Question

Let ${\cos }^{-1}(4x^3-3x)=a+b{\cos }^{-1}x$. If $xϵ[-\frac{1}{2},-1)$ then the value of $a+b\pi$ is

• A. $2\pi$
• B. $3\pi$
• C. $\pi$
• D. $-2\pi$

Answer 1

The correct option is C $\pi$

As
$co{s}^{-1}(4x^3-3x)=3co{s}^{-1}x-2\pi -1<x<-\frac{1}{2}$

Gives $a=-2\pi$ and $b=3$

Therefore,
$a+b\pi =-2\pi +3\pi =\pi$