Question

If f (x) = $\sqrt{1 - \sin 2x}$, then f ' (x) is equals to –

• A. –(cos x + sin x), for x Ī ($\frac{\pi}{4}$,$\frac{\pi}{2}$)
• B. (cos x + sin x), for x Ī (0, $\frac{\pi}{4}$)
• C. – (cos x + sin x), for x Ī (0, $\frac{\pi}{4}$)
• D. (cos x – sin x), for x ($\frac{\pi}{4}$,$\frac{\pi}{2}$)

Answer 1

Answer: (C)

– (cos x + sin x), for x Ī (0, $\frac{\pi}{4}$)

Answer 2

f (x) = $\sqrt{1 - \sin 2x}$

Ž f(x) = | sin x – cos x|

Now see, in the intervals, which one of sin x and cos x is greater .