Question

The value of c in Lagrange's mean value theorem for $f\left(x\right)=lnx$ on $[1,e]$ is

• A. $\frac{e}{2}$
• B. $e-1$
• C. $e-2$
• D. $1-e$

Answer 1

Answer: (B)

$e-1$

$f\left(x\right)=\ln \left(x\right)$ is continuous on $[1,e]$ and differentiable on $(1,e)$.

Hence application of Lagrange's mean value theorem gives us

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ where $a<c<b$.

Here $b=e$ and $a=1$.

Hence,

$f^{\prime }\left(c\right)=\frac{\ln e-\ln 1}{e-1}$

$\frac{1}{c}=\frac{1}{e-1}$

$\Rightarrow c=e-1$