If y= (2 ^ -1 x ), then (1-x^2 ) y_2 is equal to - Vidyadip

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
$-x y_1-4 y$
C
$x y_1-4 y$
D
$x y_1+4 y$

✓

Answer

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

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