Q87 / 150 · MHT CET 2021 (20 Sep Shift 1) · Mathematics

If $x \in\left(0, \frac{\pi}{2}\right)$ and $x$ satisfies the equation $\sin x \cos x=\frac{1}{4}$, then the values of $x$ are

  • Correct answer
    A
    $\frac{\pi}{12}, \frac{5 \pi}{12}$
  • B
    $\frac{\pi}{8}, \frac{3 \pi}{8}$
  • C
    $\frac{\pi}{8}, \frac{\pi}{4}$
  • D
    $\frac{\pi}{6}, \frac{\pi}{12}$

Solution

$\begin{aligned} & \sin x \cos x=\frac{1}{4} \\ & \therefore 2 \sin x \cos x=\frac{2}{4}=\frac{1}{2} \Rightarrow \sin 2 x=\frac{1}{2}=\sin 30^{\circ} \\ & \therefore 2 x=30^{\circ}=\frac{\pi}{6} \Rightarrow x=\frac{\pi}{12} \\ & \therefore x=\frac{\pi}{2} \pm \frac{\pi}{12} \Rightarrow x=\frac{\pi}{2}+\frac{\pi}{12} \text { or } x=\frac{\pi}{2}-\frac{\pi}{12} \\ & \therefore x=\frac{7 \pi}{12} \text { or } \frac{5 \pi}{12} \Rightarrow x=\frac{\pi}{12}, \frac{\pi}{12} \quad \ldots\left[\because x \in\left(0, \frac{\pi}{2}\right)\right]\end{aligned}$

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

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