{
  "index": "1979-B-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Problem B-2\nLet \\( 0<a<b \\). Evaluate\n\\[\n\\lim _{t \\rightarrow 0}\\left\\{\\int_{0}^{1}[b x+a(1-x)]^{\\prime} d x\\right\\}^{1 / t}\n\\]\n[The final answer should not involve any operations other than addition, subtraction, multiplication, division, and exponentiation.]",
  "solution": "B-2.\nLet \\( u=b x+a(1-x) \\); then the definite integral becomes\n\\[\nI(t)=\\frac{1}{b-a} \\int_{a}^{b} u^{t} d u=\\frac{b^{t+1}-a^{t+1}}{(1+t)(b-a)}\n\\]\n\nUsing standard calculus methods for evaluating limits of indeterminate expressions, one finds that\n\\[\n[I(t)]^{1 / t} \\rightarrow e^{-1}\\left(b^{b} / a^{a}\\right)^{1 /(b-a)} \\text { as } t \\rightarrow 0\n\\]",
  "vars": [
    "x",
    "t",
    "u",
    "I"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "indvarx",
        "t": "limtime",
        "u": "auxvalue",
        "I": "intvalue",
        "a": "lowerconst",
        "b": "upperconst"
      },
      "question": "Problem B-2\nLet \\( 0<lowerconst<upperconst \\). Evaluate\n\\[\n\\lim _{limtime \\rightarrow 0}\\left\\{\\int_{0}^{1}[upperconst\\,indvarx+lowerconst(1-indvarx)]^{\\prime} d indvarx\\right\\}^{1 / limtime}\n\\]\n[The final answer should not involve any operations other than addition, subtraction, multiplication, division, and exponentiation.]",
      "solution": "B-2.\nLet \\( auxvalue = upperconst\\,indvarx + lowerconst(1-indvarx) \\); then the definite integral becomes\n\\[\nintvalue(limtime)=\\frac{1}{upperconst-lowerconst} \\int_{lowerconst}^{upperconst} auxvalue^{limtime} d auxvalue=\\frac{upperconst^{limtime+1}-lowerconst^{limtime+1}}{(1+limtime)(upperconst-lowerconst)}\n\\]\n\nUsing standard calculus methods for evaluating limits of indeterminate expressions, one finds that\n\\[\n[intvalue(limtime)]^{1 / limtime} \\rightarrow e^{-1}\\left(upperconst^{upperconst} / lowerconst^{lowerconst}\\right)^{1 /(upperconst-lowerconst)} \\text { as } limtime \\rightarrow 0\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "candlewick",
        "t": "riverstone",
        "u": "meadowlark",
        "I": "lanternfly",
        "a": "quartzite",
        "b": "sandstorm"
      },
      "question": "Problem B-2\nLet \\( 0<quartzite<sandstorm \\). Evaluate\n\\[\n\\lim _{riverstone \\rightarrow 0}\\left\\{\\int_{0}^{1}[sandstorm candlewick+quartzite(1-candlewick)]^{\\prime} d candlewick\\right\\}^{1 / riverstone}\n\\]\n[The final answer should not involve any operations other than addition, subtraction, multiplication, division, and exponentiation.]",
      "solution": "B-2.\nLet \\( meadowlark=sandstorm candlewick+quartzite(1-candlewick) \\); then the definite integral becomes\n\\[\nlanternfly(riverstone)=\\frac{1}{sandstorm-quartzite} \\int_{quartzite}^{sandstorm} meadowlark^{riverstone} d meadowlark=\\frac{sandstorm^{riverstone+1}-quartzite^{riverstone+1}}{(1+riverstone)(sandstorm-quartzite)}\n\\]\n\nUsing standard calculus methods for evaluating limits of indeterminate expressions, one finds that\n\\[\n[lanternfly(riverstone)]^{1 / riverstone} \\rightarrow e^{-1}\\left(sandstorm^{sandstorm} / quartzite^{quartzite}\\right)^{1 /(sandstorm-quartzite)} \\text { as } riverstone \\rightarrow 0\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "t": "timeless",
        "u": "knownvalue",
        "I": "differential",
        "a": "uppermost",
        "b": "lowermost"
      },
