A) \[-\frac{2}{\pi }\]
B) \[\frac{2}{\pi }\]
C) \[-\frac{\pi }{2}\]
D) \[\frac{\pi }{2}\]
Correct Answer: C
Solution :
Here, \[{{x}^{2}}-2x+2={{(x-1)}^{2}}+1\ge 1\] But \[-1\le ({{x}^{2}}-2x+2)\le 1\] which is possible only when \[{{x}^{2}}-2x+2=1\] \[\therefore \] \[x=1\] Then, \[a{{(1)}^{2}}+{{\sin }^{-1}}(1)+co{{s}^{-1}}(1)=0\] \[\Rightarrow \] \[a+\frac{\pi }{2}+0=0\] \[\therefore \] \[a=-\frac{\pi }{2}\]You need to login to perform this action.
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