• # question_answer The value of $'a'$ for which $a{{x}^{2}}+\sin {{~}^{-1}}({{x}^{2}}-2x+2)+{{\cos }^{-1}}({{x}^{2}}-2x+2)=0$has a real solution, is A) $\frac{\pi }{2}$ B) $-\frac{\pi }{2}$ C) $\frac{2}{\pi }$ D) None of these

 The given is$a{{x}^{2}}+{{\sin }^{-1}}\{(x-1)+1\}+{{\cos }^{-1}}\{{{(x-1)}^{2}}+1\}=0$ $\because$$-1\le {{(x-1)}^{2}}+1\le 1\Rightarrow x=1$ So, we have $a+\frac{\pi }{2}=0\Rightarrow a=-\frac{\pi }{2}$