{
  "index": "1962-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "2. Find every real-valued function \\( f \\) whose domain is an interval \\( I \\) (finite or infinite) having 0 as a left-hand endpoint, such that for every positive member \\( x \\) of \\( I \\) the average of \\( f \\) over the closed interval \\( [0, x] \\) is equal to the geometric mean of the numbers \\( f(0) \\) and \\( f(x) \\).",
  "solution": "Solution. Since the geometric mean is defined only for positive numbers, such a function must be everywhere positive; moreover, it must be integrable on \\( [0, x] \\) for every \\( x \\) in its domain.\n\nPut \\( f(0)=a \\) for convenience and let\n\\[\nF(x)=\\int_{0}^{x} f(t) d t\n\\]\n\nThen the condition of the problem is equivalent to\n\\[\na f(x)=\\left(\\frac{1}{x} F(x)\\right)^{2}\n\\]\nfor all positive \\( x \\in I \\).\nNow (1) shows that \\( F \\) is continuous, so for \\( x>0, f \\) is continuous by (2) and \\( F \\) is then differentiable by (1). Then (2) becomes\n\\[\na F^{\\prime}(x)=\\frac{1}{x^{2}} F(x)^{2} .\n\\]\n\nWe can integrate this differential equation by separating the variables, and we find\n\\[\nF(x)=\\frac{a x}{1-c x},\n\\]\nwhere \\( c \\) is the constant of integration. Therefore\n\\[\nf(x)=F^{\\prime}(x)=\\frac{a}{(1-c x)^{2}} .\n\\]\nfor \\( x>0 \\). We note that this formula remains valid for \\( \\boldsymbol{x}=0 \\). If \\( \\boldsymbol{c}>0 \\), then \\( f \\) is not integrable on \\( [0,1 / c] \\), hence we see that every solution is given by\n\\[\nf(x)=\\frac{a}{(1-c x)^{2}}\\left\\{\\begin{array}{ll}\n\\text { for } 0 \\leq x<\\frac{1}{c}, & \\text { if } c>0 \\\\\n\\text { for } 0 \\leq x<\\infty, & \\text { if } c \\leq 0\n\\end{array}\\right.\n\\]\nwhere \\( a>0 \\).\nIt is readily checked that all functions defined by (3) are solutions of the problem.\n\nRemark. If \\( \\approx<0 \\). then (3) gives solutions of (2) that are not solutions of the problem. If \\( a=0 \\), then (2) has many more solutions, namely, any function that is zero almost everywhere on \\( \\mid 0, \\infty) \\).",
  "vars": [
    "f",
    "x",
    "t",
    "F"
  ],
  "params": [
    "a",
    "c",
    "I"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "realfunc",
        "x": "variable",
        "t": "dummyvar",
        "F": "integral",
        "a": "constanta",
        "c": "constantc",
        "I": "intervali"
      },
      "question": "2. Find every real-valued function \\( realfunc \\) whose domain is an interval \\( intervali \\) (finite or infinite) having 0 as a left-hand endpoint, such that for every positive member \\( variable \\) of \\( intervali \\) the average of \\( realfunc \\) over the closed interval \\( [0, variable] \\) is equal to the geometric mean of the numbers \\( realfunc(0) \\) and \\( realfunc(variable) \\).",
      "solution": "Solution. Since the geometric mean is defined only for positive numbers, such a function must be everywhere positive; moreover, it must be integrable on \\( [0, variable] \\) for every \\( variable \\) in its domain.\n\nPut \\( realfunc(0)=constanta \\) for convenience and let\n\\[\nintegral(variable)=\\int_{0}^{variable} realfunc(dummyvar) \\, d dummyvar\n\\]\nThen the condition of the problem is equivalent to\n\\[\nconstanta \\, realfunc(variable)=\\left(\\frac{1}{variable} integral(variable)\\right)^{2}\n\\]\nfor all positive \\( variable \\in intervali \\).\nNow (1) shows that \\( integral \\) is continuous, so for \\( variable>0, realfunc \\) is continuous by (2) and \\( integral \\) is then differentiable by (1). Then (2) becomes\n\\[\nconstanta \\, integral^{\\prime}(variable)=\\frac{1}{variable^{2}} integral(variable)^{2} .