JEE Advanced 2020 — Paper 1 — Question 31

$f(x)=\left\{\begin{array}{cc}0 & , \quad x \leq 1\\ e^{x-1}-e^{1-x} & , \quad x>1\end{array}\right.$

$g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)$

$f(x)=g(x)$ $e^{x-1}-e^{1-x}=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)$

$e^{x-1}=3 e^{1-x}$ $e^{x-1}=\sqrt{3}$

$\mathrm{e}^{\alpha-1}=\sqrt{3}$. Let $\mathrm{x}=\alpha$ be P.O.I

Required area $=\int_{0}^{1} \frac{1}{2}\left(e^{x-1}+e^{1-x}\right) d x+\int_{1}^{\alpha}\left(\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)-\left(e^{x-1}-e^{1-x}\right)\right) d x$

$=\frac{1}{2}\left|e^{x-1}-e^{1-x}\right|_{0}^{1}+\left|-\frac{1}{2} e^{x-1}-\frac{3}{2} e^{1-x}\right|_{1}^{\alpha}$

$=\frac{1}{2}\left\{\left(e^{0}-e^{0}\right)-\left(e^{-1}-e^{1}\right)\right\}-\frac{1}{2}\left\{e^{\alpha-1}+3 e^{1-\alpha}-\left(e^{0}+3 e^{0}\right)\right\}$

$=\frac{1}{2}\left(e-\frac{1}{e}\right)-\frac{1}{2}\left(\sqrt{3}+\frac{3}{\sqrt{3}}-4\right)$

$=-(\sqrt{3}-2)+\frac{1}{2}\left(e-\frac{1}{e}\right)$ $=(2-\sqrt{3})+\frac{1}{2}\left(e-\frac{1}{e}\right)$