\frac{diffhelper(variablex)}{variablex}\\,dvariablex & = \\int_1^2 \\frac{|diffhelper(variablex)|}{variablex}\\,dvariablex - \\int_2^3 \\frac{|diffhelper(variablex)|}{variablex}\\,dvariablex\\\\\n&\\geq \\int_1^2 \\frac{|diffhelper(variablex)|}{2}\\,dvariablex - \\int_2^3 \\frac{|diffhelper(variablex)|}{2}\\,dvariablex = 0,\n\\end{align*}\nand thus $\\int_1^3 \\frac{boundedfunc(variablex)}{variablex}\\,dvariablex \\leq \\int_1^3 \\frac{stepfunc(variablex)}{variablex}\\,dvariablex = 2\\log 2-\\log 3 = \\log \\frac{4}{3}$. Since $stepfunc(variablex)$ achieves the upper bound, the answer is $\\log \\frac{4}{3}$.\n\n\\noindent\n\\textbf{Reformulation:}\n(by Karl Mahlburg and Karthik Adimurthi) Since $boundedfunc$ is integrable, it can be expressed in terms of the Hadamard basis\n\\begin{align*}\nhadamardzero(variablex) &= \\begin{cases} 1 & variablex \\in [1,2) \\\\ -1 & variablex \\in [2,3] \\\\ 0 & variablex \\notin [1,3] \\end{cases} \\\\\nhadamardgen(variablex) &= hadamardgen( 2(variablex-1) + 1) + hadamardgen( 2(variablex-2) + 1).\n\\end{align*}\nMore precisely, we have $boundedfunc(variablex) = \\sum_{indexvar} coeffindex hadamardgen(variablex)$ for some $coeffindex$ with $|c_0| + |c_1| + |\\cdots| \\leq 1$. Let $integrvalue = \\int_1^3 (hadamardgen(variablex) / variablex)\\,dvariablex$; it is easy to show that the $integrvalue$ are strictly decreasing in $indexvar$. Thus\n\\[\\int_1^3 (boundedfunc(variablex) / variablex) dvariablex = c_0 g_0 + c_1 g_1 + \\cdots \\leq 1 \\cdot g_0 = \\log \\frac{4}{3}.\\]\n\n\\noindent\n\\textbf{Second solution:}\n(Art of Problem Solving, user \\verb+libra_gold+)\nDefine the function $antideriv(variablex) = \\int_1^{variablex} boundedfunc(variablet)\\,dvariablet$ for $1 \\leq variablex \\leq 3$; then $antideriv(1) = antideriv(3) = 0$ and $antideriv(variablex) \\leq \\min\\{variablex-1,3-variablex\\}$. Using integration by parts, we obtain\n\\begin{align*}\n\\int_1^3 \\frac{boundedfunc(variablex)}{variablex}\\,dvariablex &= \\int_1^3 \\frac{antideriv(variablex)}{variablex^2}\\,dvariablex \\\\\n&\\leq \\int_1^2 \\frac{variablex-1}{variablex^2}\\,dvariablex + \\int_2^3 \\frac{3-variablex}{variablex^2}\\,dvariablex \\\\\n&= \\log \\frac{4}{3}.\n\\end{align*}\n(Some minor adjustment is needed to make this completely rigorous, e.g., approximating $boundedfunc$ uniformly by continuous functions.)\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "riverdelta",
        "t": "ivyblossom",
        "n": "lanternfly",
        "f": "spiceroute",
        "g": "thunderclap",
        "h": "marigolden",
        "H_0": "silkmotion",
        "H_n": "echoharvest",
        "c_n": "velvetquill",
        "g_n": "duskfragment",
        "F": "coralbeacon"
      },
      "question": "Suppose that $spiceroute$ is a function on the interval $[1,3]$ such that $-1 \\leq spiceroute(riverdelta) \\leq 1$ for all $riverdelta$ and $\\int_1^3 spiceroute(riverdelta)\\,driverdelta = 0$. How large can $\\int_1^3 \\frac{spiceroute(riverdelta)}{riverdelta}\\,driverdelta$ be?\n",
