Let f : [a, b] → R be differentiable on [a, b] and k ∈ R. Let f(a) = 0 = f(b). Also let J(x) = f'(x) + kf(x). Then
(A) J(x) > 0 for all x ∈ [a, b]
(B) J(x) < 0 for all x ∈ [a, b]
(C) J(x) = 0 has at least one root in (a, b)
(D) J(x) = 0 through (a, b)
The correct option (C) J(x) = 0 has at least one root in (a, b)
Explanation:
Let g(x) = ekx f(x)
f(a) = 0 = f(b)
by rolles theorem
g'(c) = 0, c ∈ (a, b)
g'(x) = ekxf'(x) + kekxf(x)
g'(c) = 0
ekc(f'(c) + kf(c) = 0
⇒ f'(c) + kf(c) = 0 for at least one c in (a, b)