CUET Preparation Today
Let $f(x)=e^x, x \in[0,1]$ then a number $c$ of the Lagrange's mean value theorem is |
$\log _e(e-1)$ $\log _e(e+1)$ $\log _e e$ none of these |
$\log _e(e-1)$ |
Clearly, $f(x)$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Therefore, there exists $c \in(0,1)$ such that $f^{\prime}(c)=\frac{f(1)-f(0)}{1-0} \Rightarrow e^c=e-1 \Rightarrow c=\log _e(e-1)$ |
