Question

In each of the following options, a function and an interval are given. Choose the option containing the function and the interval for which Lagrange’s Mean Value Theorem is not applicable.

• A. \( f(x) = |x| \), \( 1 \leq x \leq 5 \)
• B. \( f(x) = [x] \), \( [\sqrt{2}, \sqrt{3}] \)
• C. \( f(x) = \log (x^2 - 1) \), \( \left[\frac{1}{e}, e-2\right] \)
• D. \( f(x) = e^x \), \( [-e, e] \)

Hint:

For LMVT, check both continuity and differentiability. Discontinuities or singularities violate the conditions.

Answer 1

The Correct Option is C

Step 1: Conditions for Lagrange’s Mean Value Theorem (LMVT) LMVT is applicable if: 1. The function is continuous in the given interval. 2. The function is differentiable in the open interval. Step 2: Check continuity and differentiability 1. \( f(x) = |x| \) is continuous and differentiable in \( 1 \leq x \leq 5 \). 2. \( f(x) = [x] \) (greatest integer function) is discontinuous at non
-integer points. 3. \( f(x) = \log (x^2
- 1) \) is undefined for \( x^2
- 1 \leq 0 \Rightarrow x \in (
-1,1) \).
- Since \( x^2
- 1 = 0 \) at \( x = \pm 1 \), it is not continuous in \( \left[\frac{1}{e}, e
-2\right] \). 4. \( f(x) = e^x \) is continuous and differentiable for all real numbers. Step 3: Conclusion Since \( f(x) = \log (x^2
- 1) \) is not defined at \( x = 1 \), LMVT is not applicable in the given interval.