Question

$f\left(x\right)=(2x-3\pi {)}^5+\frac{4}{3}x+\cos x$ and $g$ is the inverse function of $f.$ Then $g^{\prime }\left(2\pi \right)$ is equal to

• A. $\frac{7}{3}$
• B. $\frac{3}{7}$
• C. $\frac{30{\pi }^4+4}{3}$
• D. $\frac{3}{30{\pi }^4+4}$

Answer 1

The correct option is B $\frac{3}{7}$

Since, $g$ is the inverse function of $f,$
$\Rightarrow g(f\left(x\right))=x$
Differentiate w.r.t. $x,$ we get
$g^{\prime }(f\left(x\right))\cdot f^{\prime }\left(x\right)=1$
$\Rightarrow g^{\prime }(f\left(x\right))=\frac{1}{f^{\prime }\left(x\right)}$

Now, $f\left(x\right)=(2x-3\pi {)}^5+\frac{4}{3}x+\cos x$
$f^{\prime }\left(x\right)=5(2x-3\pi {)}^4+\frac{4}{3}-\sin x$
$\therefore g^{\prime }(f\left(x\right))=\frac{1}{5(2x-3\pi {)}^4+\frac{4}{3}-\sin x}$

When $f\left(x\right)=2\pi ,$
$2\pi =(2x-3\pi {)}^5+\frac{4}{3}x+\cos x$
Since, in L.H.S., the power of $\pi$ is $1$ and in R.H.S. the power of $(2x-3\pi )$ is $5,$ so solution exists only when $x=\frac{3\pi }{2}$

$\therefore g^{\prime }\left(2\pi \right)=\frac{1}{\frac{4}{3}+1}=\frac{3}{7}$