Q. Let $f(x)=\cos ^{-1}(\cos 2)+\sin ^{-1}(\sin 1) x-x^2$. If $\operatorname{sgn}(f(x))$ is maximum for true set of values of $x \in(a, b)$, then $(a+b)$ is equal to
Inverse Trigonometric Functions
Solution:
$ f ( x )=2+ x - x ^2=-\left( x ^2- x -2\right)=-( x -2)( x +1)$
$\Theta sgn ( f ( x ))=1 $
$\Rightarrow f ( x )>0$
$\therefore x \in(-1,2) $
$\therefore \text { sum }=1$
