Answer

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Hint: In this question, we will use the quotient rule of differentiation. It is given by,

$\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{\dfrac{d\left( u \right)}{dx}.v-\dfrac{d\left( v \right)}{dx}.u}{{{\left( v \right)}^{2}}}$

Then we will find the maximum value of $\dfrac{\log x}{x}$ in interval $\left( 2,\infty \right)$.

For obtaining maximum value, we will equate $\dfrac{dy}{dx}=0$.

We will also use some basic logarithm formulas to make the solution simple and short.

Complete step-by-step answer:

We will put the obtained value of x by equating $\dfrac{dy}{dx}=0$ in the given function to get the maximum value of $\dfrac{\log x}{x}$ in $\left( 2,\infty \right)$.

It is given in the question that we have to find out the maximum value of $\dfrac{\log x}{x}$ an interval $\left( 2,\infty \right)$.

Let us assume that,

$y=\dfrac{\log x}{x}..............\left( 1 \right)$

We know the differentiation of log (x) with the respect of x is $\dfrac{1}{x}$ and differentiation of $\left( \dfrac{u}{v} \right)$ is

$\begin{align}

& \dfrac{u'v-v'u}{{{v}^{2}}} \\

& or \\

& \dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{\dfrac{d\left( u \right)}{dx}.v-\dfrac{d\left( v \right)}{dx}.u}{{{\left( v \right)}^{2}}} \\

\end{align}$

Using this basic differentiation on equation (1) with respect to x, we get,

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{\dfrac{d\left( \log x \right)}{dx}.x-\dfrac{d\left( x \right)}{dx}.\log x}{{{\left( x \right)}^{2}}} \\

& \dfrac{dy}{dx}=\dfrac{\dfrac{1}{x}.x-1.\log x}{{{\left( x \right)}^{2}}} \\

\end{align}$

Cancelling out like terms in the above equation, we get,

$\dfrac{dy}{dx}=\dfrac{1-\log x}{{{\left( x \right)}^{2}}}..........\left( 2 \right)$

To get maximum point, we will equate equation (2) with 0 as, for maximum value $\dfrac{dy}{dx}=0$

$\begin{align}

& \Rightarrow \dfrac{1-\log x}{{{\left( x \right)}^{2}}}=0 \\

& \Rightarrow 1-\log x=0 \\

& \Rightarrow -\log x=-1 \\

\end{align}$

On multiplying (-1) both the sides, we get,

$\Rightarrow \log x=1...........\left( 3 \right)$

We know that $\log e=1$ from the basic logarithm. So, replacing 1 with $\log e$ in equation (3), we get,

$\Rightarrow \log \left( x \right)=\log \left( e \right).........\left( 4 \right)$

So, from this, we get x = e.

Now, from the calculus, we know that to know which point is maximum and which point is minimum we have to double differentiate the function. If f”(x) is positive then it is a point of minimum and if the sign is negative then it is a point of maximum.

Also, we have calculated that the value of x = e to get maximum value from $\dfrac{\log x}{x}$ but in question the function is restricted in the interval $\left( 2,\infty \right)$.

Also, we know that the value of e is 2.303 and we are also getting the maximum value of $\dfrac{\log x}{x}$ as e, which is lying in the given interval $\left( 2,\infty \right)$.

So, putting the value of x as e in a given function $y=\dfrac{\log x}{x}$, we get,

$\begin{align}

& y=\dfrac{\log x}{x} \\

& y=\dfrac{\log e}{e}............\left( 5 \right) \\

\end{align}$

We know that the value of log e is 1. So, putting the value of log e =1 in equation (5), we get,

$y=\dfrac{1}{e}$

Thus, the maximum value of $\dfrac{\log x}{x}$ in interval $\left( 2,\infty \right)$is $\dfrac{1}{e}$.

Therefore, option (D) is the correct answer.

Note: You can solve this question in just two steps. First, differentiate the given function and if you get $\dfrac{dy}{dx}=0$. Then find the value of the unknown directly if the value of the unknown is positive. Then it is a point of minimum and if it is negative then it is a point of maximum.

