Question

If $f\left(x\right)=(2x-\pi {)}^3+2x-\cos x$, then $\frac{d}{dx}\left(f^{-1}\right(x\left)\right)$ at $x=\pi$ is equal to

• A. $1$
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$

If $(a,b)$ is a point on the graph of $y=f\left(x\right)$, then

${\left[f^{-1}\right]}^{\prime }\left(b\right)=\frac{1}{f^{\prime }\left(a\right)}$

Now, $b=\pi$

Then,

$\pi ={(2a-\pi )}^3+2a-\cos a$

Using Hit and Trial Method,

$a=\frac{\pi }{2}$

Now,

${\left[f^{-1}\right]}^{\prime }\left(\pi \right)=\frac{1}{f^{\prime }\left(\frac{\pi }{2}\right)}$

$f^{\prime }\left(x\right)=6{(2x-\pi )}^2+2+\sin x$

$f^{\prime }\left(\frac{\pi }{2}\right)=6{(\pi -\pi )}^2+2+\sin \left(\frac{\pi }{2}\right)$

$=2+1=3$

Therefore,

${\left[f^{-1}\right]}^{\prime }\left(\pi \right)=\frac{1}{3}$