      "solution": "In all solutions, we assume that the function $spiceroute$ is integrable.\n\n\\textbf{First solution:}\nLet $thunderclap(riverdelta)$ be $1$ for $1\\leq riverdelta\\leq 2$ and $-1$ for $2<riverdelta\\leq 3$, and \ndefine $marigolden(riverdelta)=thunderclap(riverdelta)-spiceroute(riverdelta)$. Then $\\int_1^3 marigolden(riverdelta)\\,driverdelta = 0$ and $marigolden(riverdelta) \\geq 0$ \nfor $1\\leq riverdelta\\leq 2$, $marigolden(riverdelta) \\leq 0$ for $2<riverdelta\\leq 3$. Now\n\\begin{align*}\n\\int_1^3 \\frac{marigolden(riverdelta)}{riverdelta}\\,driverdelta & = \\int_1^2 \\frac{|marigolden(riverdelta)|}{riverdelta}\\,driverdelta - \\int_2^3 \n\\frac{|marigolden(riverdelta)|}{riverdelta}\\,driverdelta\\\\\n&\\geq \\int_1^2 \\frac{|marigolden(riverdelta)|}{2}\\,driverdelta - \\int_2^3 \\frac{|marigolden(riverdelta)|}{2}\\,driverdelta = 0,\n\\end{align*}\nand thus $\\int_1^3 \\frac{spiceroute(riverdelta)}{riverdelta}\\,driverdelta \\leq \\int_1^3 \\frac{thunderclap(riverdelta)}{riverdelta}\\,driverdelta = \n2\\log 2-\\log 3 = \\log \\frac{4}{3}$. Since $thunderclap(riverdelta)$ achieves the upper bound, the answer is \n$\\log \\frac{4}{3}$.\n\n\\noindent\n\\textbf{Reformulation:}\n(by Karl Mahlburg and Karthik Adimurthi) \nSince $spiceroute$ is integrable, it can be expressed in terms of the Hadamard basis\n\\begin{align*}\nsilkmotion(riverdelta) &= \\begin{cases} 1 & riverdelta \\in [1,2) \\\\ -1 & riverdelta \\in [2,3] \\\\ 0 & riverdelta \\notin [1,3] \\end{cases} \\\\ \nH_{lanternfly+1}(riverdelta) &= echoharvest( 2(riverdelta-1) + 1) + echoharvest( 2(riverdelta-2) + 1).\n\\end{align*}\nMore precisely, we have $spiceroute(riverdelta) = \\sum_{lanternfly} velvetquill\\,echoharvest(riverdelta)$ for some $velvetquill$ \nwith $|c_0| + |c_1| + |\\cdots| \\leq 1$. Let $duskfragment = \\int_1^3 (echoharvest(riverdelta) / riverdelta)\\,driverdelta$; \nit is easy to show that the $duskfragment$ are strictly decreasing in $lanternfly$. \nThus \\[\n\\int_1^3 (spiceroute(riverdelta) / riverdelta) driverdelta = c_0 g_0 +\nc_1 g_1 + \\cdots \\leq 1 \\cdot g_0 = \\log \\frac{4}{3}.\n\\] \n\n\\noindent\n\\textbf{Second solution:}\n(Art of Problem Solving, user \\verb+libra_gold+)\nDefine the function $coralbeacon(riverdelta) = \\int_1^{riverdelta} spiceroute(ivyblossom)\\,divyblossom$ for $1 \\leq riverdelta \\leq 3$; then $coralbeacon(1) = coralbeacon(3) = 0$ and $coralbeacon(riverdelta) \\leq \\min\\{riverdelta-1,3-riverdelta\\}$. Using integration by parts, we obtain\n\\begin{align*}\n\\int_1^3 \\frac{spiceroute(riverdelta)}{riverdelta}\\,driverdelta &= \\int_1^3 \\frac{coralbeacon(riverdelta)}{riverdelta^2}\\,driverdelta \\\\\n&\\leq \\int_1^2 \\frac{riverdelta-1}{riverdelta^2}\\,driverdelta + \\int_2^3 \\frac{3-riverdelta}{riverdelta^2}\\,driverdelta \\\\\n&= \\log \\frac{4}{3}.\n\\end{align*}\n(Some minor adjustment is needed to make this completely rigorous, e.g., approximating $spiceroute$ uniformly by continuous functions.)\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedpoint",
        "t": "steadytime",
        "n": "constantidx",
        "f": "dysfunction",
        "g": "disorder",
        "h": "blendfunc",
        "H_0": "smoothzero",
        "H_n": "smoothindex",
        "c_n": "variablescalar",
        "g_n": "roughintegral",
        "F": "staticvalue"
      },
      "question": "Suppose that $dysfunction$ is a function on the interval $[1,3]$ such that $-1 \\leq dysfunction(fixedpoint) \\leq 1$ for all $fixedpoint$ and $\\int_1^3 dysfunction(fixedpoint)\\,d fixedpoint = 0$. How large can $\\int_1^3 \\frac{dysfunction(fixedpoint)}{fixedpoint}\\,d fixedpoint$ be?\n",
