Question

Let $f(x)=(2 x-\pi)^3+2 x-\cos x$. If the value of $\frac{d}{d x}\left(f^{-1}(x)\right)$ 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

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 $\frac{d x}{d y}=\frac{1}{\frac{d y}{d x}}$ and $\frac{d y}{d x}=6(2 x-\pi)^2+2+\sin x$.
Now $\left.\frac{ dy }{ dx }\right]_{ x =\pi / 2}=0+2+1=3$.
$\therefore p =1 \text { and } q =3$
Hence $(p+q)=4$.