Question

If $f(x)=\frac{(x+1)\sinh x}{e^{2x}\tan x}$ and $\frac{f^{\prime }(x)}{f(x)}=\frac{1}{x+1}+\text{coth}x+g(x)$, then $g(x)=$

• A. $-2+\frac{1}{\sin x\cos x}$
• B. $2-2\text{cosec}2x$
• C. $-2(1+\text{cosec}2x)$
• D. $2-\frac{1}{\sin x\cos x}$

Answer 1

Answer: (C)

We have
$f(x)=\frac{(x+1)\sinh x}{e^{2x}\tan x}=(x+1)\sinh x{e}^{-2x}\cot x$
$\therefore f^{\prime }(x)=\sinh x{e}^{-2x}\cot x+(x+1)\cosh x{e}^{-2x}\cot x$
$+(x+1)\sin x{e}^{-2x}(-2)\cot x+(x+1)\sinh x{e}^{-2x}\left(-{\text{cosec}}^{2}x\right)$
$\Rightarrow f^{\prime }(x)=(x+1)\sinh x{e}^{-2x}\cot x$
$\left[\frac{1}{x+1}+\text{coth}x-2-\frac{{\text{cosec}}^{2}x}{\cot x}\right]$
$\Rightarrow f^{\prime }(x)=f(x)\left[\frac{1}{x+1}+\text{coth}x-2-\frac{\sin x}{{\sin }^{2}x\cos x}\right]$
$\Rightarrow \frac{f^{\prime }(x)}{f(x)}=\frac{1}{x+1}+\text{coth}x-2-2\text{cosec}2x$
$\therefore g(x)=-2(1+\text{cosec}2x)$