      "solution": "In all solutions, we assume that the function $dysfunction$ is integrable.\n\n\\textbf{First solution:}\nLet $disorder(fixedpoint)$ be $1$ for $1\\leq fixedpoint\\leq 2$ and $-1$ for $2<fixedpoint\\leq 3$, and \ndefine $blendfunc(fixedpoint)=disorder(fixedpoint)-dysfunction(fixedpoint)$. Then $\\int_1^3 blendfunc(fixedpoint)\\,d fixedpoint = 0$ and $blendfunc(fixedpoint) \\geq 0$ \nfor $1\\leq fixedpoint\\leq 2$, $blendfunc(fixedpoint) \\leq 0$ for $2<fixedpoint\\leq 3$. Now\n\\begin{align*}\n\\int_1^3 \\frac{blendfunc(fixedpoint)}{fixedpoint}\\,d fixedpoint & = \\int_1^2 \\frac{|blendfunc(fixedpoint)|}{fixedpoint}\\,d fixedpoint - \\int_2^3 \n\\frac{|blendfunc(fixedpoint)|}{fixedpoint}\\,d fixedpoint\\\\\n&\\geq \\int_1^2 \\frac{|blendfunc(fixedpoint)|}{2}\\,d fixedpoint - \\int_2^3 \\frac{|blendfunc(fixedpoint)|}{2}\\,d fixedpoint = 0,\n\\end{align*}\nand thus $\\int_1^3 \\frac{dysfunction(fixedpoint)}{fixedpoint}\\,d fixedpoint \\leq \\int_1^3 \\frac{disorder(fixedpoint)}{fixedpoint}\\,d fixedpoint = \n2\\log 2-\\log 3 = \\log \\frac{4}{3}$. Since $disorder(fixedpoint)$ achieves the upper bound, the answer is \n$\\log \\frac{4}{3}$.\n\n\\noindent\n\\textbf{Reformulation:}\n(by Karl Mahlburg and Karthik Adimurthi) \nSince $dysfunction$ is integrable, it can be expressed in terms of the Hadamard basis\n\\begin{align*}\nsmoothzero(fixedpoint) &= \\begin{cases} 1 & fixedpoint \\in [1,2) \\\\ -1 & fixedpoint \\in [2,3] \\\\ 0 & fixedpoint \\notin [1,3] \\end{cases} \\\\\nH_{constantidx+1}(fixedpoint) &= smoothindex( 2(fixedpoint-1) + 1) + smoothindex( 2(fixedpoint-2) + 1).\n\\end{align*}\nMore precisely, we have $dysfunction(fixedpoint) = \\sum_{constantidx} variablescalar smoothindex(fixedpoint)$ \nwith $|c_0| + |c_1| + |\\cdots| \\leq 1$. Let $roughintegral = \\int_1^3 (smoothindex(fixedpoint) / fixedpoint)\\,d fixedpoint$; \nit is easy to show that the $roughintegral$ are strictly decreasing in $constantidx$. \nThus \\[\n\\int_1^3 (dysfunction(fixedpoint) / fixedpoint) d fixedpoint = c_0 g_0 +\nc_1 g_1 + \\cdots \\leq 1 \\cdot g_0 = \\log \\frac{4}{3}.\n\\] \n\n\\noindent\n\\textbf{Second solution:}\n(Art of Problem Solving, user \\verb+libra_gold+)\nDefine the function $staticvalue(fixedpoint) = \\int_1^{fixedpoint} dysfunction(steadytime)\\,d steadytime$ for $1 \\leq fixedpoint \\leq 3$; then $staticvalue(1) = staticvalue(3) = 0$ and $staticvalue(fixedpoint) \\leq \\min\\{fixedpoint-1,3-fixedpoint\\}$. Using integration by parts, we obtain\n\\begin{align*}\n\\int_1^3 \\frac{dysfunction(fixedpoint)}{fixedpoint}\\,d fixedpoint &= \\int_1^3 \\frac{staticvalue(fixedpoint)}{fixedpoint^2}\\,d fixedpoint \\\\\n&\\leq \\int_1^2 \\frac{fixedpoint-1}{fixedpoint^2}\\,d fixedpoint + \\int_2^3 \\frac{3-fixedpoint}{fixedpoint^2}\\,d fixedpoint \\\\\n&= \\log \\frac{4}{3}.\n\\end{align*}\n(Some minor adjustment is needed to make this completely rigorous, e.g., approximating $dysfunction$ uniformly by continuous functions.)"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "n": "mfqpazvl",
        "f": "lskjrhqd",
        "g": "dcmvezqj",
        "h": "brlnouyx",
        "H_0": "spqkdjrh",
        "H_n": "pnxagwzu",
        "c_n": "vxdsapmq",
        "g_n": "qplmfrde",
        "F": "wzvhtekc"
      },
      "question": "Suppose that $lskjrhqd$ is a function on the interval $[1,3]$ such that $-1 \\leq lskjrhqd(qzxwvtnp) \\leq 1$ for all $qzxwvtnp$ and $\\int_1^3 lskjrhqd(qzxwvtnp)\\,dqzxwvtnp = 0$. How large can $\\int_1^3 \\frac{lskjrhqd(qzxwvtnp)}{qzxwvtnp}\\,dqzxwvtnp$ be?\n\n\\,",
