Question

A value of $c$ for which the conclusion of mean value theorem holds for the function $f\left( x \right)={{\log }_{e}}x$ on individual $\left[ 1,3 \right]$is:
(a) $\dfrac{1}{2}{{\log }_{e}}3$
(b) ${{\log }_{3}}e$
(c) $\log 3$
(d) $2{{\log }_{3}}e$

Answer 1

Hint: In this question, we will apply the mean value theorem on a given function and solve it using properties of logarithmic functions.

Complete step-by-step answer:
Mean value theorem states that, if a function $f$ is continuous on the closed interval $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right)$, then there exist a point $c$ in the interval $\left( a,b \right)$ such that $f'\left( c \right)$ is equal to the functions average rate of change over $\left[ a,b \right]$.
That is,
$f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}...........(i)$
Now in a given equation, we have a function on intervals. $\left[ 1,3 \right]$
Here, ${{\log }_{e}}x$ function is continuous on the interval of $\left[ 1,3 \right]$ and also, the function is different suitable in the interval. $\left( 1,3 \right)$
So, we can apply the mean value theorem here.
Using equation(i), we have $a=1$ and $b=3$
So, from equation (i), we get,

Putting $f\left( x \right)={{\log }_{e}}x$, we get,
$f'\left( c \right)=\dfrac{{{\log }_{e}}3-{{\log }_{e}}1}{2}..........(ii)$
Also, differentiating $f'\left( x \right)={{\log }_{e}}x$, with respect of $x$, we get,

& f'\left( x \right)=\dfrac{d}{dx}\left( {{\log }_{e}}x \right) \\\ & =\dfrac{1}{x} \\\ \end{aligned}$$ Putting $$x=c$$ we get, $$f'\left( c \right)=\dfrac{1}{c}$$ Putting this value of $$f'\left( c \right)$$ in equation (ii), we get, $$\dfrac{1}{c}=\dfrac{{{\log }_{e}}3-{{\log }_{e}}1}{2}$$ We know that, $${{\log }_{e}}m-{{\log }_{e}}n={{\log }_{e}}\dfrac{m}{n}$$ Using this, we get, $$\dfrac{1}{c}=\dfrac{{{\log }_{e}}3\left( \dfrac{3}{1} \right)}{2}$$ Cross multiplying this equation we get, $$\begin{aligned} & 2\times 1=c\times {{\log }_{e}}\dfrac{3}{1} \\\ & =c{{\log }_{e}}\left( \dfrac{3}{1} \right)=2 \\\ \end{aligned}$$ Dividing $${{\log }_{e}}\left( 3 \right)$$ from both sides of the equation, we get, $$c=2\times \dfrac{1}{{{\log }_{e}}3}$$ From properties of log function, we have, $$\dfrac{1}{{{\log }_{n}}m}={{\log }_{m}}n$$ Therefore, $$\dfrac{1}{{{\log }_{e}}3}={{\log }_{3}}e$$ Using this value in above equation, we get, $$c=2{{\log }_{3}}e$$ Hence, the correct answer is option(d). Note:In this type of question, always first check the condition of mean value theorem. Like in this question, if the interval was $\left[ -1,1 \right]$, then we would have not applied this theorem.