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: Concept Review:
Lagrange's Mean Value Theorem (MVT) requires a function \(f\) to be continuous on \([a, b]\) and differentiable on \((a, b)\). If these conditions are met, there exists \(c \in (a, b)\) such that \(f'(c) = \frac{f(b)-f(a)}{b-a}\). We must verify these conditions for each given function and interval.Step 2: Analysis of Each Case:

(A) f(x) = |x| in [-1, 2]: Continuous on [-1, 2]. Not differentiable at x = 0, which is in (-1, 2). MVT does not apply. Statement (A) is false.
(B) f(x) = cosx in [0, \(\pi\)/2]: Continuous and differentiable on [0, \(\pi\)/2] and (0, \(\pi\)/2) respectively, as it's a standard trigonometric function. MVT applies. Statement (B) is true.
(C) f(x) = \( \frac{1}{x} \) in [-1, 2]: Not defined at x = 0, which is in [-1, 2]. Thus, it's not continuous on the closed interval. MVT does not apply. Statement (C) is false.
(D) f(x) = x(x-1)(x-2) in [0, 1/2]: This polynomial function (\(f(x) = x^3 - 3x^2 + 2x\)) is continuous and differentiable everywhere. MVT applies. Statement (D) is true.
(E) f(x) = \(x^{1/3}\) in [-1, 1]: Continuous on [-1, 1]. Derivative \(f'(x) = \frac{1}{3\sqrt[3]{x^2}}\) is undefined at x = 0, which is in (-1, 1). MVT does not apply due to lack of differentiability. Statement (E) is false.
Step 3: Conclusion:
Statements (B) and (D) are true. The correct option is (C), which implies that the set of true statements is {B, D}.