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,
\[\begin{align}
  & f'\left( x \right)=\dfrac{d}{dx}\left( {{\log }_{e}}x \right) \\
 & =\dfrac{1}{x} \\
\end{align}\]
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,
\[\begin{align}
  & 2\times 1=c\times {{\log }_{e}}\dfrac{3}{1} \\
 & =c{{\log }_{e}}\left( \dfrac{3}{1} \right)=2 \\
\end{align}\]
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.