Question

The number of real solutions of; ${\cos }^{-1}x+{\cos }^{-1}2x=-\pi$ is :

• A. 0
• B. 1
• C. 2
• D. infinitely many.

Answer 1

Answer: (A)

0

We know, ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(2x\right)$
$={\cos }^{-1}(x.2x-\sqrt{1-x^2}\sqrt{1-4x^2})$
$=-\pi$
$x.2x-\sqrt{1-x^2}\sqrt{1-4x^2}=-1$
$2x^2+1=\left(\sqrt{1-x^2}\sqrt{1-4x^2}\right)$
$4x^4+4x^2+1=1+4x^4-6x^2$
$10x^2=0$
$x=0$
Hence there would be no real solution for the above equation.