Let a differentiable function f satisfy the equation 036 f ( tx36 )dt = 4 f(x). If y = f(x) is a standard parabola passing through the points (2, 1) and (-4, ), then is equal to .
Answer
64
Solution
Put tx36 = u, so dt = 36xdu and the limits t: 0 36 become u: 0 x. The equation becomes
36x 0x f(u) du = 4 f(x)
9 0x f(u) du = x f(x)
Differentiating w.r.t. x:
9f(x) = f(x) + x f'(x)
f'(x)f(x) = 9 - 1x
f(x) = cx 9 -
Since it is a standard parabola y = cx2, 9 - = 2, so = 3.
It passes through (2, 1), so 1 = 4c, i.e. c = 14 and the parabola is x2 = 4y; through (-4, ) this gives = 164 = 4.
= 43 = 64
Answer: 64