MCQ
If $\int\left( e ^{2 x }+2 e ^{ x }- e ^{- x }-1\right) e ^{\left( e ^{ x }+ e ^{- x }\right)} d x$ $=g(x) e^{\left(e^{x}+e^{-x}\right)}+c,$ where $c$ is a constant of integration, then $g (0)$ is equal to
- ✓$2$
- B$e^{2}$
- C$e$
- D$1$
$= e ^{ x }\left( e ^{ x }+1\right)- e ^{- x }\left( e ^{ x }+1\right)+ e ^{ x }$
$=\left[\left( e ^{ x }+1\right)\left( e ^{ x }- e ^{- x }\right)+ e ^{ x }\right]$
so $I =\int\left( e ^{ x }+1\right)\left( e ^{ x }- e ^{- x }\right) e ^{ e ^{ x }+ e ^{- x }}+\int e ^{ x } \cdot e ^{ e ^{ x }+ e ^{- x }} d x$
$=\left(e^{x}+1\right) e^{e^{x}+e^{-x}}-\int e^{x} \cdot e^{e^{x}+e^{-x}} d x+\int e^{x} \cdot e^{e^{x}+e^{-x}} d x$
$=\left(e^{x}+1\right) e^{e^{x}+e^{-x}}+C$
$\therefore g(x)=e^{x}+1 \Rightarrow g(0)=2$
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If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\cot^{-1}(\pi\text{x})\end{bmatrix}$ and $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\tan^{-1}(\pi\text{x})\end{bmatrix}$ then A - B is:
$\text{I}$
$0$
$2\text{I}$
$\frac{1}{2}\text{I}$