2012/030) x^2 = at^2 + b. Acceleration is proportional to:?

a) 1/x^3

b) 1/x - 1/x^2

c) - t/x^2

d) None

Answer:
acceleration is d^2x/dt^2

so differentiate both sides to get

xdx/dt = 2at

so differetiate again to get

x d^2x/dt^2 + (dx/dt)^2 = 2a

or x d2x/dt^2 + (2at/x)^2 = 2a
or d^2x/dt^2 + 4at^2/x^3 = 2a/x
or d^x /dt^2 = (2ax^2 - 4a^2t^2)/x^2
= 2a (x^2-2at^2)/x^3
= 2ab/x^3

hence ans is a)