Problem 46 - A Complete Resource Book in Mathematics for JEE Main 2017

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- \(\mathbf{2}: f\) 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) Statement-1 is true, Statement-2 is true, and Statement-2 is a correct explanation for Statement-1.

Step 1: Differentiate the Function

To find the critical points of \(f(x)\), differentiate \(f(x) = x^2 - 8x + 12\). The derivative is \(f'(x) = 2x - 8\).

Step 2: Solve for Critical Points

Set the derivative equal to zero to find critical points: \(2x - 8 = 0\). Solving for \(x\), we get \(x = 4\).

Step 3: Verify Critical Point Within Given Interval

Check if \(x = 4\) lies within the interval \((2, 6)\). Since \(4\) is within \( (2,6) \), Statement-1 is true.

Step 4: Analyze Continuity and Differentiability

The function \(f(x) = x^2 - 8x + 12\) is a polynomial, which is continuous on \([2,6]\) and differentiable on \((2,6)\).

Step 5: Check Function Values at Interval Endpoints

Calculate \(f(2)\) and \(f(6)\): \(f(2) = 2^2 - 8\cdot2 + 12 = 4\) and \(f(6) = 6^2 - 8\cdot6 + 12 = 4\). Since \(f(2) = f(6)\), Statement-2 is true.

Step 6: Apply Rolle's Theorem

Rolle's Theorem can be applied because \(f(x)\) is continuous on \([2,6]\), differentiable on \((2,6)\), and \(f(2) = f(6)\). Therefore, there exists some \(c\) where \(f'(c) = 0\). This confirms that Statement-2 is a correct explanation for Statement-1.

Key Concepts

Continuity

Continuity is a crucial concept when discussing Rolle's Theorem and other fundamental calculus ideas. Simply put, a function is continuous on a given interval if there are no breaks, jumps, or gaps in the graph of the function across that interval. For a function to be continuous at a particular point, say \(x = a\), three conditions must be satisfied:

• The function \(f(x)\) must be defined at \(x = a\).
• The limit of \(f(x)\) as \(x\) approaches \(a\) must exist.
• The limit of \(f(x)\) as \(x\) approaches \(a\) must equal the function value \(f(a)\).

Polynomials, like the function \(f(x) = x^2 - 8x + 12\) from the exercise, are always continuous across any interval because they do not have any breaks or discontinuities. This means that on the interval \([2, 6]\), the function is continuous, satisfying a key requirement of Rolle's Theorem. Understanding continuity helps us know that if we travel from one endpoint of an interval to the other without lifting a pencil from the graph, the function is continuous.

Differentiability

Differentiability is another core aspect of Rolle's Theorem. For a function to be differentiable at a particular point, it must have a defined derivative at that point. In simpler terms, the function must have a tangent line that we can compute at that point. Differentiability is generally a sign of smoothness in a graph, meaning no sharp corners or cusps exist.Polynomials, such as our function \(f(x) = x^2 - 8x + 12\), are not only continuous everywhere but also differentiable everywhere. This is because polynomials are smooth functions without sharp turns. In the interval \((2, 6)\), the function \(f(x)\) is differentiable, confirming that it meets another criterion for applying Rolle's Theorem. Differentiability ensures that we can find the rate of change smoothly across the interval, which is possible for polynomials.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are important because they can indicate maximums, minimums, or points of inflection on the function's graph. For the function \(f(x) = x^2 - 8x + 12\), we found the derivative to be \(f'(x) = 2x - 8\).Setting the derivative equal to zero, \(2x - 8 = 0\), and solving for \(x\) gives a critical point at \(x = 4\). Since this point is within the interval \((2, 6)\), it fulfills the conditions of Rolle's Theorem, which guarantees that there exists at least one \(c\) in the interval such that \(f'(c) = 0\). Understanding critical points is essential in calculus for identifying where functions change direction and their overall behavior.