{
  "index": "2006-B-5",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "For each continuous function $f: [0,1] \\to \\mathbb{R}$, let $I(f) =\n\\int_0^1 x^2 f(x)\\,dx$ and $J(x) = \\int_0^1 x \\left(f(x)\\right)^2\\,dx$.\nFind the maximum value of $I(f) - J(f)$ over all such functions $f$.",
  "solution": "The answer is $1/16$. We have\n\\begin{align*}\n&\\int_0^1 x^2 f (x)\\,dx - \\int_0^1 x f(x)^2\\,dx \\\\\n&= \\int_0^1 (x^3/4 - x\n( f(x)-x/2)^2)\\,dx \\\\\n&\\leq \\int_0^1 x^3/4\\,dx = 1/16,\n\\end{align*}\nwith equality when $f(x) = x/2$.",
  "vars": [
    "f",
    "x"
  ],
  "params": [
    "I",
    "J"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "function",
        "x": "variable",
        "I": "integralone",
        "J": "integraltwo"
      },
      "question": "For each continuous function $function: [0,1] \\to \\mathbb{R}$, let $integralone(function) =\n\\int_0^1 variable^2 function(variable)\\,d variable$ and $integraltwo(variable) = \\int_0^1 variable \\left(function(variable)\\right)^2\\,d variable$.\nFind the maximum value of $integralone(function) - integraltwo(function)$ over all such functions $function$.",
      "solution": "The answer is $1/16$. We have\n\\begin{align*}\n&\\int_0^1 variable^2 function (variable)\\,d variable - \\int_0^1 variable function(variable)^2\\,d variable \\\\\n&= \\int_0^1 (variable^3/4 - variable\n( function(variable)-variable/2)^2)\\,d variable \\\\\n&\\leq \\int_0^1 variable^3/4\\,d variable = 1/16,\n\\end{align*}\nwith equality when $function(variable) = variable/2$."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "meadowlark",
        "x": "buttercup",
        "I": "chestnut",
        "J": "kingfisher"
      },
      "question": "For each continuous function $meadowlark: [0,1] \\to \\mathbb{R}$, let $chestnut(meadowlark) =\n\\int_0^1 buttercup^2 meadowlark(buttercup)\\,d buttercup$ and $kingfisher(buttercup) = \\int_0^1 buttercup \\left(meadowlark(buttercup)\\right)^2\\,d buttercup$.\nFind the maximum value of $chestnut(meadowlark) - kingfisher(meadowlark)$ over all such functions $meadowlark$.",
      "solution": "The answer is $1/16$. We have\n\\begin{align*}\n&\\int_0^1 buttercup^2 meadowlark (buttercup)\\,d buttercup - \\int_0^1 buttercup meadowlark(buttercup)^2\\,d buttercup \\\\\n&= \\int_0^1 (buttercup^3/4 - buttercup\n( meadowlark(buttercup)-buttercup/2)^2)\\,d buttercup \\\\\n&\\leq \\int_0^1 buttercup^3/4\\,d buttercup = 1/16,\n\\end{align*}\nwith equality when $meadowlark(buttercup) = buttercup/2$.}"
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "discretefn",
        "x": "constant",
        "I": "derivative",
        "J": "difference"
      },
      "question": "For each continuous function $discretefn: [0,1] \\to \\mathbb{R}$, let $derivative(discretefn) =\n\\int_0^1 constant^2 discretefn(constant)\\,dconstant$ and $difference(constant) = \\int_0^1 constant \\left(discretefn(constant)\\right)^2\\,dconstant$.\nFind the maximum value of $derivative(discretefn) - difference(discretefn)$ over all such functions $discretefn$.",
      "solution": "The answer is $1/16$. We have\n\\begin{align*}\n&\\int_0^1 constant^2 discretefn (constant)\\,dconstant - \\int_0^1 constant discretefn(constant)^2\\,dconstant \\\\\n&= \\int_0^1 (constant^3/4 - constant\n( discretefn(constant)-constant/2)^2)\\,dconstant \\\\\n&\\leq \\int_0^1 constant^3/4\\,dconstant = 1/16,\n\\end{align*}\nwith equality when $discretefn(constant) = constant/2$. "
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "x": "hjgrksla",
        "I": "mnbvcxzq",
        "J": "plokmjin"
      },
      "question": "For each continuous function $qzxwvtnp: [0,1] \\to \\mathbb{R}$, let $mnbvcxzq(qzxwvtnp) =\n\\int_0^1 hjgrksla^2 qzxwvtnp(hjgrksla)\\,dhjgrksla$ and $plokmjin(hjgrksla) = \\int_0^1 hjgrksla \\left(qzxwvtnp(hjgrksla)\\right)^2\\,dhjgrksla$.\nFind the maximum value of $mnbvcxzq(qzxwvtnp) - plokmjin(qzxwvtnp)$ over all such functions $qzxwvtnp$.",