\n\\]\nWe can integrate this differential equation by separating the variables, and we find\n\\[\nintegral(variable)=\\frac{constanta \\, variable}{1-constantc \\, variable},\n\\]\nwhere \\( constantc \\) is the constant of integration. Therefore\n\\[\nrealfunc(variable)=integral^{\\prime}(variable)=\\frac{constanta}{(1-constantc \\, variable)^{2}} .\n\\]\nfor \\( variable>0 \\). We note that this formula remains valid for \\( \\boldsymbol{variable}=0 \\). If \\( \\boldsymbol{constantc}>0 \\), then \\( realfunc \\) is not integrable on \\( [0,1 / constantc] \\); hence we see that every solution is given by\n\\[\nrealfunc(variable)=\\frac{constanta}{(1-constantc \\, variable)^{2}}\\left\\{\\begin{array}{ll}\n\\text { for } 0 \\leq variable<\\frac{1}{constantc}, & \\text { if } constantc>0 \\\\\n\\text { for } 0 \\leq variable<\\infty, & \\text { if } constantc \\leq 0\n\\end{array}\\right.\n\\]\nwhere \\( constanta>0 \\).\nIt is readily checked that all functions defined by (3) are solutions of the problem.\n\nRemark. If \\( constantc<0 \\), then (3) gives solutions of (2) that are not solutions of the problem. If \\( constanta=0 \\), then (2) has many more solutions, namely, any function that is zero almost everywhere on \\( \\mid 0, \\infty) \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "marigold",
        "x": "constellation",
        "t": "lemonade",
        "F": "thunderbolt",
        "a": "porcelain",
        "c": "willowtree",
        "I": "sunflower"
      },
      "question": "2. Find every real-valued function \\( marigold \\) whose domain is an interval \\( sunflower \\) (finite or infinite) having 0 as a left-hand endpoint, such that for every positive member \\( constellation \\) of \\( sunflower \\) the average of \\( marigold \\) over the closed interval \\( [0, constellation] \\) is equal to the geometric mean of the numbers \\( marigold(0) \\) and \\( marigold(constellation) \\).",
      "solution": "Solution. Since the geometric mean is defined only for positive numbers, such a function must be everywhere positive; moreover, it must be integrable on \\( [0, constellation] \\) for every \\( constellation \\) in its domain.\n\nPut \\( marigold(0)=porcelain \\) for convenience and let\n\\[\nthunderbolt(constellation)=\\int_{0}^{constellation} marigold(lemonade) d lemonade\n\\]\n\nThen the condition of the problem is equivalent to\n\\[\nporcelain \\, marigold(constellation)=\\left(\\frac{1}{constellation} thunderbolt(constellation)\\right)^{2}\n\\]\nfor all positive \\( constellation \\in sunflower \\).\nNow (1) shows that \\( thunderbolt \\) is continuous, so for \\( constellation>0, marigold \\) is continuous by (2) and \\( thunderbolt \\) is then differentiable by (1). Then (2) becomes\n\\[\nporcelain \\, thunderbolt^{\\prime}(constellation)=\\frac{1}{constellation^{2}} thunderbolt(constellation)^{2} .\n\\]\n\nWe can integrate this differential equation by separating the variables, and we find\n\\[\nthunderbolt(constellation)=\\frac{porcelain \\, constellation}{1-willowtree \\, constellation},\n\\]\nwhere \\( willowtree \\) is the constant of integration. Therefore\n\\[\nmarigold(constellation)=thunderbolt^{\\prime}(constellation)=\\frac{porcelain}{(1-willowtree \\, constellation)^{2}} .