Q41 / 160 · AP EAMCET 2003 · Mathematics

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

  • Correct answer
    A
    $\log (3+\sqrt{8})$
  • B
    $\log (3-\sqrt{8})$
  • C
    $\log (2+\sqrt{18})$
  • D
    $\log (\sqrt{8}+\sqrt{27})$

Solution

We have $\sin h^{-1} x=\log \left\{x+\sqrt{x^2+1}\right\}$ $\sin h^{-1}(2)^{3 / 2}=\log \left\{2^{3 / 2}+\sqrt{\left(2^{3 / 2}\right)^2+1}\right\}$ $=\log \{\sqrt{8}+\sqrt{8+1}\}=\log \{3+\sqrt{8}\}$

Hence, the correct answer is option: 'A'.

Relevant Questions