Question

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

Answer 1

Hint: To find the answer to the question, first you have to consider $\dfrac{{\log x}}{x}$ 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 $\dfrac{{\log x}}{x}$ as a function and rewrite it,
$ \Rightarrow f(x) = \dfrac{{\log x}}{x}$
Now, differentiate our function with equal to zero to find the value for $x$ and we will get,
$ \Rightarrow f'(x) = 0$
$ \Rightarrow f'(x) = \dfrac{{x \times \dfrac{1}{x} - \log x}}{{{x^2}}} = 0$
From further simplification we will get,
$ \Rightarrow f'(x) = \dfrac{{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) = \dfrac{{{x^2}( - \dfrac{1}{x}) - 2x(1 - \log x)}}{{{x^4}}}$
Now, put value for $x$ that we find from first differentiation,
$ \Rightarrow f''(e) = \dfrac{{{e^2}( - \dfrac{1}{e}) - 2e(1 - \log e)}}{{{e^4}}}$
From further simplification we will get,
$ \Rightarrow f''(e) = \dfrac{{ - 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) = \dfrac{{\log e}}{e}$
But $\log e = 1$ so,
$ \Rightarrow f(e) = \dfrac{1}{e}$
Therefore, the local maximum value of the function $\dfrac{{\log x}}{x}$ is $\dfrac{1}{e}$ 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.