Question

The interval in which the function $ f(x)=\sin x-\cos x,\,\,\,\,0\le x\le 2\pi $ is strictly decreasing, is

• 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} $

Answer 1

Answer: (C)

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} $