JEE Main 2022 (Online) 24th June Morning ShiftQuestion 1
Let $\max_{0\le x\le2}\left\{\frac{9-x^{2}}{5-x}\right\}=\alpha$ and $\min_{0\le x\le2}\left\{\frac{9-x^{2}}{5-x}\right\}=\beta$. If $\int_{\beta-\frac{8}{3}}^{2\alpha-1}\max\left\{\frac{9-x^{2}}{5-x},x\right\}dx=\alpha_1+\alpha_2\log_e\left(\frac{8}{15}\right)$ then $\alpha_1+\alpha_2$ is equal to $\_\_\_\_\_\_$ .
Definite Integration
Answer$34$
Solution
Concept. First find $\alpha,\beta$ with the derivative of $g=\frac{9-x^2}{5-x}$. Then compare $g$ with $x$ on $[-1,3]$ and integrate the larger one. Formulas used. (1) $g'(x)=\dfrac{-2x(5-x)+(9-x^2)}{(5-x)^2}=\dfrac{(x-1)(x-9)}{(5-x)^2}$(2) $g(x)=\dfrac{x^2-9}{x-5}=x+5+\dfrac{16}{x-5}$Step 1 - $\alpha,\beta$. By (1), $g$ increases on $[0,1]$ and decreases on $[1,2]$. $g(0)=\frac95$, $g(1)=2$, $g(2)=\frac53$. So $\alpha=2$, $\beta=\frac53$. Step 2 - Limits. $\beta-\frac83=-1$ and $2\alpha-1=3$. Step 3 - Compare. $g(x)=x\iff9-x^2=5x-x^2\iff x=\frac95$. For $x<\frac95$, $g>x$; for $\frac95<x\le3$, $x>g$.Step 4 - Integrate. By (2), $$\int_{-1}^{9/5}g\,dx=\Big[\frac{x^2}2+5x+16\ln|x-5|\Big]_{-1}^{9/5}=\frac{81}{50}+\frac{27}2+16\ln\frac{16/5}{6}.$$ $$\int_{9/5}^3x\,dx=\frac12\Big(9-\frac{81}{25}\Big)=\frac{72}{25}.$$ Step 5 - Total. Constants: $\frac{81+675+144}{50}=18$; log part: $16\ln\frac8{15}$. So $\alpha_1=18$, $\alpha_2=16$, $\alpha_1+\alpha_2=34$. Answer: 34