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Question

Question: The Local maximum value of the function \(\dfrac{{\log x}}{x}\) is A) \(e\) B) \(1\) C) \(\df...

The Local maximum value of the function logxx\dfrac{{\log x}}{x} is
A) ee
B) 11
C) 1e\dfrac{1}{e}
D) 2e2e

Explanation

Solution

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

Complete step by step answer:
So, let’s consider logxx\dfrac{{\log x}}{x} as a function and rewrite it,
f(x)=logxx\Rightarrow f(x) = \dfrac{{\log x}}{x}
Now, differentiate our function with equal to zero to find the value for xx and we will get,
f(x)=0\Rightarrow f'(x) = 0
f(x)=x×1xlogxx2=0\Rightarrow f'(x) = \dfrac{{x \times \dfrac{1}{x} - \log x}}{{{x^2}}} = 0
From further simplification we will get,
f(x)=1logxx2=0\Rightarrow f'(x) = \dfrac{{1 - \log x}}{{{x^2}}} = 0
Find the value for xx and we will get,
1logx=0\Rightarrow 1 - \log x = 0
x=e\Rightarrow x = e
So, we find value for xx and that is x=ex = e .
Now, do second differentiation,
f(x)=x2(1x)2x(1logx)x4\Rightarrow f''(x) = \dfrac{{{x^2}( - \dfrac{1}{x}) - 2x(1 - \log x)}}{{{x^4}}}
Now, put value for xx that we find from first differentiation,
f(e)=e2(1e)2e(1loge)e4\Rightarrow f''(e) = \dfrac{{{e^2}( - \dfrac{1}{e}) - 2e(1 - \log e)}}{{{e^4}}}
From further simplification we will get,
f(e)=1e3\Rightarrow f''(e) = \dfrac{{ - 1}}{{{e^3}}}
See our second differentiation is negative in value so xx is maximum at ee .
Now, just put value of xx in our main function and we will get our final answer,
f(e)=logee\Rightarrow f(e) = \dfrac{{\log e}}{e}
But loge=1\log e = 1 so,
f(e)=1e\Rightarrow f(e) = \dfrac{1}{e}
Therefore, the local maximum value of the function logxx\dfrac{{\log x}}{x} is 1e\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 xx function have local minimum point else if value is negative then at That value for xx function have local maximum point.