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B
−12at2
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C
−1t3
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D
−12at3
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Solution
The correct option is D−12at3 We have x=at2,y=2at ⇒dxdt=2at=y,dydt=2a Thus dydx=dy/dtdx/dt=2ay ⇒d2ydx2=2a(−1y2)dydx=2a(−1y2)2ay=−(2a)2y3=−(2a)2(2at)3=−12at3