Question

If mean value theorem holds for the function $f\left(x\right)=logx$ on[1,e], then c=

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

Answer 1

The correct option is A e-1

Given, the mean value theorem holds for the function $f\left(x\right)=\log x$ on $[1,e]$

$\therefore f\left(1\right)=\log 1=0$

$f\left(e\right)=\log e=1$

& $f^{{}^{\prime }}\left(x\right)=\frac{1}{x}$

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

By Lagrange's mean value theorem we know,

$\frac{f\left(e\right)-f\left(1\right)}{e-1}=\frac{1}{c}[\because \frac{f\left(b\right)-f\left(a\right)}{b-a}=f^{{}^{\prime }}(c\left)\right]\Rightarrow \frac{1}{e-1}=\frac{1}{c}\Rightarrow c=e-1$