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, $f^{\prime }(x)=\cos x+\sin x.$
For strictly decreasing of
$f(x);$ put $f^{\prime }(x)<0$
$\Rightarrow$ $\sin x+\cos x<0$
$\Rightarrow$ $\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x<0$
$\Rightarrow$ $\left(\sin x\cos \frac{\pi }{4}+\cos x\sin \frac{\pi }{4}\right)<0$
$\Rightarrow$ $\sin \left(x+\frac{\pi }{4}\right)<0$
$\because$ $\sin x<0$ when $\pi <x<2\pi$
$\therefore$ $\sin \left(x+\frac{\pi }{4}\right)<0$
when $\pi <x+\frac{\pi }{4}<2\pi$
$\Rightarrow$ $\pi -\frac{\pi }{4}<x<2\pi -\frac{\pi }{4}$
$\Rightarrow$ $\frac{3\pi }{4}<x<\frac{7\pi }{4}$