Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)=\frac{(x+1) \sinh x}{e^{2 x} \tan x}$ and $\frac{f^{\prime}(x)}{f(x)}=\frac{1}{x+1}+\operatorname{coth} x+g(x)$, then $g(x)=$
Options:
Solution:
1678 Upvotes
Verified Answer
The correct answer is:
$-2(1+\operatorname{cosec} 2 x)$
We have,
$f(x)=\frac{(x+1) \sinh x}{e^{2 x} \tan x}=(x+1) \sinh x e^{-2 x} \cot x$
$\begin{array}{r}\therefore f^{\prime}(x)=\sinh x e^{-2 x} \cot x+(x+1) \cosh x e^{-2 x} \cot x \\ +(x+1) \sinh x e^{-2 x}(-2) \cot x \\ +(x+1) \sinh x e^{-2 x}\left(-\operatorname{coscc}^2 x\right)\end{array}$
$\begin{aligned} & \Rightarrow f^{\prime}(x)=(x+1) \sinh x^{-2 x} \cot x \\ & \quad\left[\frac{1}{x+1}+\operatorname{coth} x-2-\frac{\operatorname{cosec}^2 x}{\cot x}\right] \\ & \Rightarrow f^{\prime}(x)=f(x)\left[\frac{1}{x+1}+\operatorname{coth} x-2-\frac{\sin x}{\sin ^2 x \cos x}\right]\end{aligned}$
$\Rightarrow \frac{f^{\prime}(x)}{f(x)}=\frac{1}{x+1}+\operatorname{coth} x-2-2 \operatorname{cosec} 2 x$
$\therefore \quad g(x)=-2(1+\operatorname{cosec} 2 x)$
$f(x)=\frac{(x+1) \sinh x}{e^{2 x} \tan x}=(x+1) \sinh x e^{-2 x} \cot x$
$\begin{array}{r}\therefore f^{\prime}(x)=\sinh x e^{-2 x} \cot x+(x+1) \cosh x e^{-2 x} \cot x \\ +(x+1) \sinh x e^{-2 x}(-2) \cot x \\ +(x+1) \sinh x e^{-2 x}\left(-\operatorname{coscc}^2 x\right)\end{array}$
$\begin{aligned} & \Rightarrow f^{\prime}(x)=(x+1) \sinh x^{-2 x} \cot x \\ & \quad\left[\frac{1}{x+1}+\operatorname{coth} x-2-\frac{\operatorname{cosec}^2 x}{\cot x}\right] \\ & \Rightarrow f^{\prime}(x)=f(x)\left[\frac{1}{x+1}+\operatorname{coth} x-2-\frac{\sin x}{\sin ^2 x \cos x}\right]\end{aligned}$
$\Rightarrow \frac{f^{\prime}(x)}{f(x)}=\frac{1}{x+1}+\operatorname{coth} x-2-2 \operatorname{cosec} 2 x$
$\therefore \quad g(x)=-2(1+\operatorname{cosec} 2 x)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.