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Q.
The value of '$a$' for which $a x^{2}+\sin ^{-1}\left(x^{2}-2 x+2\right)$ $+\cos ^{-1}\left(x^{2}-2 x+2\right)=0$ has a real solution, is
Inverse Trigonometric Functions
Solution:
We have,
$x^{2}-2 x+2=(x-1)^{2}+1 \geq 1$
So, the given equation is meaningful for $x=1$.
Putting $x=1$, we have
$a+\sin ^{-1}(1)+\cos ^{-1}(1)=0$
$\Rightarrow a=-\frac{\pi}{2}$