Question

Find local maximum value of function $\left( x \right)=\dfrac{\ln x}{x}$
A. $\dfrac{1}{e}$
B. $-\dfrac{1}{e}$
C. $\dfrac{2}{{{e}^{2}}}$
D. None of these

Answer 1

Hint: To solve this question, we need to use the concept of first derivative. To find the local maximum or minimum, we need to use differentiation and set the derivative of that function to 0. Then solving for the value of x gives us the point of local maximum or minimum. Then by substituting this in the function, we get the maximum value of the function.

Complete step by step solution:
We are required to find the local maximum of the function given by $\dfrac{\ln x}{x}.$ Let us assume this function is represented by $f\left( x \right).$
$\Rightarrow f\left( x \right)=\dfrac{\ln x}{x}$
Now in order to find the local maximum, we use the concept of differentiation and set the derivative of this function to 0.
$\Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( \dfrac{\ln x}{x} \right)$
To differentiate this, we need to use quotient rule given as,
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\ldots \ldots \left( 1 \right)$
Here, u is ln x and v is taken as x.
We know the differentiation of ln x with respect to x is given as,
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$
We know the differentiation of x with respect to x is given as,
$\Rightarrow \dfrac{dv}{dx}=\dfrac{d}{dx}\left( x \right)=1$
Using all these in the above equation 1,
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{\ln x}{x} \right)=\dfrac{x\dfrac{d\left( \ln x \right)}{dx}-\ln x\dfrac{dx}{dx}}{{{x}^{2}}}$
Substituting the differentiation values for ln x and x,
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{\ln x}{x} \right)=\dfrac{x.\dfrac{1}{x}-\ln x.1}{{{x}^{2}}}$
In the first term, x and $\dfrac{1}{x}$ cancel giving us,
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{\ln x}{x} \right)=\dfrac{1-\ln x}{{{x}^{2}}}$
Splitting this as a product of two terms,
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{\ln x}{x} \right)=\dfrac{1}{{{x}^{2}}}.\left( 1-\ln x \right)$
Now we equate the right-hand side of this derivative to 0.
$\Rightarrow \dfrac{1}{{{x}^{2}}}.\left( 1-\ln x \right)=0$
This gives us two equations,
$\Rightarrow \dfrac{1}{{{x}^{2}}}=0$
This equation has no finite solution and we can ignore this. Taking the second equation,
$\Rightarrow \left( 1-\ln x \right)=0$
Taking the ln x term to the right-hand side,
$\Rightarrow 1=\ln x$
We know that $\ln a=b$ can be given in terms of e as ${{e}^{b}}=a.$ Using this,
$\Rightarrow x={{e}^{1}}=e$
Hence, the maximum exists at x=e. Substituting this value in the function equation,
$\Rightarrow f\left( x \right)=\dfrac{\ln e}{e}$
We know that ln e value is 1,
$\Rightarrow f\left( x \right)=\dfrac{1}{e}$
Hence, the local maximum value of the function $f\left( x \right)=\dfrac{\ln x}{x}$ is $\dfrac{1}{e}.$

So, the correct answer is “Option A”.

Note: Students need to know the concept of maximum and minimum to solve this question easily. They also need to know the basic differentiation formulae. One important point for this problem is that since x=e is a maximum point, this means that for $x\prec e,$ slope is increasing or $\dfrac{dy}{dx}\succ 0.$ Same is true the other way around, for $x\succ e,$ slope is decreasing or $\dfrac{dy}{dx}\prec 0.$