Question

If $f(x)=\frac{(x+1) \sinh x}{e^{2 x} \tan x}$ 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. $2-2 \text{cosec} 2 x$
• C. $-2(1+\text{cosec} 2 x)$
• D. $2-\frac{1}{\sin x \cos x}$

Answer 1

Answer: (C)

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