On the interval [0,1] the function ${{x}^{25}}{{\left( 1-x \right)}^{75}}$ takes its maximum value at the point
\[\begin{align}
  & A.0 \\
 & B.\dfrac{1}{4} \\
 & C.\dfrac{1}{2} \\
 & D.\dfrac{1}{3} \\
\end{align}\]

Hint: In this question, we need to find the value of x in the interval (0,1) at which the function ${{x}^{25}}{{\left( 1-x \right)}^{75}}$ takes its maximum value. For this, we will use the first derivative test. We will suppose the given function as f(x) and then differentiate to find f'(x). Then we will put f'(x)=0 to find the value of x. One of these values of x will give us the maximum value of f(x). To find the required value, we will put all these values in f(x) and find the maximum value of f(x) and hence find the value of x which makes f(x) maximum. Properties of derivative that we will use are:
(1) Product rule: Derivative of product of two functions u and v is given by $\dfrac{d}{dx}\left( u\cdot v \right)=\dfrac{udv}{dx}+\dfrac{vdu}{dx}$.
(2) $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ where n is real number.
(3) Chain rule: $\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\cdot g'\left( x \right)$.

Complete step-by-step solution
Here we are given the function as ${{x}^{25}}{{\left( 1-x \right)}^{75}},x\in \left[ 0,1 \right]$.
Let us suppose this to be f(x). So, $f\left( x \right)={{x}^{25}}{{\left( 1-x \right)}^{75}},x\in \left[ 0,1 \right]$.
Differentiating w.r.t. x, we get:
\[\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( {{x}^{25}}{{\left( 1-x \right)}^{75}} \right)\]
Using product rule which is given as $\dfrac{d}{dx}\left( u\cdot v \right)=\dfrac{udv}{dx}+\dfrac{vdu}{dx}$ we get:
\[\Rightarrow f'\left( x \right)={{x}^{25}}\dfrac{d}{dx}{{\left( 1-x \right)}^{75}}+{{\left( 1-x \right)}^{75}}\dfrac{d}{dx}\left( {{x}^{25}} \right)\]
We know, $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\text{ and }\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\cdot \dfrac{d}{dx}g'\left( x \right)$ therefore,
\[\begin{align}
  & f'\left( x \right)={{x}^{25}}\left( 75{{\left( 1-x \right)}^{75-1}}\left( -1 \right) \right)+{{\left( 1-x \right)}^{75}}\left( 25{{x}^{25-1}} \right) \\
 & \Rightarrow {{x}^{25}}\left( -75{{\left( 1-x \right)}^{74}} \right)+25{{\left( 1-x \right)}^{75}}{{x}^{24}} \\
 & \Rightarrow -75{{x}^{25}}{{\left( 1-x \right)}^{74}}+25{{\left( 1-x \right)}^{75}}{{x}^{24}} \\
\end{align}\]
Taking $25{{x}^{24}}{{\left( 1-x \right)}^{74}}$ common we get:
\[\begin{align}
  & \Rightarrow f'\left( x \right)=25{{x}^{24}}{{\left( 1-x \right)}^{74}}\left[ \left( 1-x \right)-3x \right] \\
 & \Rightarrow f'\left( x \right)=25{{x}^{24}}{{\left( 1-x \right)}^{74}}\left( 1-4x \right) \\
\end{align}\]
For finding values of x, putting f'(x)=0, we get:
\[\Rightarrow f'\left( x \right)=25{{x}^{24}}{{\left( 1-x \right)}^{74}}\left( 1-4x \right)=0\]
Therefore, $x = 0, (1-x)=0$ and $(1-4x)=0.$
\[\Rightarrow x=0,x=1\text{ and }x=\dfrac{1}{4}\]
To check maximum value, let us put these values one by one in f(x), we get:
For $x = 0, f(0) = 0.$
For $x = 1, f(1) = 0.$
For $x=\dfrac{1}{4},f\left( \dfrac{1}{4} \right)={{\left( \dfrac{1}{4} \right)}^{25}}{{\left( 1-\dfrac{1}{4} \right)}^{75}}={{\left( \dfrac{1}{4} \right)}^{25}}{{\left( \dfrac{3}{4} \right)}^{75}}\text{ }>\text{ }0$.
Thus, putting $x=\dfrac{1}{4}$ gives us the maximum value of f(x).
Therefore, option B is the correct answer.

Note: Students can make mistakes while calculating derivatives of a given function because the function involves huge numbers. Students can also attain the maximum value of any function using a second derivative test where they put the value of x in f''(x) and if f''(x) is negative then the value of x at which f''(x)< 0 gives the maximum value of f(x). Make sure that the found solution lies between (0,1).