Let f be a differentiable function satisfying f(x) = 1 - 2x + 0x e(x-t) f(t) dt, x R and let g(x) = 0x (f(t) + 2)15(t - 4)6(t + 12)17 dt, x R. If p and q are respectively the points of local minima and local maxima of g, then the value of |p + q| is equal to .
Answer
9
Solution
f(x) = 1 - 2x + ex 0x e-t f(t) dt, and f(0) = 1
Differentiating: f'(x) = -2 + ex 0x e-t f(t) dt + f(x)
= -2 + [f(x) - 1 + 2x] + f(x)
f'(x) - 2f(x) = 2x - 3
Try f(x) = ax + b: a - 2ax - 2b = 2x - 3 a = -1, b = 1
General solution f(x) = 1 - x + Ce2x; f(0) = 1 C = 0
f(x) = 1 - x
g'(x) = (3 - x)15(x - 4)6(x + 12)17
(x - 4)6 0 (no sign change at x = 4); sign changes only at x = -12 and x = 3:
x < -12: g' < 0; -12 < x < 3: g' > 0; x > 3: g' < 0
local minimum at x = -12, local maximum at x = 3
p = -12, q = 3 |p + q| = 9
Answer: 9