Question

The value of 'c' in Lagrange's mean value theorem for f(x) = logx on [1, e] is

• A. e/2
• B. e-1
• C. e-2
• D. 1-e

Answer 1

The correct option is A e/2

Given $f\left(x\right)-\log x{f}^{{}^{\prime }}\left(x\right)=\frac{1}{x}$

using Langrage's mean value theorm
$f^{{}^{\prime }}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\Rightarrow \frac{1}{c}=\frac{\log e-\log 1}{e-1}\Rightarrow \frac{1}{c}=\frac{1}{e-1}\Rightarrow c=e-1$

so option B is correct