\n\\]\nfor \\( constellation>0 \\). We note that this formula remains valid for \\( \\boldsymbol{constellation}=0 \\). If \\( \\boldsymbol{willowtree}>0 \\), then \\( marigold \\) is not integrable on \\( [0,1 / willowtree] \\), hence we see that every solution is given by\n\\[\nmarigold(constellation)=\\frac{porcelain}{(1-willowtree \\, constellation)^{2}}\\left\\{\\begin{array}{ll}\n\\text { for } 0 \\leq constellation<\\frac{1}{willowtree}, & \\text { if } willowtree>0 \\\\\n\\text { for } 0 \\leq constellation<\\infty, & \\text { if } willowtree \\leq 0\n\\end{array}\\right.\n\\]\nwhere \\( porcelain>0 \\).\nIt is readily checked that all functions defined by (3) are solutions of the problem.\n\nRemark. If \\( \\approx<0 \\). then (3) gives solutions of (2) that are not solutions of the problem. If \\( porcelain=0 \\), then (2) has many more solutions, namely, any function that is zero almost everywhere on \\( \\mid 0, \\infty) \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "constant",
        "x": "steadfast",
        "t": "permanent",
        "F": "derivative",
        "a": "negative",
        "c": "variable",
        "I": "pointset"
      },
      "question": "2. Find every real-valued function \\( constant \\) whose domain is an interval \\( pointset \\) (finite or infinite) having 0 as a left-hand endpoint, such that for every positive member \\( steadfast \\) of \\( pointset \\) the average of \\( constant \\) over the closed interval \\( [0, steadfast] \\) is equal to the geometric mean of the numbers \\( constant(0) \\) and \\( constant(steadfast) \\).",
      "solution": "Solution. Since the geometric mean is defined only for positive numbers, such a function must be everywhere positive; moreover, it must be integrable on \\( [0, steadfast] \\) for every \\( steadfast \\) in its domain.\n\nPut \\( constant(0)=negative \\) for convenience and let\n\\[\nderivative(steadfast)=\\int_{0}^{steadfast} constant(permanent) d permanent\n\\]\nThen the condition of the problem is equivalent to\n\\[\nnegative\\; constant(steadfast)=\\left(\\frac{1}{steadfast} derivative(steadfast)\\right)^{2}\n\\]\nfor all positive \\( steadfast \\in pointset \\).\nNow (1) shows that \\( derivative \\) is continuous, so for \\( steadfast>0, constant \\) is continuous by (2) and \\( derivative \\) is then differentiable by (1). Then (2) becomes\n\\[\nnegative\\; derivative^{\\prime}(steadfast)=\\frac{1}{steadfast^{2}} derivative(steadfast)^{2} .\n\\]\nWe can integrate this differential equation by separating the variables, and we find\n\\[\nderivative(steadfast)=\\frac{negative\\; steadfast}{1-variable\\; steadfast},\n\\]\nwhere \\( variable \\) is the constant of integration. Therefore\n\\[\nconstant(steadfast)=derivative^{\\prime}(steadfast)=\\frac{negative}{(1-variable\\; steadfast)^{2}} .\n\\]\nfor \\( steadfast>0 \\). We note that this formula remains valid for \\( \\boldsymbol{steadfast}=0 \\). If \\( \\boldsymbol{variable}>0 \\), then \\( constant \\) is not integrable on \\( [0,1 / variable] \\), hence we see that every solution is given by\n\\[\nconstant(steadfast)=\\frac{negative}{(1-variable\\; steadfast)^{2}}\\left\\{\\begin{array}{ll}\n\\text { for } 0 \\leq steadfast<\\frac{1}{variable}, & \\text { if } variable>0 \\\\\n\\text { for } 0 \\leq steadfast<\\infty, & \\text { if } variable \\leq 0\n\\end{array}\\right.\n\\]\nwhere \\( negative>0 \\).\nIt is readily checked that all functions defined by (3) are solutions of the problem.\n\nRemark. If \\( \\approx<0 \\). then (3) gives solutions of (2) that are not solutions of the problem. If \\( negative=0 \\), then (2) has many more solutions, namely, any function that is zero almost everywhere on \\( \\mid 0, \\infty) \\)."
