JEE Advanced 2020 · Paper 1, Mathematics question 3

The two curves.

• For $x<1$, $f(x)=e^{x-1}-e^{x-1}=0$; for $x\ge1$, $f(x)=e^{x-1}-e^{1-x}$.
• $g(x)=\cosh(x-1)>0$.

Intersection (for $x>1$):
$$e^{x-1}-e^{1-x}=\tfrac12\left(e^{x-1}+e^{1-x}\right)\iff e^{2(x-1)}=3\iff x_0=1+\tfrac12\ln3.$$

Area between $g$ above and $f$ below, from $x=0$ to $x_0$:
$$\int_0^{x_0}g\,dx=\sinh(\tfrac12\ln3)+\sinh1=\frac1{\sqrt3}+\tfrac12(e-e^{-1}),$$
$$\int_1^{x_0}f\,dx=\Big[e^{x-1}+e^{1-x}\Big]_1^{x_0}=\sqrt3+\frac1{\sqrt3}-2.$$

$$\text{Area}=\frac1{\sqrt3}+\tfrac12(e-e^{-1})-\frac4{\sqrt3}+2=(2-\sqrt3)+\tfrac12(e-e^{-1}).$$
(A)