sinh^{-1}left(2^{3/2}right) is equal to : | vedclass

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

Question

(A) We know that the formula for the inverse hyperbolic sine function is $\sinh^{-1}(x) = \log\{x + \sqrt{x^2 + 1}\}$.Substituting $x = 2^{3/2}$ into

Vedclass · Accepted Answer

$\log(3+\sqrt{8})$

Vedclass · Answer

(A) We know that the formula for the inverse hyperbolic sine function is $\sinh^{-1}(x) = \log\{x + \sqrt{x^2 + 1}\}$.Substituting $x = 2^{3/2}$ into the formula:$\sinh^{-1}(2^{3/2}) = \log\{2^{3/2} + \sqrt{(2^{3/2})^2 + 1\}}$.Since $2^{3/2} = \sqrt{2^3} = \sqrt{8}$, we have:$\sinh^{-1}(2^{3/2}) = \log\{\sqrt{8} + \sqrt{8 + 1\}}$.$= \log\{\sqrt{8} + \sqrt{9\}}$.$= \log\{3 + \sqrt{8\}}$.Thus, the correct option is $A$.

$\sinh^{-1}\left(2^{3/2}\right)$ is equal to :

Similar Questions

If $\theta = \frac{\pi}{12}$ and $x = \log \left(\cot \left(\frac{\pi}{4} + \theta\right)\right)$,then $\cosh x =$

$\sqrt{12-\sqrt{68+48 \sqrt{2}}}$ is equal to :

The real part of $(1 - i)^{-i}$ is

$\sqrt{-8 - 6i} = $

The value of $\sqrt{12 - \sqrt{68 + 48\sqrt{2}}}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module