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← JEE Main 2022 · 24 Jun · Shift 1
MathematicsInteger+4 / -1Question 27 of 90

JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 27

Let max⁡0≤x≤2{9−x25−x}=α\displaystyle\max_{0\le x\le2}\left\{\frac{9-x^2}{5-x}\right\}=\alpha and min⁡0≤x≤2{9−x25−x}=β\displaystyle\min_{0\le x\le2}\left\{\frac{9-x^2}{5-x}\right\}=\beta.

If ∫β−832α−1max⁡{9−x25−x,x}dx=α1+α2log⁡e(815)\displaystyle\int_{\beta-\frac83}^{2\alpha-1}\max\left\{\frac{9-x^2}{5-x},x\right\}dx=\alpha_1+\alpha_2\log_e\left(\frac8{15}\right) then α1+α2\alpha_1+\alpha_2 is equal to ______.

Answer and worked solution

Official answer: 34

Find α,β\alpha,\beta. g(x)=9−x25−xg(x)=\dfrac{9-x^2}{5-x} has g′=(x−1)(x−9)(5−x)2g'=\dfrac{(x-1)(x-9)}{(5-x)^2}, so on [0,2][0,2] the maximum is g(1)=2g(1)=2 and the minimum is min⁡(g(0),g(2))=min⁡(95,53)=53\min(g(0),g(2))=\min\left(\frac95,\frac53\right)=\frac53. So α=2\alpha=2, β=53\beta=\frac53, and the limits are −1-1 to 33.

Which is larger. g(x)=xg(x)=x at x=95x=\frac95; g>xg>x before and g<xg<x after (on this range).

Integrate. g(x)=x+5+16x−5g(x)=x+5+\dfrac{16}{x-5}:
∫−19/5g=[x22+5x+16ln⁡∣x−5∣]−19/5=37825+16ln⁡815,∫9/53x dx=7225.\int_{-1}^{9/5}g=\left[\frac{x^2}2+5x+16\ln|x-5|\right]_{-1}^{9/5}=\frac{378}{25}+16\ln\frac8{15},\qquad\int_{9/5}^3x\,dx=\frac{72}{25}.

Total 18+16ln⁡81518+16\ln\frac8{15}, so α1+α2=34\alpha_1+\alpha_2=34.

Answer: 34

Topics:Properties of definite integralsMaxima and minima
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