Question

If $f(x)=(2x−3π{)}^5+\frac{4}{3}x+\cos x$ and $g$ is the inverse function of $f$, then $g^{′}(2π)$ is equal to

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

Answer 1

Answer: (B)

$y=(2x−3π{)}^5+\frac{4}{3}x+\cos x$
$\frac{dy}{dx}=5(2x−3π{)}^4+\frac{4}{3}−\sin x$
$∴g^{′}(y)=\frac{dx}{dy}=\frac{1}{5(2x−3π{)}^4+\frac{4}{3}x−\sin x}$
when $y=2Ï€$
$2π=(2x−3π{)}^5+\frac{4}{3}x+\cos x$
$x=\frac{3Ï€}{2}$ is the only solution
$∴g^{′}(2π)=\frac{1}{1+\frac{4}{3}}=\frac{3}{7}$
Alternatively: $(g∘f)(x)=x⇒g^{′}(f(x)â‹\dots f^{′}(x)=1$
$g^{′}(f(x)=\frac{1}{f^{′}(x)}$
Putting $f(x)=2Ï€$
$2π=(2x−3π{)}^5+\frac{4}{3}x+\cos x$
$⇒x=\frac{3π}{2}\text{ (It is only }x=\frac{3π}{2}\text{ as f is one-one function) }$
$∴f\left(\frac{3π}{2}\right)=2π$
$⇒g^{′}(2Ï€)=\frac{1}{f^{′}\left(\frac{3Ï€}{2}\right)}=\frac{1}{7/3}=\frac{3}{7}\left(âˆ\mu f^{′}(x)=5(2x−3Ï€{)}^4+4/3−\sin x\right)$