If for |x|1,tanh ^{-1}left(frac{1}{x}right)+ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that,$	anh ^{-1}\left(\frac{1}{x}ight)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$ for $|x|>1$. We know that $	anh ^{-1}(u) = \frac{1}{2}

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(C) Given that,$	anh ^{-1}\left(\frac{1}{x}ight)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$ for $|x|>1$. We know that $	anh ^{-1}(u) = \frac{1}{2} \log _e\left(\frac{1+u}{1-u}ight)$. Substituting $u = \frac{1}{x}$,we get $	anh ^{-1}\left(\frac{1}{x}ight) = \frac{1}{2} \log _e\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}ight) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}ight)$. Also,$\operatorname{coth}^{-1}(x) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}ight)$ for $|x|>1$. Adding these two expressions: $	anh ^{-1}\left(\frac{1}{x}ight)+\operatorname{coth}^{-1}(x) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}ight) + \frac{1}{2} \log _e\left(\frac{x+1}{x-1}ight) = \log _e\left(\frac{x+1}{x-1}ight)$. Comparing this with $\log _e(f(x))$,we get $f(x) = \frac{x+1}{x-1}$. Now,substituting $x = -5$: $f(-5) = \frac{-5+1}{-5-1} = \frac{-4}{-6} = \frac{2}{3}$.

If for $|x|>1$,$\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$,then $f(-5)=$

Similar Questions

If the sum of all the solutions of $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3}$ for $-1 < x < 1$ and $x \neq 0$ is $\alpha-\frac{4}{\sqrt{3}}$,then $\alpha$ is equal to $..........$.

$\tan^{-1}(-2) - \tan^{-1}(3)$ is equal to

The value of $\sin^{-1} \left( \frac{12}{13} \right) - \sin^{-1} \left( \frac{3}{5} \right)$ is equal to

Prove $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}$,where $x \in\left(0, \frac{\pi}{4}\right)$.

$S = \tan^{-1}\left( \frac{1}{n^2 + n + 1} \right) + \tan^{-1}\left( \frac{1}{n^2 + 3n + 3} \right) + \dots + \tan^{-1}\left( \frac{1}{1 + (n + 19)(n + 20)} \right)$,then $\tan S$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module