$\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{\dfrac{d\left( u \right)}{dx}.v-\dfrac{d\left( v \right)}{dx}.u}{{{\left( v \right)}^{2}}}$

Then we will find the maximum value of $\dfrac{\log x}{x}$ in interval $\left( 2,\infty \right)$.

For obtaining maximum value, we will equate $\dfrac{dy}{dx}=0$.

We will also use some basic logarithm formulas to make the solution simple and short.

Complete step-by-step answer:

We will put the obtained value of x by equating $\dfrac{dy}{dx}=0$ in the given function to get the maximum value of $\dfrac{\log x}{x}$ in $\left( 2,\infty \right)$.

It is given in the question that we have to find out the maximum value of $\dfrac{\log x}{x}$ an interval $\left( 2,\infty \right)$.

Let us assume that,

$y=\dfrac{\log x}{x}..............\left( 1 \right)$

We know the differentiation of log (x) with the respect of x is $\dfrac{1}{x}$ and differentiation of $\left( \dfrac{u}{v} \right)$ is

$\begin{align}

& \dfrac{u'v-v'u}{{{v}^{2}}} \\

& or \\

& \dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{\dfrac{d\left( u \right)}{dx}.v-\dfrac{d\left( v \right)}{dx}.u}{{{\left( v \right)}^{2}}} \\

\end{align}$

Using this basic differentiation on equation (1) with respect to x, we get,

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{\dfrac{d\left( \log x \right)}{dx}.x-\dfrac{d\left( x \right)}{dx}.\log x}{{{\left( x \right)}^{2}}} \\

& \dfrac{dy}{dx}=\dfrac{\dfrac{1}{x}.x-1.\log x}{{{\left( x \right)}^{2}}} \\

\end{align}$

Cancelling out like terms in the above equation, we get,

$\dfrac{dy}{dx}=\dfrac{1-\log x}{{{\left( x \right)}^{2}}}..........\left( 2 \right)$

To get maximum point, we will equate equation (2) with 0 as, for maximum value $\dfrac{dy}{dx}=0$

$\begin{align}

& \Rightarrow \dfrac{1-\log x}{{{\left( x \right)}^{2}}}=0 \\

& \Rightarrow 1-\log x=0 \\

& \Rightarrow -\log x=-1 \\

\end{align}$

On multiplying (-1) both the sides, we get,

$\Rightarrow \log x=1...........\left( 3 \right)$

We know that $\log e=1$ from the basic logarithm. So, replacing 1 with $\log e$ in equation (3), we get,

$\Rightarrow \log \left( x \right)=\log \left( e \right).........\left( 4 \right)$

So, from this, we get x = e.

Now, from the calculus, we know that to know which point is maximum and which point is minimum we have to double differentiate the function. If f”(x) is positive then it is a point of minimum and if the sign is negative then it is a point of maximum.

Also, we have calculated that the value of x = e to get maximum value from $\dfrac{\log x}{x}$ but in question the function is restricted in the interval $\left( 2,\infty \right)$.

Also, we know that the value of e is 2.303 and we are also getting the maximum value of $\dfrac{\log x}{x}$ as e, which is lying in the given interval $\left( 2,\infty \right)$.

So, putting the value of x as e in a given function $y=\dfrac{\log x}{x}$, we get,

$\begin{align}

& y=\dfrac{\log x}{x} \\

& y=\dfrac{\log e}{e}............\left( 5 \right) \\

\end{align}$

We know that the value of log e is 1. So, putting the value of log e =1 in equation (5), we get,

$y=\dfrac{1}{e}$

Thus, the maximum value of $\dfrac{\log x}{x}$ in interval $\left( 2,\infty \right)$is $\dfrac{1}{e}$.

Therefore, option (D) is the correct answer.

Note: You can solve this question in just two steps. First, differentiate the given function and if you get $\dfrac{dy}{dx}=0$. Then find the value of the unknown directly if the value of the unknown is positive. Then it is a point of minimum and if it is negative then it is a point of maximum.

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