CUET Preparation Today
The value of $c$ for which the conclusion of mean value theorem-holds for the function $f(x)=\log _e x$ on the interval $[1,3]$, is |
$2 \log _3 e$ $\frac{1}{2} \log _e 3$ $\log _3 e$ $\log _e 3$ |
$2 \log _3 e$ |
Using Mean Value Theorem, we have $f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \Rightarrow \frac{1}{c}=\frac{1}{2}\{\log 3-\log 1\} \Rightarrow c=2 \log _3 e$ |