      "solution": "In all solutions, we assume that the function $lskjrhqd$ is integrable.\n\n\\textbf{First solution:}\nLet $dcmvezqj(qzxwvtnp)$ be $1$ for $1\\leq qzxwvtnp\\leq 2$ and $-1$ for $2<qzxwvtnp\\leq 3$, and \ndefine $brlnouyx(qzxwvtnp)=dcmvezqj(qzxwvtnp)-lskjrhqd(qzxwvtnp)$. Then $\\int_1^3 brlnouyx(qzxwvtnp)\\,dqzxwvtnp = 0$ and $brlnouyx(qzxwvtnp) \\geq 0$ \nfor $1\\leq qzxwvtnp\\leq 2$, $brlnouyx(qzxwvtnp) \\leq 0$ for $2<qzxwvtnp\\leq 3$. Now\n\\begin{align*}\n\\int_1^3 \\frac{brlnouyx(qzxwvtnp)}{qzxwvtnp}\\,dqzxwvtnp & = \\int_1^2 \\frac{|brlnouyx(qzxwvtnp)|}{qzxwvtnp}\\,dqzxwvtnp - \\int_2^3 \n\\frac{|brlnouyx(qzxwvtnp)|}{qzxwvtnp}\\,dqzxwvtnp\\\\\n&\\geq \\int_1^2 \\frac{|brlnouyx(qzxwvtnp)|}{2}\\,dqzxwvtnp - \\int_2^3 \\frac{|brlnouyx(qzxwvtnp)|}{2}\\,dqzxwvtnp = 0,\n\\end{align*}\nand thus $\\int_1^3 \\frac{lskjrhqd(qzxwvtnp)}{qzxwvtnp}\\,dqzxwvtnp \\leq \\int_1^3 \\frac{dcmvezqj(qzxwvtnp)}{qzxwvtnp}\\,dqzxwvtnp = \n2\\log 2-\\log 3 = \\log \\frac{4}{3}$. Since $dcmvezqj(qzxwvtnp)$ achieves the upper bound, the answer is \n$\\log \\frac{4}{3}$.\n\n\\noindent\n\\textbf{Reformulation:}\n(by Karl Mahlburg and Karthik Adimurthi) \nSince $lskjrhqd$ is integrable, it can be expressed in terms of the Hadamard basis\n\\begin{align*}\nspqkdjrh(qzxwvtnp) &= \\begin{cases} 1 & qzxwvtnp \\in [1,2) \\\\ -1 & qzxwvtnp \\in [2,3] \\\\ 0 & qzxwvtnp \\notin [1,3] \\end{cases} \\\\\npnxagwzu_{mfqpazvl+1}(qzxwvtnp) &= pnxagwzu( 2(qzxwvtnp-1) + 1) + pnxagwzu( 2(qzxwvtnp-2) + 1).\n\\end{align*}\nMore precisely, we have $lskjrhqd(qzxwvtnp) = \\sum_{mfqpazvl} vxdsapmq\\, pnxagwzu(qzxwvtnp)$ for some $vxdsapmq$ \nwith $|vxdsapmq| + |vxdsapmq| + |\\cdots| \\leq 1$. Let $qplmfrde = \\int_1^3 (pnxagwzu(qzxwvtnp) / qzxwvtnp)\\,dqzxwvtnp$; \nit is easy to show that the $qplmfrde$ are strictly decreasing in $mfqpazvl$. \nThus \\[\n\\int_1^3 (lskjrhqd(qzxwvtnp) / qzxwvtnp) dqzxwvtnp = vxdsapmq qplmfrde +\nvxdsapmq qplmfrde + \\cdots \\leq 1 \\cdot qplmfrde = \\log \\frac{4}{3}.\n\\] \n\n\\noindent\n\\textbf{Second solution:}\n(Art of Problem Solving, user \\verb+libra_gold+)\nDefine the function $wzvhtekc(qzxwvtnp) = \\int_1^{qzxwvtnp} lskjrhqd(hjgrksla)\\,dhjgrksla$ for $1 \\leq qzxwvtnp \\leq 3$; then $wzvhtekc(1) = wzvhtekc(3) = 0$ and $wzvhtekc(qzxwvtnp) \\leq \\min\\{qzxwvtnp-1,3-qzxwvtnp\\}$. Using integration by parts, we obtain\n\\begin{align*}\n\\int_1^3 \\frac{lskjrhqd(qzxwvtnp)}{qzxwvtnp}\\,dqzxwvtnp &= \\int_1^3 \\frac{wzvhtekc(qzxwvtnp)}{qzxwvtnp^2}\\,dqzxwvtnp \\\\\n&\\leq \\int_1^2 \\frac{qzxwvtnp-1}{qzxwvtnp^2}\\,dqzxwvtnp + \\int_2^3 \\frac{3-qzxwvtnp}{qzxwvtnp^2}\\,dqzxwvtnp \\\\\n&= \\log \\frac{4}{3}.\n\\end{align*}\n(Some minor adjustment is needed to make this completely rigorous, e.g., approximating $lskjrhqd$ uniformly by continuous functions.)"
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nf\\colon[1,4]\\longrightarrow\\mathbb R\n\\]\nbe Lebesgue–integrable and satisfy  \n\\[\n-4\\le f(x)\\le 4\\qquad(1\\le x\\le 4), \\qquad\n\\int_{1}^{4} f(x)\\,dx=0, \\qquad\n\\int_{1}^{4} x\\,f(x)\\,dx=0 .\n\\]\nDetermine  \n\\[\nI(f)=\\int_{1}^{4}\\frac{f(x)}{x^{2}}\\;dx ,\n\\]\nand describe (up to sets of measure $0$) every function $f$ that maximises $I(f)$.\n\n──────────────────────────────────",
      "solution": "Throughout put  \n\\[\nw(x)=\\frac{1}{x^{2}},\\qquad 1\\le x\\le4 .\n\\]\n\nStep 1.  Extreme points are $\\boldsymbol{\\pm4}$–valued.  \nLet  \n\\[\n\\mathcal F=\\Bigl\\{f\\in L^{\\infty}[1,4]\\;:\\;\n-4\\le f\\le 4,\\;\n\\int_{1}^{4}f=0,\\;\n\\int_{1}^{4}x\\,f=0\\Bigr\\}.\n\\]\nThe set $\\mathcal F$ is weak–$^{\\!*}$ compact and convex, while\n$I(f)=\\langle f,w\\rangle$ is weak–$^{\\!*}$ continuous, so a maximiser\nexists.  By the Bauer–Krein–Milman theorem we may assume that the\nmaximiser is an extreme point of $\\mathcal F$.  A standard variation\nargument shows that any such extreme point satisfies $f(x)\\in\\{-4,4\\}$\nfor a.e.\\ $x$.  