Q76 / 150 · MHT CET 2022 (05 Aug Shift 1) · Mathematics

$\frac{d}{d x}\left(\sin ^{-1}\left(3 x-4 x^3\right)\right), \frac{1}{2}<x<1$ is

  • A
    $\frac{1}{3 \sqrt{1-x^2}}$
  • Correct answer
    B
    $\frac{-3}{\sqrt{1-x^2}}$
  • C
    $\frac{-1}{3 \sqrt{1-x^2}}$
  • D
    $\frac{3}{\sqrt{1-x^2}}$

Solution

$\begin{aligned} & \frac{\mathrm{d}\left(\sin ^{-1}\left(3 \mathrm{x}-4 \mathrm{x}^3\right)\right)}{\mathrm{dx}} \\ & =\frac{\mathrm{d}\left(\pi-3 \sin ^{-1} \mathrm{x}\right)}{\mathrm{dx}}\left[\text { as } \frac{1}{2}<\mathrm{x}<1\right]=\frac{-3}{\sqrt{1-\mathrm{x}^2}}\end{aligned}$

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

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