Problem 56 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \(c \in(2,7)\).

Step-by-Step Solution

Answer

The expression equals Option B: \(5 f^{\prime}(c)\).

Step 1: Recognize the Mean Value Theorem for Derivatives

The Mean Value Theorem (MVT) for derivatives states that for a continuous and differentiable function \( f \) on the interval \([a,b]\), there exists at least one \( c \) in \( (a,b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). In this problem, \( a = 2 \) and \( b = 7 \).

Step 2: Apply the Mean Value Theorem

Applying the MVT, we get: \[ f'(c) = \frac{f(7) - f(2)}{7 - 2} = \frac{f(7) - f(2)}{5} \] There exists a \( c \) in \( (2,7) \) such that this equation holds.

Step 3: Analyze Given Expression

We are given the expression \((f(7) - f(2)) \frac{f(7)^{2} + (f(2))^{2} + f(2)f(7)}{3} \). This expression has the form of a product of differences and averages of squares.

Step 4: Simplify and Compare with Choices

Simplify the given expression:1. Note that \( f'(c) = \frac{f(7) - f(2)}{5} \) implies \( f(7) - f(2) = 5f'(c) \).2. Substitute in the expression:\[ (5f'(c)) \frac{f(7)^{2} + f(2)^{2} + f(2)f(7)}{3} = 5f'(c) \times \frac{f(7)^{2} + f(2)^{2} + f(2)f(7)}{3} \]3. Check the options: - Option A: \(5 f^{2}(c) f^{\prime}(C)\) - Option B: \(5 f^{\prime}(c)\) - Option C: \(f(c) f^{\prime}(C)\)Clearly, the expression transformed corresponds to Option B.

Key Concepts

Continuous function

A continuous function is one that does not have any breaks, jumps, or holes in its graph. Imagine you are drawing the graph of a continuous function without lifting your pencil from the paper. That's a great way to visualize continuity! If a function, say \( f(x) \), is continuous on an interval \([a, b]\), it essentially means that the function smoothly covers all values within that interval.
Here are a few key characteristics of continuous functions:

• You can find the value of the function at any point within the interval \([a, b]\) by simply substituting the point into \( f(x) \). There's no concern about the function being undefined.
• The Intermediate Value Theorem applies, meaning if \( f \) takes on different values at two points, then it also takes on every value in between.
• Continuous functions on a closed interval \([a, b]\) always attain a maximum and a minimum value.

Differentiable function

A differentiable function is one that has a derivative at each point within its interval. This means that not only is the function smooth (like being continuous), but its graph also doesn't have any sharp turns or cusps. The existence of a derivative implies that there's a definite slope or rate of change at every point.
Differentiable functions have several interesting properties:

• Every differentiable function over an interval is also continuous over that interval. This is crucial, as differentiability implies smoothness without any breaks or jumps.
• The Mean Value Theorem applies to differentiable functions. This theorem guarantees at least one point, \( c \), within the interval \((a, b)\), where the derivative \( f'(c) \) is equivalent to the average rate of change over \([a, b]\).
• Analytically, differentiability means that the slope between two points on the function approaches a finite number as the points get infinitely close together.

Interval notation

Interval notation is a way to describe a range of values along the real number line. It's particularly helpful in calculus and analysis for expressing domains, ranges, and certain conditions without verbose sentences.
Here's how interval notation works:

• A square bracket \([\ or \)] indicates inclusivity, meaning the endpoint is part of the interval, like \([a, b]\) includes \(a\) and \(b\).
• A parenthesis \((\ or \)) indicates exclusivity, meaning the endpoint is not part of the interval, like \((a, b)\) does not include \(a\) and \(b\).
• Interval notation often uses combinations of these symbols, such as \([a, b)\), where \(a\) is included, but \(b\) is not.

This notation is more efficient and clearer than word descriptions. For example, the interval \([2, 7]\) includes all numbers from 2 to 7, but to specify numbers strictly between 2 and 7, we use \((2, 7)\).