A) \[0<x<\frac{3\pi }{4}\]
B) \[\frac{7\pi }{4}<x<2\pi \]
C) \[\frac{3\pi }{4}<x<\frac{7\pi }{4}\]
D) \[0<x<\frac{7\pi }{4}\]
Correct Answer: C
Solution :
Given function is, \[f(x)=\sin x-\cos x,\] where \[x\in [0,\,2\pi ]\] Now, \[f'(x)=\cos x+\sin x.\] For strictly decreasing of \[f(x);\] put \[f'(x)<0\] \[\Rightarrow \] \[\sin x+\cos x<0\] \[\Rightarrow \] \[\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos \,x<0\] \[\Rightarrow \] \[\left( \sin x\cos \frac{\pi }{4}+\cos x\sin \frac{\pi }{4} \right)<0\] \[\Rightarrow \] \[\sin \left( x+\frac{\pi }{4} \right)<0\] \[\because \] \[\sin x<0\] when \[\pi <x<2\pi \] \[\therefore \] \[\sin \left( x+\frac{\pi }{4} \right)<0\] when \[\pi <x+\frac{\pi }{4}<2\pi \] \[\Rightarrow \] \[\pi -\frac{\pi }{4}<x<2\pi -\frac{\pi }{4}\] \[\Rightarrow \] \[\frac{3\pi }{4}<x<\frac{7\pi }{4}\]
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