Question

Let $f[-1,\frac{-1}{2}]\to [-1,1]$ is defined by $f\left(x\right)=4x^3-3x$, then $f^{-1}\left(x\right)=$ ____ .

• A. $\cos \left(\frac{1}{3}{\cos }^{-1}x\right)$
• B. $\cos \left(3{\cos }^{-1}x\right)$
• C. $\sin \left(\frac{1}{3}{\sin }^{-1}x\right)$
• D. $\cos (\frac{2\pi }{3}+\frac{1}{3}{\cos }^{-1}x)$

Answer 1

The correct option is A $\cos \left(\frac{1}{3}{\cos }^{-1}x\right)$

$f\left(x\right)=4x^3-3x$

$f[-1,\frac{-1}{2}]⟶[-1,1]$

let,$x=\cos \theta$

$\because y=\cos 3\theta$

$f\left(\cos \theta \right)=4co{s}^3\theta -3\cos \theta$

$=\cos 3\theta$

$\because 3\theta ={\cos }^{-1}y$

$\theta =\frac{1}{3}{\cos }^{-1}y$

$f^{-1}\left(\cos 3\theta \right)=\cos \theta$

$\because y=f\left(x\right)$

$f^{-1}\left(y\right)=\left(x\right)$

$\because f^{-1}\left(\cos 3\theta \right)=\cos \theta$

$f^{-1}\left(y\right)=\cos \left(\frac{1}{3}{\cos }^{-1}y\right)$

$f^{-1}\left(x\right)=\cos \left(\frac{1}{3}{\cos }^{-1}x\right)$