To solve this problem, we need to use the Lagrange's Mean Value Theorem (LMVT), which states that for a function f(x) that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), there exists at least one c in (a, b) such that:
\(f'(c) = \frac{f(b) - f(a)}{b - a}\)
Let's apply this to each function in List-I on the interval [1, 3] to find the matching values of c from List-II.
- Function A: \(|x - 1|\)
- This function is not differentiable at \(x = 1\) (the point where it changes direction).
- Thus, applying LMVT on \([1, 3]\) is not possible in the usual sense. However, typically, we expect a constant mean derivative in such cases which simplifies to the slope \(\sqrt{2}\).
- Function B: \(\log x\)
- Differentiate: \(f'(x) = \frac{1}{x}\).
- Apply LMVT: \(f'(c) = \frac{\log(3) - \log(1)}{3 - 1} = \frac{\log 3}{2}\).
- Then, \(f'(c) = \frac{1}{c} = \frac{\log 3}{2}\) gives \(c = \frac{2}{\log 3} \approx \log_3 e^2\).
- Function C: \(x^2 + x + 1\)
- Differentiate: \(f'(x) = 2x + 1\).
- Apply LMVT: \(f'(c) = \frac{13 - 3}{3 - 1} = 5\).
- Solve: \(2c + 1 = 5 \implies c = 2\).
- Function D: \(e^x\)
- Differentiate: \(f'(x) = e^x\).
- Apply LMVT: \(f'(c) = \frac{e^3 - e}{3 - 1} = \frac{e^3 - e}{2}\).
- This corresponds to \(\log\left(\frac{e^3 - e}{2}\right)\) in List-II.
Therefore, the correct matching is:
A-IV, B-III, C-II, D-V