A) \[\pm 1\]
B) \[\pm \frac{1}{\sqrt{2}}\]
C) \[\pm \,\,\frac{1}{2}\]
D) 0 and 1
Correct Answer: A
Solution :
\[f'(x)=\sqrt{2-{{x}^{2}}}\]\[\Rightarrow {{x}^{2}}-\sqrt{2-{{x}^{2}}}=0\] or \[{{x}^{4}}+{{x}^{2}}-2=0\] or \[({{x}^{2}}+2)({{x}^{2}}-1)=0\] \[\therefore {{x}^{2}}-1=0,\,\,\,\,\therefore x=\pm 1\].
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