Let y=y(x) be a solution of the differential equation, 1-x2dydx+1-y2=0,|x|<1 . If y12=32, then y-12 is equal to:

dy1-y2+dx1-x2=0⇒sin-1y+sin-1x=c At x=12,y=32⇒ c=π2⇒sin-1y=cos-1x Hence y12=sincos-112=12

Let y=y(x) be a solution of the differential equation, 1-x2dydx+1-y2=0,|x|<1 . If y12=32, then y-12 is equal to:

dy1-y2+dx1-x2=0⇒sin-1y+sin-1x=c At x=12,y=32⇒ c=π2⇒sin-1y=cos-1x Hence y12=sincos-112=12

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$Let y = y (x) be a solution of the differential equation, \sqrt{1 - x^{2}} \frac{dy}{dx} + \sqrt{1 - y^{2}} = 0, | x | < 1 . If$ $y (\frac{1}{2}) = \frac{\sqrt{3}}{2}, then y (\frac{- 1}{\sqrt{2}}) is equal to:$

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$- \frac{1}{\sqrt{2}}$

$\frac{\sqrt{3}}{2}$

$- \frac{\sqrt{3}}{2}$

$\frac{1}{\sqrt{2}}$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{d y}{\sqrt{1 - y^{2}}} + \frac{d x}{\sqrt{1 - x^{2}}} = 0 \Rightarrow \sin^{- 1} y + \sin^{- 1} x = c \\ At x = \frac{1}{2}, y = \frac{\sqrt{3}}{2} \Rightarrow c = \frac{π}{2} \Rightarrow \sin^{- 1} y = \cos^{- 1} x \\ Hence y (\frac{1}{\sqrt{2}}) = \sin (\cos^{- 1} \frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} \end{array}$