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "x": "hjgrksla",
        "t": "mnbvcxse",
        "F": "plmoknji",
        "a": "zxcvbnma",
        "c": "qwertyui",
        "I": "asdfghjk"
      },
      "question": "2. Find every real-valued function \\( qzxwvtnp \\) whose domain is an interval \\( asdfghjk \\) (finite or infinite) having 0 as a left-hand endpoint, such that for every positive member \\( hjgrksla \\) of \\( asdfghjk \\) the average of \\( qzxwvtnp \\) over the closed interval \\( [0, hjgrksla] \\) is equal to the geometric mean of the numbers \\( qzxwvtnp(0) \\) and \\( qzxwvtnp(hjgrksla) \\).",
      "solution": "Solution. Since the geometric mean is defined only for positive numbers, such a function must be everywhere positive; moreover, it must be integrable on \\( [0, hjgrksla] \\) for every \\( hjgrksla \\) in its domain.\n\nPut \\( qzxwvtnp(0)=zxcvbnma \\) for convenience and let\n\\[\nplmoknji(hjgrksla)=\\int_{0}^{hjgrksla} qzxwvtnp(mnbvcxse) d mnbvcxse\n\\]\n\nThen the condition of the problem is equivalent to\n\\[\nzxcvbnma \\, qzxwvtnp(hjgrksla)=\\left(\\frac{1}{hjgrksla} plmoknji(hjgrksla)\\right)^{2}\n\\]\nfor all positive \\( hjgrksla \\in asdfghjk \\).\nNow (1) shows that \\( plmoknji \\) is continuous, so for \\( hjgrksla>0, qzxwvtnp \\) is continuous by (2) and \\( plmoknji \\) is then differentiable by (1). Then (2) becomes\n\\[\nzxcvbnma \\, plmoknji^{\\prime}(hjgrksla)=\\frac{1}{hjgrksla^{2}} plmoknji(hjgrksla)^{2} .\n\\]\n\nWe can integrate this differential equation by separating the variables, and we find\n\\[\nplmoknji(hjgrksla)=\\frac{zxcvbnma \\, hjgrksla}{1-qwertyui \\, hjgrksla},\n\\]\nwhere \\( qwertyui \\) is the constant of integration. Therefore\n\\[\nqzxwvtnp(hjgrksla)=plmoknji^{\\prime}(hjgrksla)=\\frac{zxcvbnma}{(1-qwertyui \\, hjgrksla)^{2}} .\n\\]\nfor \\( hjgrksla>0 \\). We note that this formula remains valid for \\( \\boldsymbol{hjgrksla}=0 \\). If \\( \\boldsymbol{qwertyui}>0 \\), then \\( qzxwvtnp \\) is not integrable on \\( [0,1 / qwertyui] \\), hence we see that every solution is given by\n\\[\nqzxwvtnp(hjgrksla)=\\frac{zxcvbnma}{(1-qwertyui \\, hjgrksla)^{2}}\\left\\{\\begin{array}{ll}\n\\text { for } 0 \\leq hjgrksla<\\frac{1}{qwertyui}, & \\text { if } qwertyui>0 \\\\\n\\text { for } 0 \\leq hjgrksla<\\infty, & \\text { if } qwertyui \\leq 0\n\\end{array}\\right.\n\\]\nwhere \\( zxcvbnma>0 \\).\nIt is readily checked that all functions defined by (3) are solutions of the problem.\n\nRemark. If \\( \\approx<0 \\). then (3) gives solutions of (2) that are not solutions of the problem. If \\( zxcvbnma=0 \\), then (2) has many more solutions, namely, any function that is zero almost everywhere on \\( \\mid 0, \\infty) \\)."
