Question

Which of the following statements are true?
(A) For f(x) = |x|, for all x in [-1, 2]; Lagrange's mean value theorem is satisfied
(B) For f(x) = cosx, for all x in [0, \(\pi\)/2]; Lagrange's mean value theorem is satisfied
(C) For f(x) = \( \frac{1}{x} \), for all x in [-1, 2]; Lagrange's mean value theorem is satisfied
(D) For f(x) = x(x-1)(x-2), for all x in [0, 1/2]; Lagrange's mean value theorem is satisfied
(E) For f(x) = \(x^{1/3}\), for all x in [-1, 1]; Lagrange's mean value theorem is satisfied
Choose the correct answer from the options given below:

• A. (A), (D) and (E) only
• B. (A), (B) and (D) only
• C. (B) and (D) only
• D. (B), (D) and (E) only

Hint:

To quickly check for the applicability of Mean Value Theorems (Rolle's or Lagrange's), look for "problem spots" in the given interval:
Points where the function is undefined (e.g., division by zero).
Points where the function might not be differentiable (e.g., sharp corners like in |x|, or vertical tangents like in \(x^{1/3}\)).
If any such point lies within the interval, the theorem's conditions are likely violated.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Lagrange's Mean Value Theorem (MVT) states that if a function \(f\) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \(c\) in \((a, b)\) such that \(f'(c) = \frac{f(b)-f(a)}{b-a}\). We need to check if these two conditions (continuity and differentiability) hold for each function on its given interval.
Step 2: Detailed Explanation:

(A) f(x) = |x| in [-1, 2]: The function \(f(x) = |x|\) is continuous on [-1, 2]. However, it is not differentiable at x = 0, which lies in the open interval (-1, 2). Since the differentiability condition fails, the MVT is not guaranteed to apply. Statement (A) is false.
(B) f(x) = cosx in [0, \(\pi\)/2]: The function \(f(x) = \cos x\) is a trigonometric function that is continuous and differentiable for all real numbers. Therefore, it is certainly continuous on [0, \(\pi\)/2] and differentiable on (0, \(\pi\)/2). Both conditions are met. Statement (B) is true.
(C) f(x) = \( \frac{1}{x} \) in [-1, 2]: The function \(f(x) = 1/x\) is not defined at x = 0, which lies in the interval [-1, 2]. Therefore, the function is not continuous on the closed interval. The MVT does not apply. Statement (C) is false.
(D) f(x) = x(x-1)(x-2) in [0, 1/2]: This is a polynomial function (\(f(x) = x^3 - 3x^2 + 2x\)). Polynomials are continuous and differentiable everywhere. Thus, the conditions for MVT are satisfied on the interval [0, 1/2]. Statement (D) is true.
(E) f(x) = \(x^{1/3}\) in [-1, 1]: The function \(f(x) = x^{1/3}\) is continuous on [-1, 1]. Its derivative is \(f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}\). This derivative is not defined at x = 0, which lies in the open interval (-1, 1). The function has a vertical tangent at x=0. Since the differentiability condition fails, the MVT is not guaranteed to apply. Statement (E) is false.
Step 3: Final Answer:
Only statements (B) and (D) are true. Therefore, the correct option is (C).