Problem 46 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=x^{2}-8 x+12, x \in[2,6]\). Statement-1: \(f^{\prime}(c)=0\) for some \(c \in(2,6)\) Statement-2: \(f\) is continuous on \([2,6]\) and differentiable on \((2,6)\) with \(f(2)=f(6)\) (A) Statement-1 is true, Statement- 2 is true, Statement- 2 is a correct explanantion for Statement- 1 (B) Statement-1 is true, Statement-2 is true, Statement- 2 is not a correct expalantion for Statement-1 (C) Statement-1 is true, Statement- 2 is false (D) Statement-1 is false, Statement- 2 is true

Step-by-Step Solution

Answer

A: Both statements are true, and Statement-2 explains Statement-1.

Step 1: Verify Continuity and Differentiability of f(x)

The function \(f(x) = x^2 - 8x + 12\) is a polynomial. Polynomial functions are continuous and differentiable everywhere on the real line. Therefore, \(f(x)\) is continuous on \([2,6]\) and differentiable on \((2,6)\).

Step 2: Check Values at the Endpoints and Find f(2) and f(6)

Calculate \(f(2)\) and \(f(6)\):\[f(2) = 2^2 - 8 \cdot 2 + 12 = 4 - 16 + 12 = 0.\]\[f(6) = 6^2 - 8 \cdot 6 + 12 = 36 - 48 + 12 = 0.\]Since \(f(2) = f(6) = 0\), statement-2 is verified as true.

Step 3: Apply Rolle's Theorem

Rolle's Theorem states that if a function is continuous on \([a, b]\), differentiable on \((a, b)\), and \(f(a) = f(b)\), then there exists at least one \(c \in (a, b)\) such that \(f'(c) = 0\).Given \(f(2) = f(6)\), and that it's continuous and differentiable as per Step 1, Rolle's Theorem guarantees the existence of some \(c \in (2, 6)\) such that \(f'(c) = 0\).

Step 4: Find f'(x) and Solve for c

Compute the derivative \(f'(x)\):\[f'(x) = \frac{d}{dx}(x^2 - 8x + 12) = 2x - 8.\]Set \(f'(x) = 0\) to find \(c\):\[2x - 8 = 0 \2x = 8 \x = 4.\]Since \(4 \in (2, 6)\), the value satisfies the condition in Rolle's Theorem, confirming statement-1 as true.

Key Concepts

Polynomial Continuity

Understanding polynomial continuity is essential, especially when dealing with theories like Rolle's Theorem. Polynomials are some of the most reliable functions we encounter in calculus because they are continuous everywhere on their domain.
This means you can draw the graph of a polynomial without lifting your pencil from the paper, implying no gaps, jumps, or interruptions.
Within any closed interval, such as \([2, 6]\) in this exercise, polynomials maintain their continuity, ensuring a smooth graph from start to finish.

• Polynomials: defined as expressions involving terms made up of variables and coefficients.
• Continuity: no abrupt changes or interruptions in the function.
• Relevance: guarantees that conditions necessary for applying Rolle's Theorem are met.

In this problem, since \(f(x) = x^2 - 8x + 12\) is a polynomial, it is assuredly continuous on \([2, 6]\). This satisfies one of the crucial criteria for Rolle's Theorem.

Differentiability

Differentiability is a concept closely related to continuity. While continuity ensures a function doesn’t have breaks, differentiability tells us that the function has a well-defined tangent at every point in the interval.
For polynomials, just as with continuity, differentiability doesn’t pose issues as they can be differentiated at any point within their domains.
The significance of differentiability in the context of Rolle’s Theorem is that it ensures smooth transitions between small changes in input. To be specific:

• The function must not only remain uninterrupted but also have smooth, non-sharp turns.
• Ensures the existence of a derivative, like \(f'(x) = 2x - 8\) in this problem.
• Enables us to find such points where the slope of the tangent (derivative) is zero.

For \(f(x) = x^2 - 8x + 12\), the function is differentiable on \( (2, 6)\), meeting a condition for applying Rolle’s Theorem.

Function Endpoints Equality

For Rolle's Theorem to be applicable, a vital condition is that the function values at the endpoints of the interval must be equal. This implies that the function starts and finishes at the same height when viewed graphically.
In mathematical terms, if \(f(a) = f(b)\), then the criteria for Rolle’s Theorem checks out for this part.
This equality is significant because it sets a natural ground for expecting a zero-slope point (a point \(c\) where \(f'(c) = 0\)) between \(a\) and \(b\). In our given problem:

• Calculated that \(f(2) = 0\) and \(f(6) = 0\)
• Ensures \(f(2) = f(6)\), verifying the endpoint requirement.
• Enables us to assert the existence of a point where the derivative equals zero on \((2, 6)\).

Hence, having verified \(f(2) = f(6)\), the conditions for Rolle’s Theorem are perfectly set.