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Q.

cos−1yb=2logx2,x>0⇒x2d2ydx2+xdydx=

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a

4y

b

-4y

c

0

d

8y

answer is B.

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Detailed Solution

cos−1⁡yb=2log⁡x2⇒−11−y2/b2×1bdydx=2x⇒−1b2−y2dydx=2x⇒xdydx=−2b2−y2⇒x2y12=4b2−y2⇒x22y1y2+2xy12=−8yy1⇒x2y2+xy1=−4y
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cos−1yb=2logx2,x>0⇒x2d2ydx2+xdydx=