Question

If $f\left(x\right)=\log x$ find all numbers strictly between $e^2$ and $e^3$ such that $f^{\prime }\left(x\right)=\frac{f\left(e^3\right)-f\left(e^2\right)}{e^3-e^2}$

• A. There exists only one value and it is equal to $(e^3-e^2)$
• B. There exists only one value and it is equal to $\left(e^3\right)$
• C. There exists no value in the given interval
• D. none of these

Answer 1

The correct option is A There exists only one value and it is equal to $(e^3-e^2)$

Since $f$ is continuous and differentiable $\forall x>0$
$f$ will be continuous and differentiable in $(e^2,e^3)$
Thus using LMVT theorem there will exist some $c\in (e^2,e^3)$
Such that $f^{\prime }\left(c\right)=\frac{f\left(e^3\right)-f\left(e^2\right)}{e^3-e^2}$

$\Rightarrow \frac{d}{dx}\left(\log x\right){|}_{x=c}=\frac{\log e^3-\log e^2}{e^3-e^2}$

$\Rightarrow \frac{1}{x}{|}_{x=c}=\frac{3-2}{e^3-e^2}$

$\Rightarrow \frac{1}{c}=\frac{1}{e^3-e^2}\Rightarrow c=e^3-e^2$

Also since $f$ is a monotonic function, hence there will exist only one such $c$.