A) \[\frac{\pi }{2}\]
B) \[-\frac{\pi }{2}\]
C) \[\frac{2}{\pi }\]
D) None of these
Correct Answer: B
Solution :
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 \[a+\frac{\pi }{2}=0\Rightarrow a=-\frac{\pi }{2}\] |
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