      "solution": "The answer is $1/16$. We have\n\\begin{align*}\n&\\int_0^1 hjgrksla^2 qzxwvtnp (hjgrksla)\\,dhjgrksla - \\int_0^1 hjgrksla qzxwvtnp(hjgrksla)^2\\,dhjgrksla \\\\\n&= \\int_0^1 (hjgrksla^3/4 - hjgrksla\n( qzxwvtnp(hjgrksla)-hjgrksla/2)^2)\\,dhjgrksla \\\\\n&\\leq \\int_0^1 hjgrksla^3/4\\,dhjgrksla = 1/16,\n\\end{align*}\nwith equality when $qzxwvtnp(hjgrksla) = hjgrksla/2$.}"
    },
    "kernel_variant": {
      "question": "Let f, g be real–valued continuous functions on [0,3].  Define  \n\n  I(f,g)=∫₀³ x⁵ [ f(x)+2g(x) ] dx,     \n  J(f,g)=∫₀³ x⁴ [ 5f(x)²+4f(x)g(x)+8g(x)² ] dx.  \n\nDetermine  \n\n  M = sup{ I(f,g)−J(f,g) : (f,g)∈C([0,3],ℝ²) },  \n\nand describe all pairs (f,g) for which the supremum is attained.",
      "solution": "Step 1.  Re-writing the functionals in matrix form.  \nFor every x∈[0,3] put  \n h(x)= (f(x),g(x))ᵀ, u(x)=x (1,2)ᵀ, w(x)=x⁴,  \n\nand let  \n\n M =  [ 5  2 ]       (so that hᵀMh = 5f² + 4fg + 8g²).  \n      [ 2  8 ]\n\nThen  \n\n    I(f,g)=∫₀³ w(x) u(x)·h(x) dx,    \n    J(f,g)=∫₀³ w(x) h(x)ᵀM h(x) dx,  \n\nso that for every x  \n\n    w(x)[u(x)·h(x) − h(x)ᵀMh(x)]  \n\nis the integrand of I(f,g)−J(f,g).\n\nStep 2.  A pointwise quadratic inequality.  \nBecause the symmetric matrix M is positive–definite (det M = 36 > 0), for any vector y  \n\n    u·y − yᵀMy ≤ ¼ uᵀM⁻¹u,                               (1)\n\nwith equality iff y = ½ M⁻¹u.  \n(This is the standard “completing the square’’ or, equivalently, the Cauchy–Schwarz\ninequality in the inner product induced by M.)\n\nApply (1) with y = h(x) and u = u(x); multiply by the non-negative weight w(x); we obtain the pointwise bound\n\n    w(x)[u(x)·h(x) − h(x)ᵀM h(x)] ≤ ¼ w(x) u(x)ᵀM⁻¹u(x).    (2)\n\nStep 3.  Computing the right–hand side of (2).  \nFirst compute M⁻¹:\n\n    M⁻¹ = (1/36) [ 8  −2 ]  \n                [ −2  5 ].\n\nHence  \n\n    u(x)ᵀM⁻¹u(x) = x² (1,2)·M⁻¹(1,2)ᵀ  \n                  = x² [1,2]·(1/36)[4,8]ᵀ  \n                  = x²·(20/36) = (5/9) x².                  (3)\n\nUsing (3) in (2) gives the pointwise inequality\n\n    u(x)·h(x) − h(x)ᵀM h(x) ≤ (5/36) x².                    (4)\n\nMultiplying by w(x)=x⁴ and integrating,\n\n    I(f,g) − J(f,g) ≤ (5/36) ∫₀³ x⁶ dx  \n                     = (5/36)·(3⁷/7)= (5/36)·(2187/7)  \n                     = 1215/28.                             (5)\n\nThus M = 1215/28.\n\nStep 4.  Existence of maximisers.  \nBecause the integrand in (4) is continuous and the upper bound is integrable, the functional is bounded above; standard compactness arguments (or a direct verification with the candidate below) show the supremum is attained.\n\nStep 5.  Characterising all maximisers.  \nEquality in (1) (hence in (4) and (5)) holds exactly when\n\n    h(x) = ½ M⁻¹u(x) for every x.                         (6)\n\nWith M⁻¹u(x) = x·(1/36)[4,8]ᵀ = x (1/9, 2/9)ᵀ, (6) yields the unique pair\n\n    f₀(x) = x/18,  g₀(x) = x/9    (0 ≤ x ≤ 3).            (7)\n\nA direct substitution of (7) into I−J achieves the upper bound in (5), so\n\n    M = 1215/28,\n\nand (f₀,g₀) is the only maximising pair.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.803201",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the original problem involves a single function; the enhanced variant requires optimising over the 2-dimensional function space C([0,3],ℝ²).  \n2. Matrix structure: one must handle a quadratic form defined by a positive-definite matrix, invert that matrix, and recognise how to complete the square in this vector setting.  \n3. Variable dependence: the “linear’’ part of the integrand now carries an extra factor x, coupling the spatial variable with the vector coefficients, so the competitor must correctly keep track of powers of x when applying the inequality.  \n4. Longer calculation chain: determining M⁻¹, evaluating the pointwise bound, integrating x⁶ over [0,3], and checking equality conditions all add layers absent from the scalar prototype.  \n5. Uniqueness proof: showing that equality forces the exact vector identity (6) at every x is more delicate than in the one-function case.\n\nThese additions demand a firm grasp of quadratic forms, matrix inequalities, and functional analytic reasoning, making the task substantially deeper than either the original or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let f, g be real–valued continuous functions on [0,3].  Define  \n\n  I(f,g)=∫₀³ x⁵ [ f(x)+2g(x) ] dx,     \n  J(f,g)=∫₀³ x⁴ [ 5f(x)²+4f(x)g(x)+8g(x)² ] dx.  \n\nDetermine  \n\n  M = sup{ I(f,g)−J(f,g) : (f,g)∈C([0,3],ℝ²) },  \n\nand describe all pairs (f,g) for which the supremum is attained.",
