Question

If $x^2-10x+17<{\cos }^{-1}\left(\cos 4\right)+{\tan }^{-1}\left(\tan 5\right)\forall x\in Z,$ then the number of integral value(s) of $x$ is

Answer 1

Given : $x^2-10x+17<{\cos }^{-1}\left(\cos 4\right)+{\tan }^{-1}\left(\tan 5\right)$
We know that
${\cos }^{-1}\left(\cos 4\right)=2\pi -4;\pi \le x\le 2\pi$
${\tan }^{-1}\left(\tan 5\right)=5-2\pi ;\frac{3\pi }{2}<x<2\pi$
So, inequality becomes : $x^2-10x+17<2\pi -4+5-2\pi$
$\Rightarrow x^2-10x+16<0$
$\Rightarrow (x-2)(x-8)<0$
$\Rightarrow x\in (2,8)$
$\therefore$ Five integral values are $3,4,5,6,7.$