Question

Let $f:[2,7] \rightarrow[0, \infty)$ be a continuous and differentiable function. Then, $(f(7)-f(2)) \frac{\left(f(7)^{2}+(f(2))^{2}+f(2) f(7)\right.}{3}$ is equal to (A) $5 f^{2}(c) f^{\prime}(C)$ (B) $5 f^{\prime}(c)$ (C) $f(c) f^{\prime}(C)$ (D) None of these where $c \in(2,7)$.

Name: Problem 56, Chapter 14: A Complete Resource Book in Mathematics for JEE Main
Uploaded: 2021-10-16T07:09:22-08:00
Duration: 3 min 16 s
Channel: Varsha Aggarwal
Description: Let f:[2,7] →[0, ∞) be a continuous and differentiable function. Then, (f(7)-f(2)) ((f(7)^2+(f(2))^2+f(2) f(7).)/(3) is equal to (A) 5 f^2(c) f^'(C) (B) 5 f^'(c) (C) f(c) f^'(C) (D) None of these where c ∈(2,7).

   Let $f:[2,7] \rightarrow[0, \infty)$ be a continuous and differentiable function. Then, $(f(7)-f(2)) \frac{\left(f(7)^{2}+(f(2))^{2}+f(2) f(7)\right.}{3}$ is equal to
(A) $5 f^{2}(c) f^{\prime}(C)$
(B) $5 f^{\prime}(c)$
(C) $f(c) f^{\prime}(C)$
(D) None of these where $c \in(2,7)$.

A Complete Resource Book in Mathematics for JEE Main

Dinesh Khattar 2019 Edition

Chapter 14, Problem 56

Instant Answer

Solved by Expert Varsha Aggarwal

10/16/2021

Step 1

Since $f(x)$ is continuous and differentiable, $g(x)$ will also be continuous and differentiable on the interval $[2,7]$. Show more…

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Let $f:[2,7] \rightarrow[0, \infty)$ be a continuous and differentiable function. Then, $(f(7)-f(2)) \frac{\left(f(7)^{2}+(f(2))^{2}+f(2) f(7)\right.}{3}$ is equal to (A) $5 f^{2}(c) f^{\prime}(C)$ (B) $5 f^{\prime}(c)$ (C) $f(c) f^{\prime}(C)$ (D) None of these where $c \in(2,7)$.

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Key Concepts

Mean Value Theorem

This theorem states that for a function that is continuous on a closed interval and differentiable on the open interval, there exists a point c in the open interval where the instantaneous rate of change (the derivative) equals the average rate of change over the interval. In this problem, the theorem is applied to a function derived from f to relate the difference f(7)^3 - f(2)^3 to f(c)^2 times f'(c), thereby connecting the values at the endpoints with the derivative at an intermediate point.

Difference of Cubes Factorization

The difference of cubes identity, which expresses a^3 - b^3 as (a - b)(a^2 + ab + b^2), is used here to rewrite the expression (f(7) - f(2)) multiplied by (f(7)^2 + f(2)^2 + f(2)f(7)). This factorization simplifies the given expression to (f(7)^3 - f(2)^3), enabling the application of the Mean Value Theorem on the transformed function.

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