Question

Let ƒ : [2, 7] ® [0, ) be a continuous and differentiable function. Then

(ƒ (7) – ƒ (2)) $\frac{(ƒ(7))^{2} + (ƒ(2))^{2} + ƒ(2)ƒ(7)}{3}$ where c Ī [2, 7]

(1) 5 ƒ ² c ƒ¢ (3) (2) 5ƒ¢ (3) (3) ƒ(3) ƒ¢(3) (4) None

• A. 5 ƒ ² c ƒ¢ (3)
• B. 5 ƒ¢ (3)
• C. ƒ(3) ƒ¢(3)
• D. None

Answer 1

Answer: (A)

5 ƒ ² c ƒ¢ (3)

Answer 2

Let g(x) = ƒ³(x)

Ž g¢(x) = 3ƒ²(x) . ƒ¢ (x)

Q ƒ : [2, 7] ® [0, ) Ž g : [2, 7] ® [0, )

Using Lagrage’s Mean Value theorem on g(x) we get

g¢(3) = $\frac{g(7)–g(2)}{5}$ ; c Ī [2, 7]

Ž 5 ƒ²(3) ƒ¢(3) = (ƒ(7) – ƒ(2))

$\frac{\lbrack(ƒ(7))^{2} + (ƒ(2))^{2} + ƒ(2)ƒ(7)\rbrack}{3}$.

Hence (1) is the correct answer.