Hence\n\\[\nf=4\\bigl(\\chi_{A}-\\chi_{B}\\bigr)\\qquad\\text{a.e.},\n\\tag{1.1}\n\\]\nwith $A,B\\subset[1,4]$ measurable and disjoint.\n\nStep 2.  The two linear constraints for $\\boldsymbol{\\pm4}$–functions.  \nWrite\n\\[\n|A|=m,\\qquad \n\\alpha=\\int_{A}x\\,dx,\\qquad \n|B|,\\;\\int_{B}x\\,dx\\text{ analogous}.\n\\]\nThe equalities $\\int f=0$ and $\\int x f=0$ give\n\\[\n|A|=|B|=\\frac32, \\qquad \\int_{A}x\\,dx=\\int_{B}x\\,dx.\n\\tag{2.1}\n\\]\nFor such an $f$\n\\[\nI(f)=4\\Bigl(\\int_{A}w-\\int_{B}w\\Bigr).\n\\tag{2.2}\n\\]\n\nStep 3.  The dual linear programme.  \nFor $\\lambda=(\\lambda_{0},\\lambda_{1})\\in\\mathbb R^{2}$ define  \n\\[\ng_{\\lambda}(x)=w(x)-\\lambda_{0}-\\lambda_{1}x .\n\\]\nBecause the two moment integrals of $f$ vanish,\n\\[\nI(f)=\\int_{1}^{4} f(x)g_{\\lambda}(x)\\,dx .\n\\]\nSince $\\lvert f\\rvert\\le4$,\n\\[\nI(f)\\le4\\int_{1}^{4}\\lvert g_{\\lambda}(x)\\rvert\\,dx .\n\\tag{3.1}\n\\]\nTherefore\n\\[\nI_{\\max}\\le\n\\inf_{\\lambda\\in\\mathbb R^{2}}4\\int_{1}^{4}\\lvert g_{\\lambda}(x)\\rvert\\,dx .\n\\tag{3.2}\n\\]\n\nStep 4.  Pointwise maximisers for fixed $\\boldsymbol{\\lambda}$.  \nFor every $\\lambda$ set  \n\\[\nf_{\\lambda}(x)=4\\,\\operatorname{sgn}\\bigl(g_{\\lambda}(x)\\bigr)\n\\quad\n(\\operatorname{sgn}(0)\\text{ arbitrary in }[-1,1]).\n\\]\nThen $\\lvert f_{\\lambda}\\rvert\\le4$ and\n\\[\nI\\bigl(f_{\\lambda}\\bigr)=4\\int_{1}^{4}\\lvert g_{\\lambda}(x)\\rvert\\,dx .\n\\tag{4.1}\n\\]\nHence equality in (3.1) is achieved by $f_{\\lambda}$ if, in addition,\n$f_{\\lambda}$ meets the two constraints (2.1).  The task is to find\n$\\lambda$ such that\n\n(a) $g_{\\lambda}$ changes sign exactly as $+\\,-\\,+$ on $[1,4]$,  \n(b) the corresponding $f_{\\lambda}$ satisfies (2.1).\n\nStep 5.  Geometry of $\\boldsymbol{g_{\\lambda}}$.  \nBecause $w(x)=x^{-2}$ is strictly decreasing and convex on $[1,4]$, the\nequation\n\\[\nw(x)=\\lambda_{0}+\\lambda_{1}x\n\\tag{5.1}\n\\]\nhas at most two solutions.  To produce the desired sign pattern, we need\nexactly two distinct zeros\n\\[\n1<a<b<4 .\n\\tag{5.2}\n\\]\nConvexity implies\n\\[\ng_{\\lambda}(x)=\n\\begin{cases}\n>0,&1<x<a\\text{ or }b<x<4,\\\\\n<0,&a<x<b .\n\\end{cases}\n\\]\nConsequently\n\\[\nf_{\\lambda}(x)=\n\\begin{cases}\n\\;4,&1<x<a\\text{ or }b<x<4,\\\\[4pt]\n-4,&a<x<b .\n\\end{cases}\n\\tag{5.3}\n\\]\n\nStep 6.  Enforcing the two moment constraints.  \nFor $f_{\\lambda}$ of the form (5.3) the conditions (2.1) give  \n\n(i)  length condition  \n\\[\n(a-1)+(4-b)=\\frac32,\n\\tag{6.1}\n\\]\n\n(ii) first–moment condition  \n\\[\n\\int_{1}^{a}x\\,dx+\\int_{b}^{4}x\\,dx=\\int_{a}^{b}x\\,dx .\n\\tag{6.2}\n\\]\n\nEquation (6.1) yields  \n\\[\nb=a+\\frac32 .\n\\tag{6.3}\n\\]\nSubstituting (6.3) in (6.2) and solving gives the unique solution  \n\\[\na=\\frac74,\\qquad b=\\frac{13}{4}.\n\\tag{6.4}\n\\]\n\nStep 7.  The dual parameters $\\boldsymbol{\\lambda_{0},\\lambda_{1}}$.  \nBecause $a,b$ satisfy (5.1), the linear function $\\lambda_{0}+\\lambda_{1}x$\nmust pass through the points $\\bigl(a,w(a)\\bigr)$ and\n$\\bigl(b,w(b)\\bigr)$.  Hence\n\\[\n\\lambda_{1}\n=\\frac{w(b)-w(a)}{\\,b-a\\,}\n=\\frac{\\dfrac{16}{169}-\\dfrac{16}{49}}{\\dfrac32}\n=-\\frac{1280}{8281},\n\\tag{7.1}\n\\]\nand then\n\\[\n\\lambda_{0}\n=w(a)-\\lambda_{1}a\n=\\frac{16}{49}-\\frac{7}{4}\\lambda_{1}\n=\\frac{242\\,256}{405\\,769}\\approx0.59693.\n\\tag{7.2}\n\\]\nWith this pair $\\lambda=(\\lambda_{0},\\lambda_{1})$ the function\n$g_{\\lambda}$ has zeros exactly at $a,b$ and the sign pattern required\nin Step 5; therefore $f^{*}:=f_{\\lambda}$ is feasible.  