Question

The value of `a` for which $a{x}^2+{\sin }^{-1}\left(x^2-2x+2\right)+{\cos }^{-1}\left(x^2-2x+2\right)=0$ has real solution is

Answer 1

$a{x}^2{\sin }^{-1}(x^2-2x+2)+{\cos }^{-1}(x^2-2x+2)=0$

we know that

${\sin }^{-1}\left(f\right(x\left)\right)+{\cos }^{-1}\left(f\right(x\left)\right)=\frac{\pi }{2}$

$a{x}^2+\frac{\pi }{2}=0\Rightarrow a{x}^2=-\frac{\pi }{2}$

$a$ must be negative

$a\in (-\infty ,0)$