      "solution": "Step 1.  Re-writing the functionals in matrix form.  \nFor every x∈[0,3] put  \n h(x)= (f(x),g(x))ᵀ, u(x)=x (1,2)ᵀ, w(x)=x⁴,  \n\nand let  \n\n M =  [ 5  2 ]       (so that hᵀMh = 5f² + 4fg + 8g²).  \n      [ 2  8 ]\n\nThen  \n\n    I(f,g)=∫₀³ w(x) u(x)·h(x) dx,    \n    J(f,g)=∫₀³ w(x) h(x)ᵀM h(x) dx,  \n\nso that for every x  \n\n    w(x)[u(x)·h(x) − h(x)ᵀMh(x)]  \n\nis the integrand of I(f,g)−J(f,g).\n\nStep 2.  A pointwise quadratic inequality.  \nBecause the symmetric matrix M is positive–definite (det M = 36 > 0), for any vector y  \n\n    u·y − yᵀMy ≤ ¼ uᵀM⁻¹u,                               (1)\n\nwith equality iff y = ½ M⁻¹u.  \n(This is the standard “completing the square’’ or, equivalently, the Cauchy–Schwarz\ninequality in the inner product induced by M.)\n\nApply (1) with y = h(x) and u = u(x); multiply by the non-negative weight w(x); we obtain the pointwise bound\n\n    w(x)[u(x)·h(x) − h(x)ᵀM h(x)] ≤ ¼ w(x) u(x)ᵀM⁻¹u(x).    (2)\n\nStep 3.  Computing the right–hand side of (2).  \nFirst compute M⁻¹:\n\n    M⁻¹ = (1/36) [ 8  −2 ]  \n                [ −2  5 ].\n\nHence  \n\n    u(x)ᵀM⁻¹u(x) = x² (1,2)·M⁻¹(1,2)ᵀ  \n                  = x² [1,2]·(1/36)[4,8]ᵀ  \n                  = x²·(20/36) = (5/9) x².                  (3)\n\nUsing (3) in (2) gives the pointwise inequality\n\n    u(x)·h(x) − h(x)ᵀM h(x) ≤ (5/36) x².                    (4)\n\nMultiplying by w(x)=x⁴ and integrating,\n\n    I(f,g) − J(f,g) ≤ (5/36) ∫₀³ x⁶ dx  \n                     = (5/36)·(3⁷/7)= (5/36)·(2187/7)  \n                     = 1215/28.                             (5)\n\nThus M = 1215/28.\n\nStep 4.  Existence of maximisers.  \nBecause the integrand in (4) is continuous and the upper bound is integrable, the functional is bounded above; standard compactness arguments (or a direct verification with the candidate below) show the supremum is attained.\n\nStep 5.  Characterising all maximisers.  \nEquality in (1) (hence in (4) and (5)) holds exactly when\n\n    h(x) = ½ M⁻¹u(x) for every x.                         (6)\n\nWith M⁻¹u(x) = x·(1/36)[4,8]ᵀ = x (1/9, 2/9)ᵀ, (6) yields the unique pair\n\n    f₀(x) = x/18,  g₀(x) = x/9    (0 ≤ x ≤ 3).            (7)\n\nA direct substitution of (7) into I−J achieves the upper bound in (5), so\n\n    M = 1215/28,\n\nand (f₀,g₀) is the only maximising pair.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.614236",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the original problem involves a single function; the enhanced variant requires optimising over the 2-dimensional function space C([0,3],ℝ²).  \n2. Matrix structure: one must handle a quadratic form defined by a positive-definite matrix, invert that matrix, and recognise how to complete the square in this vector setting.  \n3. Variable dependence: the “linear’’ part of the integrand now carries an extra factor x, coupling the spatial variable with the vector coefficients, so the competitor must correctly keep track of powers of x when applying the inequality.  \n4. Longer calculation chain: determining M⁻¹, evaluating the pointwise bound, integrating x⁶ over [0,3], and checking equality conditions all add layers absent from the scalar prototype.  \n5. Uniqueness proof: showing that equality forces the exact vector identity (6) at every x is more delicate than in the one-function case.\n\nThese additions demand a firm grasp of quadratic forms, matrix inequalities, and functional analytic reasoning, making the task substantially deeper than either the original or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}