Let f be a twice differentiable function such that f(x)= 0x (t-x) dt- 0x f(t) t dt, x (- 2, 2 ). Then f'' ( 6 )+12f' (- 6 )+f ( 6 ) is equal to .
Answer
5
Solution
Putting u = t - x, 0x (t - x) dt = -x0 u du = [- e u ]-x0 = e x
so f(x) = e x - 0x f(t) t dt, and f(0) = 0. Differentiating,
f'(x) = - x - f(x) x
f'(x) + f(x) x = - x
This is linear with I.F. = e x dx = x, so
f(x) x = - x x dx = - x + C
Since f(0) = 0, 0 = -1 + C, i.e. C = 1. Hence f(x) = x - 1, f'(x) = - x and f''(x) = - x, and
f'' ( 6 ) + 12f' (- 6 ) + f ( 6 ) = - 32 + 12 12 + 32 - 1
∴ The value is 5
Answer: 5