If Rolle’s theorem is applicable to the function $f\left( x \right)=\dfrac{\ln x}{x}$, $\left( x>0 \right)$ over the interval $\left[ a,b \right]$ where $a\in I$, $b\in I$, then find the possible value of ${{a}^{2}}+{{b}^{2}}$?
(a) 20
(b) 25
(c) 45
(d) 10

Hint: We start solving the problem by recalling the definition and conditions of Rolle’s theorem. We then check whether the given conditions are satisfied to apply Rolle’s theorem or not. After checking we apply $f\left( a \right)=f\left( b \right)$ to get our first condition. We then apply ${{f}^{'}}\left( c \right)=0$ to get our second condition. Using these two conditions, we find the values of a and b and substitute in ${{a}^{2}}+{{b}^{2}}$ to get the required result.

Complete step-by-step answer:
According to the problem, we have given that the Rolle’s theorem is applicable to the function $f\left( x \right)=\dfrac{\ln x}{x}$, $\left( x>0 \right)$ over the interval $\left[ a,b \right]$ where $a\in I$, $b\in I$. We need to find the possible values for ${{a}^{2}}+{{b}^{2}}$.
Let us recall the definition of Rolle’s theorem. If the given function $f\left( x \right)$ is continuous in the closed $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right)$ such that $f\left( a \right)=f\left( b \right)$, then there exists a $c\in \left( a,b \right)$ such that ${{f}^{'}}\left( c \right)=0$.
According to the problem, the function is differentiable in the open interval $\left( a,b \right)$.
We know a and b should be equal to each other.
Now, we have $f\left( a \right)=f\left( b \right)$.
$\Rightarrow \dfrac{\ln a}{a}=\dfrac{\ln b}{b}$.
$\Rightarrow b\ln a=a\ln b$.
We know that $x\ln y=\ln {{y}^{x}}$.
$\Rightarrow \ln {{a}^{b}}=\ln {{b}^{a}}$.
We know that if $\ln x=\ln y$, then $x=y$.
$\Rightarrow {{a}^{b}}={{b}^{a}}$---(1).
Now, we have ${{f}^{'}}\left( c \right)=0$.
Let us find ${{f}^{'}}\left( x \right)$ first.
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( \dfrac{\ln x}{x} \right)\].
We know that $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$.
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{x\dfrac{d}{dx}\left( \ln x \right)-\ln x\dfrac{d}{dx}\left( x \right)}{{{x}^{2}}}\].
We know that $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$ and $\dfrac{d}{dx}\left( x \right)=1$.
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{x\left( \dfrac{1}{x} \right)-\ln x\left( 1 \right)}{{{x}^{2}}}\].
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{1-\ln x}{{{x}^{2}}}\].
\[\Rightarrow {{f}^{'}}\left( c \right)=0\].
\[\Rightarrow \dfrac{1-\ln c}{{{c}^{2}}}=0\].
Since c lies between a and b, as these values are greater than zero then the value of c should be greater than zero. This makes the denominator greater than 0.
So, we have $1-\ln c=0$.
$\Rightarrow \ln c=1$.
$\Rightarrow c={{e}^{1}}$.
$\Rightarrow c=e$.
$\Rightarrow c=2.73$ ---(2).
Here we have given that a and b are integers (positive). So, the value of a must be less than the value of c obtained which has possibilities of 1 and 2.
We know that any power of 1 is 1. In order to satisfy equation (1), b must be equal to 1, which is a contradiction as a is not equal to b.
So, the value a must be 2.
Let us substitute the value of a in equation (1).
$\Rightarrow {{2}^{b}}={{b}^{2}}$. Let us assume the value of b be 3.
So, we get ${{2}^{3}}={{3}^{2}}$
$\Rightarrow 8=9$, which is a contradiction.
Let us assume the value of b be 4.
So, we get ${{2}^{4}}={{4}^{2}}$
$\Rightarrow 16=16$, which is true.
So, the values of a and b are 2 and 4.
Let us find the value of ${{a}^{2}}+{{b}^{2}}$.
$\Rightarrow {{a}^{2}}+{{b}^{2}}={{2}^{2}}+{{4}^{2}}$.
$\Rightarrow {{a}^{2}}+{{b}^{2}}=4+16$.
$\Rightarrow {{a}^{2}}+{{b}^{2}}=20$.
∴ The possible value of ${{a}^{2}}+{{b}^{2}}$ is 20.

So, the correct answer is “Option (a)”.

Note: Here we cannot get the absolute values without trial and error for them. Because we only have one condition to solve for the values of two variables which can be solved without taking an assumption. We cannot apply Rolle’s theorem directly without checking its conditions but it is already given in this problem, we don’t need to worry. Similarly, we can expect this problem to apply other mean value theorems.