Together with\n(4.1) and (3.2) it follows that $f^{*}$ attains the global maximum and\n\\[\nI_{\\max}=4\\int_{1}^{4}\\lvert g_{\\lambda}(x)\\rvert\\,dx .\n\\]\n\nStep 8.  Evaluation of the maximal value.  \nUsing (2.2) with $a,b$ from (6.4),\n\\[\n\\begin{aligned}\nI_{\\max}\n&=4\\left[\\left(\\int_{1}^{a}+\\int_{b}^{4}\\right)\\frac{dx}{x^{2}}\n          -\\int_{a}^{b}\\frac{dx}{x^{2}}\\right]\\\\\n&=4\\left[\\left(1-\\frac{1}{a}\\right)+\\left(\\frac{1}{b}-\\frac14\\right)\n         -\\left(\\frac{1}{a}-\\frac{1}{b}\\right)\\right]\\\\\n&=4\\left[\\frac34-\\frac{2}{a}+\\frac{2}{b}\\right]\n       =4\\left[\\frac34-\\frac{8}{7}+\\frac{8}{13}\\right]\n       =\\frac{81}{91}.\n\\end{aligned}\n\\]\n\nStep 9.  Uniqueness of the maximiser.  \nLet $f$ be any maximiser.  Using (3.1)–(3.2) with the $\\lambda$ from\n(7.1)–(7.2) we have\n\\[\n\\int_{1}^{4}f(x)g_{\\lambda}(x)\\,dx\n=4\\int_{1}^{4}\\lvert g_{\\lambda}(x)\\rvert\\,dx .\n\\]\nPointwise,\n\\[\n\\lvert f(x)g_{\\lambda}(x)\\rvert\\le4\\lvert g_{\\lambda}(x)\\rvert ,\n\\]\nso equality of the integrals forces $\\lvert f(x)\\rvert=4$ and\n$f(x)g_{\\lambda}(x)=4\\lvert g_{\\lambda}(x)\\rvert$ for a.e.\\ $x$.  Hence\n$f=4\\,\\operatorname{sgn}(g_{\\lambda})=f^{*}$ a.e.; the maximiser is\nunique up to sets of measure $0$.\n\nFinal answer  \n\\[\n\\boxed{\\;\nI_{\\max}=\\frac{81}{91},\\qquad\nf^{*}(x)=\n\\begin{cases}\n\\;4,&1\\le x\\le\\dfrac74\\ \\text{or}\\ \\dfrac{13}{4}\\le x\\le4,\\\\[6pt]\n-4,&\\dfrac74<x<\\dfrac{13}{4}.\n\\end{cases}\n\\;}\n\\]\n\n──────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.833920",
        "was_fixed": false,
        "difficulty_analysis": "1.  Two independent moment constraints (total mass and first moment) were added,\n    turning the feasible set from a simple “balanced‐mass” condition to a\n    codimension-2 slice in L∞.  \n2.  Solving the problem now requires functional-analytic duality (or linear-\n    programming duality) and the theory of best uniform approximation; elementary\n    rearrangement arguments from the original problem no longer suffice.  \n3.  The maximiser is obtained through the Chebyshev alternation theorem, which\n    demands identifying the unique minimax linear polynomial approximant to\n    1/x² on [1,4]—far more sophisticated than the piece-wise constant extremal\n    function of the original.  \n4.  Algebraic manipulation leads to a transcendental equation whose solution\n    γ = (32/5)^{1/3} is not rational; the final exact value of the maximum\n    involves fractional powers, unlike the simple logarithm in the original.  \n5.  Constructing the extremal function requires analysing sign patterns of the\n    error function and verifying that the alternation conditions are met.\n\nAll of these layers (duality theory, minimax approximation, alternation,\nnon-elementary constants) make this enhanced variant markedly harder than both\nthe original problem and the previous kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nf\\colon[1,4]\\longrightarrow\\mathbb R\n\\]\nbe Lebesgue–integrable and satisfy  \n\\[\n-4\\le f(x)\\le 4\\qquad(1\\le x\\le 4), \\qquad\n\\int_{1}^{4} f(x)\\,dx=0, \\qquad\n\\int_{1}^{4} x\\,f(x)\\,dx=0 .\n\\]\nDetermine  \n\\[\nI(f)=\\int_{1}^{4}\\frac{f(x)}{x^{2}}\\;dx ,\n\\]\nand describe (up to sets of measure $0$) every function $f$ that maximises $I(f)$.\n\n──────────────────────────────────",
