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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $y=\sin \left(2 \sin ^{-1} x\right)$, then $\left(1-x^2\right) y_2$ is equal to
  • A
    $-x y_1+4 y$
  • B
    $-x y_1-4 y$
  • ✓
    $x y_1-4 y$
  • D
    $x y_1+4 y$
✓

Answer

Correct option: C.
$x y_1-4 y$
$\text { We have, } y=\sin \left(2 \sin ^{-1} x\right)$
$\Rightarrow y=\sin \left[\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)\right]$
$\Rightarrow y=2 x \sqrt{1-x^2}.........(i)$
$\Rightarrow y_1=2 x \times \frac{-2 x}{2 \sqrt{1-x^2}}+2 \sqrt{1-x^2}=\frac{-4 x^2+2}{\sqrt{1-x^2}}.......(ii)$
$\therefore y_2=\frac{\sqrt{1-x^2}(-8 x)-\left(-4 x^2+2\right) \times \frac{-2 x}{2 \sqrt{1-x^2}}}{1-x^2}$
$\quad=\frac{4 x^3-6 x}{\left(1-x^2\right) \sqrt{1-x^2}} \Rightarrow\left(1-x^2\right) y_2=\frac{4 x^3-6 x}{\sqrt{1-x^2}}$
Now, consider $x y_1-4 y$
$=\frac{-4 x^3+2 x}{\sqrt{1-x^2}}-8 x \sqrt{1-x^2}$
$=\frac{4 x^3-6 x}{\sqrt{1-x^2}}$
$[$Using $(i)$ and $(ii)]$
Thus, $\left(1-x^2\right) y_2=x y_1-4 y$

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