      "question": "Problem B-2\nLet \\( 0<uppermost<lowermost \\). Evaluate\n\\[\n\\lim _{timeless \\rightarrow 0}\\left\\{\\int_{0}^{1}[lowermost fixedvalue+uppermost(1-fixedvalue)]^{\\prime} d fixedvalue\\right\\}^{1 / timeless}\n\\]\n[The final answer should not involve any operations other than addition, subtraction, multiplication, division, and exponentiation.]",
      "solution": "B-2.\nLet \\( knownvalue=lowermost fixedvalue+uppermost(1-fixedvalue) \\); then the definite integral becomes\n\\[\ndifferential(timeless)=\\frac{1}{lowermost-uppermost} \\int_{uppermost}^{lowermost} knownvalue^{timeless} d knownvalue=\\frac{lowermost^{timeless+1}-uppermost^{timeless+1}}{(1+timeless)(lowermost-uppermost)}\n\\]\n\nUsing standard calculus methods for evaluating limits of indeterminate expressions, one finds that\n\\[\n[differential(timeless)]^{1 / timeless} \\rightarrow e^{-1}\\left(lowermost^{lowermost} / uppermost^{uppermost}\\right)^{1 /(lowermost-uppermost)} \\text { as } timeless \\rightarrow 0\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "u": "vclfnrye",
        "I": "pmdsxqro",
        "a": "kefluzjd",
        "b": "wpmagoti"
      },
      "question": "Problem B-2\nLet \\( 0<kefluzjd<wpmagoti \\). Evaluate\n\\[\n\\lim _{hjgrksla \\rightarrow 0}\\left\\{\\int_{0}^{1}[wpmagoti qzxwvtnp+kefluzjd(1-qzxwvtnp)]^{\\prime} d qzxwvtnp\\right\\}^{1 / hjgrksla}\n\\]\n[The final answer should not involve any operations other than addition, subtraction, multiplication, division, and exponentiation.]",
      "solution": "B-2.\nLet \\( vclfnrye=wpmagoti qzxwvtnp+kefluzjd(1-qzxwvtnp) \\); then the definite integral becomes\n\\[\npmdsxqro(hjgrksla)=\\frac{1}{wpmagoti-kefluzjd} \\int_{kefluzjd}^{wpmagoti} vclfnrye^{hjgrksla} d vclfnrye=\\frac{wpmagoti^{hjgrksla+1}-kefluzjd^{hjgrksla+1}}{(1+hjgrksla)(wpmagoti-kefluzjd)}\n\\]\n\nUsing standard calculus methods for evaluating limits of indeterminate expressions, one finds that\n\\[\n[pmdsxqro(hjgrksla)]^{1 / hjgrksla} \\rightarrow e^{-1}\\left(wpmagoti^{wpmagoti} / kefluzjd^{kefluzjd}\\right)^{1 /(wpmagoti-kefluzjd)} \\text { as } hjgrksla \\rightarrow 0\n\\]"
    },
    "kernel_variant": {
      "question": "Let a, b, c be three distinct positive real numbers with a < b < c.  Define  \n\n  T = {(x, y) ∈ ℝ² | x ≥ 0, y ≥ 0, x + y ≤ 1}  (the unit right-triangle).  \n\nCompute  \n\n  lim_{t→0} ⎧ 2 ∬_T [ a(1 – x – y) + b x + c y ]^t dy dx ⎫^{1/t}.  \n\n(The final answer may involve only addition, subtraction, multiplication, division and exponentiation.)",
      "solution": "Step 1.  Set  \n\n  F(t)=2 ∬_T [ a(1 – x – y)+b x+c y ]^t dy dx.  \n\n(The factor 2 makes F(0)=1, simplifying the limit.)\n\n--------------------------------------------------------------------\nStep 2.  Inner integration (fix x, integrate in y).  \n\nWrite L(x,y)=a(1–x–y)+b x+c y.  \nFor fixed x (0 ≤ x ≤ 1) let  \n\n  u = L(x,y) = a+a_x + (c–a) y, where a_x = (b–a)x.  \n\nThus du = (c–a) dy and when y runs from 0 to 1–x we have  \n\n  u_0 = a + a_x, u_1 = a + a_x + (c–a)(1–x)\n     = a + (c–a) + (b–c)x = c + (b–c)x.\n\nHence  \n\n  ∫_{0}^{1–x}L(x,y)^t dy\n  = 1/(c–a) ∫_{u_0}^{u_1} u^t du\n  = (u_1^{t+1} – u_0^{t+1}) /[(c–a)(t+1)].\n\n--------------------------------------------------------------------\nStep 3.  Outer integration in x.\n\nPut u_0 = a+(b–a)x, u_1 = c+(b–c)x.  Then  \n\n  F(t)= 2/[(c–a)(t+1)] ∫_{0}^{1} [u_1^{\\,t+1} – u_0^{\\,t+1}] dx.\n\nBecause u_0,u_1 are linear in x, the integrals can be evaluated exactly:  \n\n  ∫_{0}^{1} u_0^{\\,t+1} dx = 1/(b–a) ∫_{a}^{b} s^{t+1}ds\n            = (b^{t+2} – a^{t+2}) /[(b–a)(t+2)],\n\n  ∫_{0}^{1} u_1^{\\,t+1} dx = 1/(b–c) ∫_{c}^{b} s^{t+1}ds\n            = (b^{t+2} – c^{t+2}) /[(b–c)(t+2)].