Question

If $f\left(x\right)=\sqrt{1-sin2x},$ then $f^{\prime }\left(x\right)$ is equal to :

• A. $-(cosx+sinx),$ for $x\in (\pi /4,\pi /2)$
• B. $(cosx+sinx),$ for $x\in (0,\pi /4)$
• C. $-(cosx+sinx),$ for $x\in (0,\pi /4)$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $-(cosx+sinx),$ for $x\in (\pi /4,\pi /2)$
B $(cosx+sinx),$ for $x\in (0,\pi /4)$
C $-(cosx+sinx),$ for $x\in (0,\pi /4)$

Given

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

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

$=\sqrt{{(\cos x-\sin x)}^2}$

$\therefore f\left(x\right)=|\cos x-\sin x|$

$\therefore f^{\prime }\left(x\right)=\frac{|\cos x-\sin x|}{\cos x-\sin x}(-\sin x-\cos x)$

$xϵ(0,\frac{\pi }{4}),\sin x<\cos x$

$\therefore \cos x-\sin x>0$

$\therefore f^{\prime }\left(x\right)=-(\cos x+\sin x)$

$xϵ(\frac{\pi }{4},\frac{\pi }{2}),\sin x>\cos x$

$\therefore \cos x-\sin x<0$

$\therefore f^{\prime }\left(x\right)=(\cos x+\sin x)$