A) ? 2
B) 2
C) \[-2\sqrt{\frac{\pi }{2}}\]
D) 0
Correct Answer: D
Solution :
Since \[\frac{dy}{dx}=-\sin (\sin {{x}^{2}}).\cos {{x}^{2}}.2x\] Therefore, at \[x=\sqrt{\frac{\pi }{2}},\] \[\cos {{x}^{2}}=\cos \frac{\pi }{2}=0\] Þ\[\frac{dy}{dx}=0\].You need to login to perform this action.
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