Let f : [a, b] → R be differentiable on [a, b] and k ∈ R. Let

1.	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)
Answer» The correct option (C) J(x) = 0 has at least one root in (a, b) Explanation: Let g(x) = e^kx f(x) f(a) = 0 = f(b) by rolles theorem g'(c) = 0, c ∈ (a, b) g'(x) = e^kxf'(x) + ke^kxf(x) g'(c) = 0 e^kc(f'(c) + kf(c) = 0 ⇒ f'(c) + kf(c) = 0 for at least one c in (a, b)

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)

Answer»

The correct option (C) J(x) = 0 has at least one root in (a, b)

Explanation:

Let g(x) = e^kx f(x)

f(a) = 0 = f(b)

by rolles theorem

g'(c) = 0, c ∈ (a, b)

g'(x) = e^kxf'(x) + ke^kxf(x)

g'(c) = 0

e^kc(f'(c) + kf(c) = 0

⇒ f'(c) + kf(c) = 0 for at least one c in (a, b)

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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) &gt; 0 for all x ∈ [a, b] (B) J(x) &lt; 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)

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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)