Question

Let $f(x)=(2x-\pi {)}^3+2x-\cos x$. If the value of $\frac{d}{dx}\left(f^{-1}(x)\right)$ at $x=\pi$ can be expressed in the form of $\frac{p}{q}$ (where $p$ and $q$ are natural numbers in their lowest form), then find the value of $(p+q)$.

Answer 1

Answer: 4

When $f(x)=\pi$, then $x=\frac{\pi }{2}$.
[As $f(x)$ is an increasing function on $R$, so $f(x)$ is invertible.]
We have to find $\frac{dx}{dy}$ at $y=\pi$.
Now $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$ and $\frac{dy}{dx}=6(2x-\pi {)}^2+2+\sin x$.
Now ${\frac{dy}{dx}]}_{x=\pi /2}=0+2+1=3$.
$\therefore p=1\text{ and }q=3$
Hence $(p+q)=4$.