      "solution": "Throughout put  \n\\[\nw(x)=\\frac{1}{x^{2}},\\qquad 1\\le x\\le4 .\n\\]\n\nStep 1.  Extreme points are $\\boldsymbol{\\pm4}$–valued.  \nLet  \n\\[\n\\mathcal F=\\Bigl\\{f\\in L^{\\infty}[1,4]\\;:\\;\n-4\\le f\\le 4,\\;\n\\int_{1}^{4}f=0,\\;\n\\int_{1}^{4}x\\,f=0\\Bigr\\}.\n\\]\nThe set $\\mathcal F$ is weak–$^{\\!*}$ compact and convex, while\n$I(f)=\\langle f,w\\rangle$ is weak–$^{\\!*}$ continuous, so a maximiser\nexists.  By the Bauer–Krein–Milman theorem we may assume that the\nmaximiser is an extreme point of $\\mathcal F$.  A standard variation\nargument shows that any such extreme point satisfies $f(x)\\in\\{-4,4\\}$\nfor a.e.\\ $x$.  Hence\n\\[\nf=4\\bigl(\\chi_{A}-\\chi_{B}\\bigr)\\qquad\\text{a.e.},\n\\tag{1.1}\n\\]\nwith $A,B\\subset[1,4]$ measurable and disjoint.\n\nStep 2.  The two linear constraints for $\\boldsymbol{\\pm4}$–functions.  \nWrite\n\\[\n|A|=m,\\qquad \n\\alpha=\\int_{A}x\\,dx,\\qquad \n|B|,\\;\\int_{B}x\\,dx\\text{ analogous}.\n\\]\nThe equalities $\\int f=0$ and $\\int x f=0$ give\n\\[\n|A|=|B|=\\frac32, \\qquad \\int_{A}x\\,dx=\\int_{B}x\\,dx.\n\\tag{2.1}\n\\]\nFor such an $f$\n\\[\nI(f)=4\\Bigl(\\int_{A}w-\\int_{B}w\\Bigr).\n\\tag{2.2}\n\\]\n\nStep 3.  The dual linear programme.  \nFor $\\lambda=(\\lambda_{0},\\lambda_{1})\\in\\mathbb R^{2}$ define  \n\\[\ng_{\\lambda}(x)=w(x)-\\lambda_{0}-\\lambda_{1}x .\n\\]\nBecause the two moment integrals of $f$ vanish,\n\\[\nI(f)=\\int_{1}^{4} f(x)g_{\\lambda}(x)\\,dx .\n\\]\nSince $\\lvert f\\rvert\\le4$,\n\\[\nI(f)\\le4\\int_{1}^{4}\\lvert g_{\\lambda}(x)\\rvert\\,dx .\n\\tag{3.1}\n\\]\nTherefore\n\\[\nI_{\\max}\\le\n\\inf_{\\lambda\\in\\mathbb R^{2}}4\\int_{1}^{4}\\lvert g_{\\lambda}(x)\\rvert\\,dx .\n\\tag{3.2}\n\\]\n\nStep 4.  Pointwise maximisers for fixed $\\boldsymbol{\\lambda}$.  \nFor every $\\lambda$ set  \n\\[\nf_{\\lambda}(x)=4\\,\\operatorname{sgn}\\bigl(g_{\\lambda}(x)\\bigr)\n\\quad\n(\\operatorname{sgn}(0)\\text{ arbitrary in }[-1,1]).\n\\]\nThen $\\lvert f_{\\lambda}\\rvert\\le4$ and\n\\[\nI\\bigl(f_{\\lambda}\\bigr)=4\\int_{1}^{4}\\lvert g_{\\lambda}(x)\\rvert\\,dx .\n\\tag{4.1}\n\\]\nHence equality in (3.1) is achieved by $f_{\\lambda}$ if, in addition,\n$f_{\\lambda}$ meets the two constraints (2.1).  The task is to find\n$\\lambda$ such that\n\n(a) $g_{\\lambda}$ changes sign exactly as $+\\,-\\,+$ on $[1,4]$,  \n(b) the corresponding $f_{\\lambda}$ satisfies (2.1).\n\nStep 5.  Geometry of $\\boldsymbol{g_{\\lambda}}$.  \nBecause $w(x)=x^{-2}$ is strictly decreasing and convex on $[1,4]$, the\nequation\n\\[\nw(x)=\\lambda_{0}+\\lambda_{1}x\n\\tag{5.1}\n\\]\nhas at most two solutions.  To produce the desired sign pattern, we need\nexactly two distinct zeros\n\\[\n1<a<b<4 .\n\\tag{5.2}\n\\]\nConvexity implies\n\\[\ng_{\\lambda}(x)=\n\\begin{cases}\n>0,&1<x<a\\text{ or }b<x<4,\\\\\n<0,&a<x<b .\n\\end{cases}\n\\]\nConsequently\n\\[\nf_{\\lambda}(x)=\n\\begin{cases}\n\\;4,&1<x<a\\text{ or }b<x<4,\\\\[4pt]\n-4,&a<x<b .\n\\end{cases}\n\\tag{5.3}\n\\]\n\nStep 6.  Enforcing the two moment constraints.  \nFor $f_{\\lambda}$ of the form (5.3) the conditions (2.1) give  \n\n(i)  length condition  \n\\[\n(a-1)+(4-b)=\\frac32,\n\\tag{6.1}\n\\]\n\n(ii) first–moment condition  \n\\[\n\\int_{1}^{a}x\\,dx+\\int_{b}^{4}x\\,dx=\\int_{a}^{b}x\\,dx .\n\\tag{6.2}\n\\]\n\nEquation (6.1) yields  \n\\[\nb=a+\\frac32 .\n\\tag{6.3}\n\\]\nSubstituting (6.3) in (6.2) and solving gives the unique solution  \n\\[\na=\\frac74,\\qquad b=\\frac{13}{4}.\n\\tag{6.4}\n\\]\n\nStep 7.  The dual parameters $\\boldsymbol{\\lambda_{0},\\lambda_{1}}$.  \nBecause $a,b$ satisfy (5.1), the linear function $\\lambda_{0}+\\lambda_{1}x$\nmust pass through the points $\\bigl(a,w(a)\\bigr)$ and\n$\\bigl(b,w(b)\\bigr)$.  