Question

Let $f:R\to R$ be a function such that $f\left(f\right(x\left)\right)=x^2-x+1$ for all $x$ and the value of ${\tan }^{-1}\left(f\right(1\left)\right)+{\sec }^{-1}\left(f\right(1\left)\right)+{\cos }^{-1}\left(f\right(0\left)\right)+{\cot }^{-1}\left(f\right(0\left)\right)$ is equal to $\frac{a\pi }{b}$ where $a$ and $b$ are co-prime. Then the value of $a+b$ is

Answer 1

$f\left(f\right(x\left)\right)=x^2-x+1$
Replace $x$ by $f\left(x\right)$
$f\left(f\right(f\left(x\right)\left)\right)=\left(f\right(x){)}^2-f(x)+1$
$\Rightarrow f(x^2-x+1)=\left(f\right(x){)}^2-f(x)+1\dots (1)$
Put $x=0$ in equation $\left(1\right)$
$f\left(1\right)=\left(f\right(0){)}^2-f(0)+1\dots (2)$
Put $x=1$ in equation $\left(1\right)$
$f\left(1\right)=\left(f\right(1){)}^2-f(1)+1$
$\Rightarrow \left(f\right(1)-1{)}^2=0$
$\Rightarrow f\left(1\right)=1$
Put $f\left(1\right)=1$ in equation $\left(2\right),$ we get
$f\left(0\right)=0$ or $1$
But $f\left(0\right)=0$ is rejected because it does not satisfy $f\left(f\right(x\left)\right)=x^2-x+1$
$\therefore f\left(0\right)$ = 1 and $f\left(1\right)=1$

Now, ${\tan }^{-1}\left(1\right)+{\sec }^{-1}\left(1\right)+{\cos }^{-1}\left(1\right)+{\cot }^{-1}\left(1\right)$
$=\frac{\pi }{4}+0+0+\frac{\pi }{4}=\frac{\pi }{2}\Rightarrow a+b=1+2=3$