Question

Let the function $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=e^{x-1}-e^{-|x-1|}$ and $g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)$.
Then the area of the region in the first quadrant bounded by the curves $y=f(x), y=g(x)$ and $x=0$ is

• A. $(2-\sqrt{3})+\frac{1}{2}\left(e-e^{-1}\right)$
• B. $(2+\sqrt{3})+\frac{1}{2}\left(e-e^{-1}\right)$
• C. $(2-\sqrt{3})+\frac{1}{2}\left(e+e^{-1}\right)$
• D. $(2+\sqrt{3})+\frac{1}{2}\left(e+e^{-1}\right)$

Answer 1

Answer: (A)

$\because f(x)=e^{x-1}-e^{-|x-1|}$
$\therefore f(x)=\begin{cases}0, & x<1 \\ e^{x-1}-e^{-(x-1)}, & x \geq 1\end{cases}$
and $g(x)=\frac{e^{x-1}+e^{1-x}}{2}$
$\therefore $ If $f(x)=g(x)$ then $2 e^{x-1}-2 e^{-(x-1)}$
$=e^{x-1}+e^{1-x}$
$\Rightarrow e^{x-1}=3 e^{1-x}$
$\Rightarrow e^{2(x-1)}=3$
$\therefore x=\frac{1}{2} \ln 3+1$

Required area $=\int\limits_{0}^{\frac{1}{2} \ln 3+1} g(x) d x-\int\limits_{1}^{\frac{1}{2} \ln 3+1} f(x) d x$
$=\frac{1}{2}\left(e^{x-1}-e^{1-x}\right)_{0}^{\frac{1}{2} \ln 3+1}-\left(e^{x-1}+e^{1-x}\right)_{1}^{\frac{1}{2} \ln 3+1}$
$=\frac{1}{2}\left(e^{\frac{1}{2} \ln 3}-e^{\frac{1}{2} \ln 3}\right)-\frac{1}{2}\left(e^{-1}-e\right)-\left(e^{\frac{1}{2} \ln 3}+e^{-\frac{1}{2} \ln 3}\right)+2$
$=(2-\sqrt{3})+\frac{1}{2}\left(e-\frac{1}{e}\right)$