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Question: Let ƒ(x) and g(x) be defined and differentiable for x ³ x<sub>0</sub> and ƒ(x<sub>0</sub>) = g(x<sub...

Let ƒ(x) and g(x) be defined and differentiable for x ³ x0 and ƒ(x0) = g(x0),ƒ ¢ (x) > g ¢ (x) for x > x0, then –

A

ƒ(x) < g(x) for some x > x0

B

ƒ(x) = g(x) for some x > x0

C

ƒ(x) > g(x) for some x > x0

D

None of these

Answer

ƒ(x) > g(x) for some x > x0

Explanation

Solution

Consider the function f(x) = ƒ(x) – g(x). On the interval x0, x. Then, f(x) satisfies all the conditions of Lagrange’s mean value theorem on x0, x.

There exists at least one c Ī (x0, x) such that

f(x) – f(x0) = f ¢ (3) (x – x0)

Ž f(x) = f ¢ (3) (x – x0) (Q f(x0) = 0) … (1)

Also, f¢(x) = ƒ ¢ (x) – g ¢ (x)

Ž f ¢ (3) = ƒ ¢ (3) – g ¢ (3) > 0

(Q ƒ ¢ (x) > g ¢ (x) for x > x0)

from (1), f(x) > 0, for x > x0

Ž ƒ(x) – g(x) > 0 for x > x0

or ƒ(x) > g(x) for x > x0 .