Eliminating the arbitrary constants B and C in the expression y=2/{3C}(Cx-1)^{3/2}+B,we get [NDA II 2014]

Correct Answer is : 2x(dydx)d2ydx2=1+(dydx)2

x[1+(\frac{dy}{dx})^2]=\frac{d^2y}{dx^2}

2x(\frac{dy}{dx})\frac{d^2y}{dx^2}=1+(\frac{dy}{dx})^2

(\frac{dy}{dx})\frac{d^2y}{dx^2}=1

(\frac{dy}{dx})^2+1=\frac{d^2y}{dx^2}

UPSC NDA II 2014 Mathematics Paper

Question : 87 of 120

Marks: +1, -0

Eliminating the arbitrary constants B and C in the expression y=

(Cx−1)

+B,we get

[NDA II 2014]

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