Question

Given a function f : [0, 4] → R is differentiable, then for some a, b ∈ (0, 4) [f(4)]^2 – [f(0)]^2 =

• A. 8 f ′(2) f (1)
• B. 4 f ′(2) f (1)
• C. 2 f ′(2) f (1)
• D. f′(2) f(1)

Answer 1

Answer: (A)

8 f ′(2) f (1)

Answer 2

Since f(x) is differentiable in [0, 4], using Lagrange's Mean Value Theorem.

f′(2) = $\frac{f(4) - f(0)}{4}$, b ∈ (0, 4) …(1)

Now, {f(4)² – {f(0)}² = $\frac{4\{ f(4) - f(0)\}}{4}$ {f(4) + f(0)}

= 4f′(2) {f(4) + f(0)}… (2)

Also, from Intermediate Mean Value Theorem,

$\frac{f(4) + f(0)}{2}$ = f(1) for a ∈ (0, 4).

Hence, from (2) [f(4)²] – [f(0)]² = 8f′ (2) f(1).