Question

Let $f:R\to R$ be defined as $f(x)=(2x-3\pi {)}^3+\frac{4}{3}x+\cos x$ and $g=f^{-1}$, then

• A. $g^{\prime }(2\pi )=\frac{3}{7}$
• B. $g^{\prime }(2\pi )=\frac{7}{3}$
• C. $g^{\prime \prime }(2\pi )=0$
• D. $g^{\prime \prime }(2\pi )=\frac{-27}{343}$

Answer 1

Answer: (A), (C)

Now, $g^{\prime }(2\pi )=\frac{1}{f^{\prime }\left(\frac{3\pi }{2}\right)}=\frac{1}{\frac{7}{3}}=\frac{3}{7}$
$f^{\prime }(x)=2\times 3(2x-3\pi {)}^2+\frac{4}{3}-\sin x$
$\therefore f^{\prime }\left(\frac{3\pi }{2}\right)=\frac{4}{3}+1=\frac{7}{3}$
Also, $g^{\prime \prime }(2\pi )=\frac{-f^{\prime \prime }\left(\frac{3\pi }{2}\right)}{{\left(f^{\prime }\left(\frac{3\pi }{2}\right)\right)}^3}=0$
$f^{\prime \prime }(x)=12\times 2(2x-3\pi )+0-\cos x$
$\therefore f^{\prime \prime }\left(\frac{3\pi }{2}\right)=0$