Hence\n\\[\n\\lambda_{1}\n=\\frac{w(b)-w(a)}{\\,b-a\\,}\n=\\frac{\\dfrac{16}{169}-\\dfrac{16}{49}}{\\dfrac32}\n=-\\frac{1280}{8281},\n\\tag{7.1}\n\\]\nand then\n\\[\n\\lambda_{0}\n=w(a)-\\lambda_{1}a\n=\\frac{16}{49}-\\frac{7}{4}\\lambda_{1}\n=\\frac{242\\,256}{405\\,769}\\approx0.59693.\n\\tag{7.2}\n\\]\nWith this pair $\\lambda=(\\lambda_{0},\\lambda_{1})$ the function\n$g_{\\lambda}$ has zeros exactly at $a,b$ and the sign pattern required\nin Step 5; therefore $f^{*}:=f_{\\lambda}$ is feasible.  Together with\n(4.1) and (3.2) it follows that $f^{*}$ attains the global maximum and\n\\[\nI_{\\max}=4\\int_{1}^{4}\\lvert g_{\\lambda}(x)\\rvert\\,dx .\n\\]\n\nStep 8.  Evaluation of the maximal value.  \nUsing (2.2) with $a,b$ from (6.4),\n\\[\n\\begin{aligned}\nI_{\\max}\n&=4\\left[\\left(\\int_{1}^{a}+\\int_{b}^{4}\\right)\\frac{dx}{x^{2}}\n          -\\int_{a}^{b}\\frac{dx}{x^{2}}\\right]\\\\\n&=4\\left[\\left(1-\\frac{1}{a}\\right)+\\left(\\frac{1}{b}-\\frac14\\right)\n         -\\left(\\frac{1}{a}-\\frac{1}{b}\\right)\\right]\\\\\n&=4\\left[\\frac34-\\frac{2}{a}+\\frac{2}{b}\\right]\n       =4\\left[\\frac34-\\frac{8}{7}+\\frac{8}{13}\\right]\n       =\\frac{81}{91}.\n\\end{aligned}\n\\]\n\nStep 9.  Uniqueness of the maximiser.  \nLet $f$ be any maximiser.  Using (3.1)–(3.2) with the $\\lambda$ from\n(7.1)–(7.2) we have\n\\[\n\\int_{1}^{4}f(x)g_{\\lambda}(x)\\,dx\n=4\\int_{1}^{4}\\lvert g_{\\lambda}(x)\\rvert\\,dx .\n\\]\nPointwise,\n\\[\n\\lvert f(x)g_{\\lambda}(x)\\rvert\\le4\\lvert g_{\\lambda}(x)\\rvert ,\n\\]\nso equality of the integrals forces $\\lvert f(x)\\rvert=4$ and\n$f(x)g_{\\lambda}(x)=4\\lvert g_{\\lambda}(x)\\rvert$ for a.e.\\ $x$.  Hence\n$f=4\\,\\operatorname{sgn}(g_{\\lambda})=f^{*}$ a.e.; the maximiser is\nunique up to sets of measure $0$.\n\nFinal answer  \n\\[\n\\boxed{\\;\nI_{\\max}=\\frac{81}{91},\\qquad\nf^{*}(x)=\n\\begin{cases}\n\\;4,&1\\le x\\le\\dfrac74\\ \\text{or}\\ \\dfrac{13}{4}\\le x\\le4,\\\\[6pt]\n-4,&\\dfrac74<x<\\dfrac{13}{4}.\n\\end{cases}\n\\;}\n\\]\n\n──────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.636922",
        "was_fixed": false,
        "difficulty_analysis": "1.  Two independent moment constraints (total mass and first moment) were added,\n    turning the feasible set from a simple “balanced‐mass” condition to a\n    codimension-2 slice in L∞.  \n2.  Solving the problem now requires functional-analytic duality (or linear-\n    programming duality) and the theory of best uniform approximation; elementary\n    rearrangement arguments from the original problem no longer suffice.  \n3.  The maximiser is obtained through the Chebyshev alternation theorem, which\n    demands identifying the unique minimax linear polynomial approximant to\n    1/x² on [1,4]—far more sophisticated than the piece-wise constant extremal\n    function of the original.  \n4.  Algebraic manipulation leads to a transcendental equation whose solution\n    γ = (32/5)^{1/3} is not rational; the final exact value of the maximum\n    involves fractional powers, unlike the simple logarithm in the original.  \n5.  Constructing the extremal function requires analysing sign patterns of the\n    error function and verifying that the alternation conditions are met.\n\nAll of these layers (duality theory, minimax approximation, alternation,\nnon-elementary constants) make this enhanced variant markedly harder than both\nthe original problem and the previous kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}