    },
    "kernel_variant": {
      "question": "Fix a positive real parameter α > 0 (α need not be an integer).  \nLet I be an interval whose left–hand end-point is 1, i.e.  \n\n  I = [1, b) with 1 < b ≤ ∞ .  \n\nDetermine every continuously differentiable and strictly positive function  \n\n  f : I → (0, ∞)\n\nthat fulfils, for every x ∈ I with x > 1, the weighted–mean identity  \n\n           α                                    \n     ——————————— ∫¹ˣ t^{α−1} f(t)\\,dt = ( f(1)^{α}\\,f(x) )^{1/(α+1)} . (★)  \n     x^{α} − 1                                    \n\n(Geometrically: the α–dimensional radial average of f on the shell 1 ≤ t ≤ x\nequals the geometric mean in which f(1) appears α times and f(x) once.)\n\nDescribe explicitly all such functions and, for each of them, the maximal\nsub-interval of I on which it is defined.",
      "solution": "Put  \n\n  c := f(1) > 0,  F(x) := ∫₁ˣ t^{α−1} f(t)\\,dt  (x > 1). (1)\n\nBecause f is C¹ and positive, F is C¹ and strictly increasing with  \n\n  F′(x) = x^{α−1} f(x).               (2)\n\n--------------------------------------------------------------------\n1.  Rewriting the functional identity  \n--------------------------------------------------------------------\nEquation (★) rewrites via (1) as  \n\n  α F(x) = (x^{α} − 1) [c^{α} f(x)]^{1/(α+1)}.    (3)\n\n--------------------------------------------------------------------\n2.  Eliminating f(x)  \n--------------------------------------------------------------------\nRaise (3) to the power α+1 and use (2):\n\nα^{α+1} F(x)^{α+1}\n = (x^{α} − 1)^{α+1} c^{α} f(x)\n = (x^{α} − 1)^{α+1} c^{α} · F′(x) / x^{α-1}.   (4)\n\nHence  \n\n  F′(x) = K x^{α-1} (x^{α} − 1)^{-(α+1)} F(x)^{α+1}, (5)\n\nwith K := α^{α+1}/c^{α} > 0.\n\n--------------------------------------------------------------------\n3.  Separation of variables  \n--------------------------------------------------------------------\nBecause F > 0 we may divide by F^{α+1}:\n\n  F^{-(α+1)} dF = K x^{α-1}(x^{α} − 1)^{-(α+1)} dx. (6)\n\n--------------------------------------------------------------------\n4.  Integrating  \n--------------------------------------------------------------------\nLeft side: ∫ F^{-(α+1)}dF = −F^{−α}/α.        (7)\n\nRight side: let u = x^{α}, so du = αx^{α−1}dx:\n\n∫ x^{α-1}(x^{α} − 1)^{-(α+1)}dx\n= (1/α)∫_{1}^{x^{α}} (u − 1)^{-(α+1)}du\n= −(x^{α} − 1)^{-α}/α².            (8)\n\nInsert (7)–(8) into (6):\n\n−F^{−α}/α = −(K/α²)(x^{α} − 1)^{-α} + C.     (9)\n\n--------------------------------------------------------------------\n5.  Solving for F  \n--------------------------------------------------------------------\nMultiply by −α and write D := −αC:\n\n  F(x)^{−α} = (K/α)(x^{α} − 1)^{−α} + D.    (10)\n\nSince K/α = α^{α}/c^{α}, (10) becomes\n\n  F(x)^{−α} = (α^{α}/c^{α})[(x^{α} − 1)^{−α} + λ], (11)\n\nwhere we set  \n\n  λ := D c^{α}/α^{α}  (λ ∈ ℝ).        (12)\n\nTaking the −α-th power yields  \n\n  F(x) = (c/α)(x^{α} − 1)[1 + λ(x^{α} − 1)^{α}]^{−1/α}. (13)\n\n--------------------------------------------------------------------\n6.  Extracting f  \n--------------------------------------------------------------------\nPut s := x^{α} − 1 (so ds/dx = αx^{α−1}).  From (13)\n\nF(x) = (c/α) s (1 + λ s^{α})^{−1/α},\n\nhence  \n\n  F′(x)= c x^{α−1}(1 + λ s^{α})^{−1/α−1}.      (14)\n\nComparing with (2) furnishes  \n\n  f(x) = c [1 + λ(x^{α} − 1)^{α}]^{−1/α−1}.    (15)\n\n--------------------------------------------------------------------\n7.  Positivity and maximal domain  \n--------------------------------------------------------------------\nThe bracket in (15) must stay positive.\n\n•  If λ ≥ 0 it never vanishes, so f extends to the whole semi-infinite\n   interval [1, ∞).\n\n•  If λ < 0 we need  \n\n  1 + λ(x^{α} − 1)^{α} > 0  \n  ⇔ (x^{α} − 1)^{α} < |λ|^{−1}  \n  ⇔ x^{α} − 1 < |λ|^{−1/α}  \n  ⇔ x < (1 + |λ|^{−1/α})^{1/α}.          (16)\n\nThus  \n\n  b = ∞ if λ ≥ 0,  and  b = (1 + |λ|^{−1/α})^{1/α} if λ < 0. (17)\n\nBecause the denominator in (15) vanishes at x = b when λ < 0, the right end-point is not included: the solution lives on [1, b).\n\n--------------------------------------------------------------------\n8.  Verification  \n--------------------------------------------------------------------\nSubstituting (15) into (★) (use s = x^{α} − 1 and ds/dx = αx^{α−1}) gives an\nidentity, so every pair (c, λ) satisfying (17) indeed produces a solution.  \nConversely all solutions have been obtained, because any C¹ positive solution\nnecessarily leads to the differential equation (5) whose general solution is\n(15).\n\n--------------------------------------------------------------------\nFinal classification  \n--------------------------------------------------------------------\nAll continuously differentiable positive solutions of (★) are  \n\n  f_{c,λ}(x) =                         c  \n     ———————————————       x ∈ [1,b_{λ}),    (18)  \n    (1 + λ (x^{α} − 1)^{α})^{1 + 1/α}\n\nwhere  \n\n •  c = f(1) > 0 is arbitrary,  \n •  λ ∈ ℝ is arbitrary,  \n •  b_{λ} is given by (17).  \n\nFor λ ≥ 0 we have b_{λ}=∞; for λ < 0 the maximal interval ends at  \nb_{λ}=(1+|λ|^{−1/α})^{1/α}.  \nNo other function satisfies the weighted–mean identity (★).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.534244",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional weighting:  \n The integral average now involves the weight t^{α−1}, mimicking an α-dimensional volume element.  This forces the solver to cope with a non-uniform kernel and to recognise the substitution u = t^{α}.  \n\n2. Non-integer parameter α and fractional exponents:  \n α is an arbitrary positive real, so the solution must work equally for irrational values.  Handling the power (x^{α}−1)^{−α} and the exponent −1/α requires comfort with general real exponents, not just integers.\n\n3. Non-linear first-order ODE with variable coefficients:  \n Eliminating the integral leads to the nonlinear equation (5) whose right-hand side depends simultaneously on x and F in a non-trivial way.  Solving it demands separation of variables, a change of variables, and careful tracking of constants.\n\n4. Domain analysis:  \n Because the denominator 1+λ (x^{α}−1)^{α} may vanish, one must analyse where the solution remains positive and hence admissible.  This extra case-work is absent in the original problem.\n\n5. Parameter family of solutions:  \n The final answer involves two independent real parameters (f(1) and λ) and a sharp description of the maximal interval of validity, which is subtler than in the original setting.\n\nAll these layers go beyond the original cubic-root identity on a flat interval, making the enhanced variant significantly harder and requiring a deeper, multi-step argument."
