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

$\operatorname{Tan}^{-1} \left( \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1} \right) + \operatorname{Tan}^{-1} \left( \frac{1}{\sqrt{5}} \right) =$

If $\sin ^{-1} x + \sin ^{-1} y = \frac{\pi}{2}$,then $x^{2}$ is equal to

If $\sec ^{-1}\left(\frac{5}{x}\right)+\sin ^{-1} \left(\frac{4}{5}\right)=\frac{\pi}{2}$,where $x \neq 0$,then $x=$ . . . . . . .

If $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2},$ then the value of $x$ is

The range of the function $f(x) = \sqrt{|\sin^{-1}|\sin x|| - |\cos^{-1}|\cos x||}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module