\n\nTherefore  \n\n  F(t)= 2/[(c–a)(t+1)]·{ (b^{t+2} – c^{t+2}) /[(b–c)(t+2)] – \n                           (b^{t+2} – a^{t+2}) /[(b–a)(t+2)] }.\n\n--------------------------------------------------------------------\nStep 4.  First-order expansion about t = 0.\n\nWrite G(t)=F(t).  Because the integrand equals 1 when t=0, we have G(0)=1.  \nPut t→0 and expand every factor to first order:\n\n (b^{t+2} – c^{t+2}) = (b² – c²) + t(b² ln b – c² ln c) + O(t²),  \n (b^{t+2} – a^{t+2}) = (b² – a²) + t(b² ln b – a² ln a) + O(t²),  \n\n 1/(t+1) = 1 – t + O(t²),  1/(t+2)=½(1 – t/2) + O(t²).\n\nSubstituting and keeping only the O(t) terms gives  \n\n G(t)=1 + t·D + O(t²),  where  \n\n D = –3/2 + [b² ln b – c² ln c] /[(b–c)(c–a)] – [b² ln b – a² ln a] /[(b–a)(c–a)].\n\n--------------------------------------------------------------------\nStep 5.  Passage to the limit.\n\nSince G(t)=1 + tD + O(t²),  \n\n  lim_{t→0} [G(t)]^{1/t} = e^{D}.\n\nExplicitly\n\n  lim = exp{ –3/2\n             + [b² ln b – c² ln c] /[(b–c)(c–a)]\n             – [b² ln b – a² ln a] /[(b–a)(c–a)] }.\n\n--------------------------------------------------------------------\nStep 6.  Remove the logarithms (only powers allowed).\n\nBecause exp( x ln x ) = x^x, we rewrite each logarithmic term:\n\n e^{\\,b² ln b – c² ln c} = b^{b²} / c^{c²}, etc.\n\nHence the result can be expressed using only the permitted operations:\n\n lim_{t→0} { 2 ∬_T [ a(1–x–y)+b x+c y ]^t dy dx }^{1/t}\n = e^{–3/2}·\n     ( b^{b²} / c^{c²} )^{1/[(b–c)(c–a)]} ·\n     ( a^{a²} / b^{b²} )^{1/[(b–a)(c–a)]}.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.641301",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem moves from a single integral on [0,1] to a double integral over the two-dimensional simplex T.  \n2. More variables: three independent parameters (a, b, c) instead of two.  \n3. Additional algebra: evaluating the inner integral produces two new linear limits depending on x; the outer integral introduces two separate β-type integrals that must be handled simultaneously.  \n4. Deeper expansion: to extract the first-order term of F(t) one must expand several coupled rational factors, keep track of cancellations and confirm that the zero-order term is exactly 1.  \n5. Final expression management: logarithms arising from differentiation have to be eliminated so that the answer uses only elementary arithmetic and exponentiation, requiring an extra non-trivial rewriting step.  \n\nThese layers demand multivariable integration techniques, careful asymptotic expansion, and sophisticated algebraic manipulation—altogether a substantially harder task than either the original problem or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let a, b, c be three distinct positive real numbers with a < b < c.  Define  \n\n  T = {(x, y) ∈ ℝ² | x ≥ 0, y ≥ 0, x + y ≤ 1}  (the unit right-triangle).  \n\nCompute  \n\n  lim_{t→0} ⎧ 2 ∬_T [ a(1 – x – y) + b x + c y ]^t dy dx ⎫^{1/t}.  \n\n(The final answer may involve only addition, subtraction, multiplication, division and exponentiation.)",
