Question

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

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

Answer 1

The correct option is A $(2-\sqrt{3})+\frac{1}{2}(e-e^{-1})$

$f\left(x\right)=e^{x-1}-e^{-|x-1|}$
$f\left(x\right)=$$\{\begin{array}{c}{e}^{x-1}-e^{-(x-1)},x\ge 1\\ e^{x-1}-e^{x-1},x<1\end{array}$
$f\left(x\right)=$$\{\begin{array}{c}{e}^{x-1}-\frac{1}{e^{(x-1)}},x\ge 1\\ 0,x<1\end{array}$
& $g\left(x\right)=\frac{1}{2}(e^{x-1}+\frac{1}{e^{x-1}})=\frac{1}{2}(e^{x-1}+e^{1-x})$
Now $f\left(x\right)=g\left(x\right)$
$e^{x-1}-\frac{1}{e^{x-1}}=\frac{1}{2}(e^{x-1}+\frac{1}{e^{x-1}})$
$2e^{x-1}-\frac{2}{e^{x-1}}=e^{x-1}+\frac{1}{e^{x-1}}c$
$e^{x-1}-\frac{3}{e^{x-1}}=0\Rightarrow e^{x-1}=\sqrt{3}$
$x=1+\frac{\ln 3}{2}=1+\ln \sqrt{3}$
Area$=\underset{0}{\overset{1}{\int }}g\left(x\right)dx+\underset{1}{\overset{1+\frac{\ln 3}{2}}{\int }}\left[g\right(x)-f(x\left)\right]dx$
$=\underset{0}{\overset{1}{\int }}\frac{1}{2}(e^{x-1}+\frac{1}{e^{x-1}})dx+\underset{1}{\overset{1+\frac{\ln 3}{2}}{\int }}[\frac{1}{2}(e^{x-1}+\frac{1}{e^{x-1}})-(e^{x-1}-\frac{1}{e^{x-1}})]dx$
$=\frac{e-e^{-1}}{2}+2-\sqrt{3}$