A) \[[5\pi /6,\,3\pi /4]\]
B) \[[\pi /4,\,\pi /2]\]
C) \[[3\pi /2,\,5\pi /2]\]
D) None of these
Correct Answer: D
Solution :
\[f(x)=\sin x-\cos x\] \[{f}'(x)=\cos x+\sin x=\sqrt{2}\left[ \cos \,\,\left( x-\frac{\pi }{4} \right) \right]\]=\[\sqrt{2}\cos \left( x-\frac{\pi }{4} \right)\] For \[f(x)\] decreasing, \[{f}'(x)<0\] \[\frac{\pi }{2}<\left( x-\frac{\pi }{4} \right)<\frac{3\pi }{2}\], (within \[0\le x\le 2\pi \]). Þ \[\frac{3\pi }{4}<x\le \frac{7\pi }{4}\].You need to login to perform this action.
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