Question

Let the function $f:R\to R$ and $g:R\to R$ 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)

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

Required area $=\underset{0}{\overset{\frac{1}{2}\ln 3+1}{\int }}g(x)dx-\underset{1}{\overset{\frac{1}{2}\ln 3+1}{\int }}f(x)dx$
$=\frac{1}{2}{\left(e^{x-1}-e^{1-x}\right)}_0^{\frac{1}{2}\ln 3+1}-{\left(e^{x-1}+e^{1-x}\right)}_1^{\frac{1}{2}\ln 3+1}$
$=\frac{1}{2}\left(e^{\frac{1}{2}\ln 3}-e^{\frac{1}{2}\ln 3}\right)-\frac{1}{2}\left(e^{-1}-e\right)-\left(e^{\frac{1}{2}\ln 3}+e^{-\frac{1}{2}\ln 3}\right)+2$
$=(2-\sqrt{3})+\frac{1}{2}\left(e-\frac{1}{e}\right)$