A) \[p<\frac{1}{2}\]
B) \[p>\frac{1}{2}\]
C) \[p<2\]
D) \[p>2\]
Correct Answer: B
Solution :
\[f(x)\]will be monotonically decreasing, if\[f'(x)<0\]. Þ \[f'(x)=-\sin x-2p<0\]Þ\[\frac{1}{2}\sin x+p>0\] Þ \[p>\frac{1}{2}\,,\,\,[\because -1\le \sin x\le 1]\].You need to login to perform this action.
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