Answer
Verified
415.2k+ views
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.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE