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
Putting x = 1 gives 0 = 3f(1) + 1 - 4, so 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
Since f(1) = 1, 1 = -1 + c, so c = 2 and f(x) = 2x - x2.
Thus f(2) = 0 and f(3) = 6 - 9 = -3.
f(2) - f(3) = 0 - (-3) = 3
Answer: (2)