Question

The Local maximum value of the function ${\displaystyle \frac{\log x}{x}}$ is
A) $e$
B) $1$
C) ${\displaystyle \frac{1}{e}}$
D) $2e$

Answer 1

Hint: To find the answer to the question, first you have to consider ${\displaystyle \frac{\log x}{x}}$ as a function. Then you have to differentiate function with equal to zero. Then find the value of $x$ from the first differentiate. Then do a second differentiation and put the value of $x$ In it and check whether the coming answer is negative or positive. If it’s positive then put that value of $x$ in our main function and you will find the answer.

Complete step by step answer:
So, let’s consider ${\displaystyle \frac{\log x}{x}}$ as a function and rewrite it,
$\Rightarrow f(x)={\displaystyle \frac{\log x}{x}}$
Now, differentiate our function with equal to zero to find the value for $x$ and we will get,
$\Rightarrow f^{\prime }(x)=0$
$\Rightarrow f^{\prime }(x)={\displaystyle \frac{x\times {\displaystyle \frac{1}{x}}-\log x}{x^2}}=0$
From further simplification we will get,
$\Rightarrow f^{\prime }(x)={\displaystyle \frac{1-\log x}{x^2}}=0$
Find the value for $x$ and we will get,
$\Rightarrow 1-\log x=0$
$\Rightarrow x=e$
So, we find value for $x$ and that is $x=e$ .
Now, do second differentiation,
$\Rightarrow f^{″}(x)={\displaystyle \frac{x^2(-{\displaystyle \frac{1}{x}})-2x(1-\log x)}{x^4}}$
Now, put value for $x$ that we find from first differentiation,
$\Rightarrow f^{″}(e)={\displaystyle \frac{e^2(-{\displaystyle \frac{1}{e}})-2e(1-\log e)}{e^4}}$
From further simplification we will get,
$\Rightarrow f^{″}(e)={\displaystyle \frac{-1}{e^3}}$
See our second differentiation is negative in value so $x$ is maximum at $e$ .
Now, just put value of $x$ in our main function and we will get our final answer,
$\Rightarrow f(e)={\displaystyle \frac{\log e}{e}}$
But $\log e=1$ so,
$\Rightarrow f(e)={\displaystyle \frac{1}{e}}$
Therefore, the local maximum value of the function ${\displaystyle \frac{\log x}{x}}$ is ${\displaystyle \frac{1}{e}}$ and that is option (C).

Note:
In this problem we have to find our local maximum point, but what do they ask for a local minimum point? so there is nothing new for that. You just have to do a second differentiation and check whether the coming value is positive or negative. If value is positive then at That value for $x$ function have local minimum point else if value is negative then at That value for $x$ function have local maximum point.