      }
    },
    "original_kernel_variant": {
      "question": "Fix a positive real parameter α > 0 (α need not be an integer).  \nLet I be an interval whose left–hand end-point is 1, i.e.  \n\n  I = [1, b) with 1 < b ≤ ∞ .  \n\nDetermine every continuously differentiable and strictly positive function  \n\n  f : I → (0, ∞)\n\nthat fulfils, for every x ∈ I with x > 1, the weighted–mean identity  \n\n           α                                    \n     ——————————— ∫¹ˣ t^{α−1} f(t)\\,dt = ( f(1)^{α}\\,f(x) )^{1/(α+1)} . (★)  \n     x^{α} − 1                                    \n\n(Geometrically: the α–dimensional radial average of f on the shell 1 ≤ t ≤ x\nequals the geometric mean in which f(1) appears α times and f(x) once.)\n\nDescribe explicitly all such functions and, for each of them, the maximal\nsub-interval of I on which it is defined.",
      "solution": "Put  \n\n  c := f(1) > 0,  F(x) := ∫₁ˣ t^{α−1} f(t)\\,dt  (x > 1). (1)\n\nBecause f is C¹ and positive, F is C¹ and strictly increasing with  \n\n  F′(x) = x^{α−1} f(x).               (2)\n\n--------------------------------------------------------------------\n1.  Rewriting the functional identity  \n--------------------------------------------------------------------\nEquation (★) rewrites via (1) as  \n\n  α F(x) = (x^{α} − 1) [c^{α} f(x)]^{1/(α+1)}.    (3)\n\n--------------------------------------------------------------------\n2.  Eliminating f(x)  \n--------------------------------------------------------------------\nRaise (3) to the power α+1 and use (2):\n\nα^{α+1} F(x)^{α+1}\n = (x^{α} − 1)^{α+1} c^{α} f(x)\n = (x^{α} − 1)^{α+1} c^{α} · F′(x) / x^{α-1}.   (4)\n\nHence  \n\n  F′(x) = K x^{α-1} (x^{α} − 1)^{-(α+1)} F(x)^{α+1}, (5)\n\nwith K := α^{α+1}/c^{α} > 0.\n\n--------------------------------------------------------------------\n3.  Separation of variables  \n--------------------------------------------------------------------\nBecause F > 0 we may divide by F^{α+1}:\n\n  F^{-(α+1)} dF = K x^{α-1}(x^{α} − 1)^{-(α+1)} dx. (6)\n\n--------------------------------------------------------------------\n4.  Integrating  \n--------------------------------------------------------------------\nLeft side: ∫ F^{-(α+1)}dF = −F^{−α}/α.        (7)\n\nRight side: let u = x^{α}, so du = αx^{α−1}dx:\n\n∫ x^{α-1}(x^{α} − 1)^{-(α+1)}dx\n= (1/α)∫_{1}^{x^{α}} (u − 1)^{-(α+1)}du\n= −(x^{α} − 1)^{-α}/α².            (8)\n\nInsert (7)–(8) into (6):\n\n−F^{−α}/α = −(K/α²)(x^{α} − 1)^{-α} + C.     (9)\n\n--------------------------------------------------------------------\n5.  Solving for F  \n--------------------------------------------------------------------\nMultiply by −α and write D := −αC:\n\n  F(x)^{−α} = (K/α)(x^{α} − 1)^{−α} + D.    (10)\n\nSince K/α = α^{α}/c^{α}, (10) becomes\n\n  F(x)^{−α} = (α^{α}/c^{α})[(x^{α} − 1)^{−α} + λ], (11)\n\nwhere we set  \n\n  λ := D c^{α}/α^{α}  (λ ∈ ℝ).        (12)\n\nTaking the −α-th power yields  \n\n  F(x) = (c/α)(x^{α} − 1)[1 + λ(x^{α} − 1)^{α}]^{−1/α}. (13)\n\n--------------------------------------------------------------------\n6.  Extracting f  \n--------------------------------------------------------------------\nPut s := x^{α} − 1 (so ds/dx = αx^{α−1}).  From (13)\n\nF(x) = (c/α) s (1 + λ s^{α})^{−1/α},\n\nhence  \n\n  F′(x)= c x^{α−1}(1 + λ s^{α})^{−1/α−1}.      (14)\n\nComparing with (2) furnishes  \n\n  f(x) = c [1 + λ(x^{α} − 1)^{α}]^{−1/α−1}.    (15)\n\n--------------------------------------------------------------------\n7.  Positivity and maximal domain  \n--------------------------------------------------------------------\nThe bracket in (15) must stay positive.\n\n•  If λ ≥ 0 it never vanishes, so f extends to the whole semi-infinite\n   interval [1, ∞).\n\n•  If λ < 0 we need  \n\n  1 + λ(x^{α} − 1)^{α} > 0  \n  ⇔ (x^{α} − 1)^{α} < |λ|^{−1}  \n  ⇔ x^{α} − 1 < |λ|^{−1/α}  \n  ⇔ x < (1 + |λ|^{−1/α})^{1/α}.          (16)\n\nThus  \n\n  b = ∞ if λ ≥ 0,  and  b = (1 + |λ|^{−1/α})^{1/α} if λ < 0. (17)\n\nBecause the denominator in (15) vanishes at x = b when λ < 0, the right end-point is not included: the solution lives on [1, b).\n\n--------------------------------------------------------------------\n8.  Verification  \n--------------------------------------------------------------------\nSubstituting (15) into (★) (use s = x^{α} − 1 and ds/dx = αx^{α−1}) gives an\nidentity, so every pair (c, λ) satisfying (17) indeed produces a solution.  \nConversely all solutions have been obtained, because any C¹ positive solution\nnecessarily leads to the differential equation (5) whose general solution is\n(15).\n\n--------------------------------------------------------------------\nFinal classification  \n--------------------------------------------------------------------\nAll continuously differentiable positive solutions of (★) are  \n\n  f_{c,λ}(x) =                         c  \n     ———————————————       x ∈ [1,b_{λ}),    (18)  \n    (1 + λ (x^{α} − 1)^{α})^{1 + 1/α}\n\nwhere  \n\n •  c = f(1) > 0 is arbitrary,  \n •  λ ∈ ℝ is arbitrary,  \n •  b_{λ} is given by (17).  \n\nFor λ ≥ 0 we have b_{λ}=∞; for λ < 0 the maximal interval ends at  \nb_{λ}=(1+|λ|^{−1/α})^{1/α}.  \nNo other function satisfies the weighted–mean identity (★).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.445143",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional weighting:  \n The integral average now involves the weight t^{α−1}, mimicking an α-dimensional volume element.  This forces the solver to cope with a non-uniform kernel and to recognise the substitution u = t^{α}.  \n\n2. Non-integer parameter α and fractional exponents:  \n α is an arbitrary positive real, so the solution must work equally for irrational values.  Handling the power (x^{α}−1)^{−α} and the exponent −1/α requires comfort with general real exponents, not just integers.\n\n3. Non-linear first-order ODE with variable coefficients:  \n Eliminating the integral leads to the nonlinear equation (5) whose right-hand side depends simultaneously on x and F in a non-trivial way.  Solving it demands separation of variables, a change of variables, and careful tracking of constants.\n\n4. Domain analysis:  \n Because the denominator 1+λ (x^{α}−1)^{α} may vanish, one must analyse where the solution remains positive and hence admissible.  This extra case-work is absent in the original problem.\n\n5. Parameter family of solutions:  \n The final answer involves two independent real parameters (f(1) and λ) and a sharp description of the maximal interval of validity, which is subtler than in the original setting.\n\nAll these layers go beyond the original cubic-root identity on a flat interval, making the enhanced variant significantly harder and requiring a deeper, multi-step argument."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}