Question

Let the functions:$R⟶R$ and $g:R⟶R$ 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

$(2-\sqrt{3})+\frac{1}{2}(e-e^{-1})$

Explanation for the correct option:

Finding the area of the region in the first quadrant bounded by the curve:

Given that,

$$\begin{gathered} f\left(x\right)=e^{x-1}-e^{-|x-1|} \\ f\left(x\right)=\left\{\begin{array}{ll}{e}^{x-1}-e^{-(x-1)}& x\ge 1\\ e^{x-1}-e^{(x-1)}=0& x<1\end{array}\right. \\ f\left(x\right)=\left\{\begin{array}{ll}{e}^{x-1}-e^{1-x}& x\ge 1\\ 0& x<1\end{array}\right. \end{gathered}$$

and

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

Now, $f\left(x\right)=g\left(x\right)$

$$\begin{gathered} \Rightarrow e^{x-1}-\frac{1}{e^{x-1}}=\frac{1}{2}(e^{x-1}+\frac{1}{e^{x-1}}) \\ \Rightarrow 2e^{x-1}-\frac{2}{e^{x-1}}=e^{x-1}+\frac{1}{e^{x-1}} \\ \Rightarrow e^{x-1}-\frac{3}{e^{x-1}}=0 \\ \Rightarrow e^{x-1}=\sqrt{3} \\ \Rightarrow x=1+\frac{\ln \left(3\right)}{2}\mathbf{\left[}\mathbf{\because }{\mathbf{e}}^{\mathbf{x}}\mathbf{=}{\mathbf{a}}^{\mathbf{y}}\mathbf{}\mathbf{}\mathbf{\Rightarrow }\mathbf{}\mathbf{x}\mathbf{=}\mathbf{y}\mathbf{}\mathbf{l}\mathbf{n}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{\right]} \\ \Rightarrow x=1+\ln \left(\sqrt{3}\right) \end{gathered}$$

Now, area of the region is,

$$\begin{gathered} Area={\int }_0^{1}g\left(x\right)dx+{\int }_1^{1+\frac{\ln \left(3\right)}{2}}\left\{g\right(x)-f(x\left)\right\}dx \\ ={\int }_0^1\frac{1}{2}(e^{x-1}+\frac{1}{e^{x-1}})dx+{\int }_1^{1+\frac{\ln \left(3\right)}{2}}\frac{1}{2}(e^{x-1}+\frac{1}{e^{x-1}})-(e^{x-1}+\frac{1}{e^{x-1}})dx \\ =\frac{e-e^{-1}}{2}+2-\sqrt{3} \end{gathered}$$

Therefore, the correct answer is option (A).