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 a real solution, is

• A. $\frac{\pi }{2}$
• B. $-\frac{\pi }{2}$
• C. $\frac{2}{\pi }$
• D. $-\frac{2}{\pi }$

Answer 1

Answer: (B)

We have,
$x^2-2x+2=(x-1{)}^2+1\ge 1$
So, the given equation is meaningful for $x=1$.
Putting $x=1$, we have
$a+{\sin }^{-1}(1)+{\cos }^{-1}(1)=0$
$\Rightarrow a=-\frac{\pi }{2}$