Question

On $\left[ {1,e} \right]$, the greatest value of ${x^2}\log x$
(A) ${e^2}$
(B) $\dfrac{1}{e}\log \left( {\dfrac{1}{{\sqrt e }}} \right)$
(C) ${e^2}\log \left( {\sqrt e } \right)$
(D) None of these

Answer 1

Hint: To find the greatest value of the function, we will find the derivative of the function. We can judge the behavior of the function in the interval given to us, $\left[ {1,e} \right]$. Then, if the function is increasing, the maximum value occurs at the right end of the interval and if the function is decreasing, the maximum value occurs at the left end of the interval.

Complete step by step solution: 
Now, $f\left( x \right) = {x^2}\log x$
Differentiating both sides with respect to x, we get,
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2}\log x} \right)$
Using product rule of differentiation $\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right) + g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right)$, we get,
$ \Rightarrow f'\left( x \right) = {x^2}\dfrac{d}{{dx}}\left( {\log x} \right) + \log x\dfrac{d}{{dx}}\left( {{x^2}} \right)$
Substituting the derivative of $\log x$ and ${x^2}$ as $\dfrac{1}{x}$ and $2x$, we get,
$ \Rightarrow f'\left( x \right) = {x^2} \times \dfrac{1}{x} + \log x \times 2x$
$ \Rightarrow f'\left( x \right) = x + 2x\log x$
So, in the interval $\left[ {1,e} \right]$, logarithmic function $\log x$ is positive. Also, x is positive.
So, $f'\left( x \right) = x + 2x\log x > 0$
So, the function $f\left( x \right) = {x^2}\log x$ is increasing in the interval $\left[ {1,e} \right]$.
So, the maximum value occurs at the right end of the interval.
So, $f\left( e \right) = {e^2}\log e = {e^2}$
Hence, the maximum value of ${x^2}\log x$ in $\left[ {1,e} \right]$ is ${e^2}$. Option (a) is the correct answer.

Note: In an increasing function, the value of function increases with the increasing value of variable whereas, in a decreasing function, the value of function decreases with the increasing value of variable. The function ${x^2}\log x$ may have a critical point, but it is not in the interval $\left[ {1,e} \right]$, given to us. In this interval, the function increases with the increasing value of x.