      "solution": "Step 1.  Set  \n\n  F(t)=2 ∬_T [ a(1 – x – y)+b x+c y ]^t dy dx.  \n\n(The factor 2 makes F(0)=1, simplifying the limit.)\n\n--------------------------------------------------------------------\nStep 2.  Inner integration (fix x, integrate in y).  \n\nWrite L(x,y)=a(1–x–y)+b x+c y.  \nFor fixed x (0 ≤ x ≤ 1) let  \n\n  u = L(x,y) = a+a_x + (c–a) y, where a_x = (b–a)x.  \n\nThus du = (c–a) dy and when y runs from 0 to 1–x we have  \n\n  u_0 = a + a_x, u_1 = a + a_x + (c–a)(1–x)\n     = a + (c–a) + (b–c)x = c + (b–c)x.\n\nHence  \n\n  ∫_{0}^{1–x}L(x,y)^t dy\n  = 1/(c–a) ∫_{u_0}^{u_1} u^t du\n  = (u_1^{t+1} – u_0^{t+1}) /[(c–a)(t+1)].\n\n--------------------------------------------------------------------\nStep 3.  Outer integration in x.\n\nPut u_0 = a+(b–a)x, u_1 = c+(b–c)x.  Then  \n\n  F(t)= 2/[(c–a)(t+1)] ∫_{0}^{1} [u_1^{\\,t+1} – u_0^{\\,t+1}] dx.\n\nBecause u_0,u_1 are linear in x, the integrals can be evaluated exactly:  \n\n  ∫_{0}^{1} u_0^{\\,t+1} dx = 1/(b–a) ∫_{a}^{b} s^{t+1}ds\n            = (b^{t+2} – a^{t+2}) /[(b–a)(t+2)],\n\n  ∫_{0}^{1} u_1^{\\,t+1} dx = 1/(b–c) ∫_{c}^{b} s^{t+1}ds\n            = (b^{t+2} – c^{t+2}) /[(b–c)(t+2)].\n\nTherefore  \n\n  F(t)= 2/[(c–a)(t+1)]·{ (b^{t+2} – c^{t+2}) /[(b–c)(t+2)] – \n                           (b^{t+2} – a^{t+2}) /[(b–a)(t+2)] }.\n\n--------------------------------------------------------------------\nStep 4.  First-order expansion about t = 0.\n\nWrite G(t)=F(t).  Because the integrand equals 1 when t=0, we have G(0)=1.  \nPut t→0 and expand every factor to first order:\n\n (b^{t+2} – c^{t+2}) = (b² – c²) + t(b² ln b – c² ln c) + O(t²),  \n (b^{t+2} – a^{t+2}) = (b² – a²) + t(b² ln b – a² ln a) + O(t²),  \n\n 1/(t+1) = 1 – t + O(t²),  1/(t+2)=½(1 – t/2) + O(t²).\n\nSubstituting and keeping only the O(t) terms gives  \n\n G(t)=1 + t·D + O(t²),  where  \n\n D = –3/2 + [b² ln b – c² ln c] /[(b–c)(c–a)] – [b² ln b – a² ln a] /[(b–a)(c–a)].\n\n--------------------------------------------------------------------\nStep 5.  Passage to the limit.\n\nSince G(t)=1 + tD + O(t²),  \n\n  lim_{t→0} [G(t)]^{1/t} = e^{D}.\n\nExplicitly\n\n  lim = exp{ –3/2\n             + [b² ln b – c² ln c] /[(b–c)(c–a)]\n             – [b² ln b – a² ln a] /[(b–a)(c–a)] }.\n\n--------------------------------------------------------------------\nStep 6.  Remove the logarithms (only powers allowed).\n\nBecause exp( x ln x ) = x^x, we rewrite each logarithmic term:\n\n e^{\\,b² ln b – c² ln c} = b^{b²} / c^{c²}, etc.\n\nHence the result can be expressed using only the permitted operations:\n\n lim_{t→0} { 2 ∬_T [ a(1–x–y)+b x+c y ]^t dy dx }^{1/t}\n = e^{–3/2}·\n     ( b^{b²} / c^{c²} )^{1/[(b–c)(c–a)]} ·\n     ( a^{a²} / b^{b²} )^{1/[(b–a)(c–a)]}.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.509369",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem moves from a single integral on [0,1] to a double integral over the two-dimensional simplex T.  \n2. More variables: three independent parameters (a, b, c) instead of two.  \n3. Additional algebra: evaluating the inner integral produces two new linear limits depending on x; the outer integral introduces two separate β-type integrals that must be handled simultaneously.  \n4. Deeper expansion: to extract the first-order term of F(t) one must expand several coupled rational factors, keep track of cancellations and confirm that the zero-order term is exactly 1.  \n5. Final expression management: logarithms arising from differentiation have to be eliminated so that the answer uses only elementary arithmetic and exponentiation, requiring an extra non-trivial rewriting step.  \n\nThese layers demand multivariable integration techniques, careful asymptotic expansion, and sophisticated algebraic manipulation—altogether a substantially harder task than either the original problem or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}