Let f : [1, ) R be a differentiable function. If 6 1x f(t) dt = 3xf(x) + x3 - 4 for all x 1, then the value of f(2) - f(3) is :
- 4
- 3
- -4
- -3
Answer
(2) 3
Solution
Put x = 1: 0 = 3f(1) + 1 - 4
f(1) = 1
Differentiating w.r.t. x (Leibnitz rule):
6f(x) = 3f(x) + 3xf'(x) + 3x2
xf'(x) - f(x) = -x2
xf'(x) - f(x)x2 = -1
ddx ( f(x)x ) = -1
f(x)x = -x + c
f(1) = 1 1 = -1 + c c = 2
f(x) = 2x - x2
f(2) = 0, f(3) = 6 - 9 = -3
f(2) - f(3) = 0 - (-3) = 3
Answer: (2)