Q246 / 270 · NDA 2018 (Phase 2) · Mathematics

What is $\frac{\mathrm{d} \sqrt{1-\sin 2 \mathrm{x}}}{\mathrm{dx}}$ equal to, where $\frac{\pi}{4} \lt \mathrm{x} \lt \frac{\pi}{2}$ ?

  • Correct answer
    A
    $\cos x+\sin x$
  • B
    $-(\cos x+\sin x)$
  • C
    $\pm(\cos x+\sin x)$
  • D
    None of the above

Solution

$\begin{aligned} & \text { For } \frac{\pi}{4} \lt x \lt \frac{\pi}{2} \\ & \sqrt{1-\sin 2 x}=\left(\sin ^2 x+\cos ^2 x+2 \sin x \cdot \cos x\right)^{1 / 2} \\ & =(\sin x-\cos x) \\ & \text {Differentiation }=\cos x+\sin x\end{aligned}$

Hence, the correct answer is option: 'A'.

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