On the interval $\left[ 0,1 \right]$ the function ${{x}^{25}}{{\left( 1-x \right)}^{75}}$ takes its maximum value at the point?
(a) $0$
(b) $\dfrac{1}{4}$
(c) \[\dfrac{1}{2}\]
(d) \[\dfrac{1}{3}\]

Hint: Assume the given function as $f\left( x \right)$. Differentiate the function with respect to x and use the product rule of derivative for the derivative of product of two functions (u and v) given as $\dfrac{d\left( u\times v \right)}{dx}=\left( u\times \dfrac{dv}{dx} \right)+\left( v\times \dfrac{du}{dx} \right)$. Use the formula \[\dfrac{d\left[ {{f}^{n}}\left( x \right) \right]}{dx}=n{{f}^{n-1}}\left( x \right)f'\left( x \right)\]. Substitute $f'\left( x \right)=0$ and find the values of x in the given interval $\left[ 0,1 \right]$. Again differentiate the function to find $f''\left( x \right)$ by using the product rule for three functions (u, v and w) given as $\dfrac{d\left( u\times v\times w \right)}{dx}=\left( uv\times \dfrac{dw}{dx} \right)+\left( uw\times \dfrac{dv}{dx} \right)+\left( vw\times \dfrac{du}{dx} \right)$ and substitute the values of x that were obtained earlier. If at x = a $f''\left( a \right)>0$ then that point is the point of minima and if $f''\left( a \right)<0$ then that point is the point of maxima. In case $f''\left( a \right)=0$ then x = a is an inflection point and there may not be a minimum or maximum at that point.

Complete step-by-step solution:
Here we have been provided with the function ${{x}^{25}}{{\left( 1-x \right)}^{75}}$ and we are asked to find the point of maxima in the interval $\left[ 0,1 \right]$. Assuming the function as $f\left( x \right)$ and differentiating both the sides with respect to x we get,
$\Rightarrow f'\left( x \right)=\dfrac{d\left( {{x}^{25}}{{\left( 1-x \right)}^{75}} \right)}{dx}$
We can consider the function as the product of two functions u and v, so using the product rule of derivative given as $\dfrac{d\left( u\times v \right)}{dx}=\left( u\times \dfrac{dv}{dx} \right)+\left( v\times \dfrac{du}{dx} \right)$ we get,
$\Rightarrow f'\left( x \right)={{x}^{25}}\times \dfrac{d\left( {{\left( 1-x \right)}^{75}} \right)}{dx}+{{\left( 1-x \right)}^{75}}\times \dfrac{d\left( {{\left( x \right)}^{25}} \right)}{dx}$
Using the formula \[\dfrac{d\left[ {{f}^{n}}\left( x \right) \right]}{dx}=n{{f}^{n-1}}\left( x \right)f'\left( x \right)\] we get,
$\begin{align}
  & \Rightarrow f'\left( x \right)={{x}^{25}}\times \left( -75{{\left( 1-x \right)}^{74}} \right)+{{\left( 1-x \right)}^{75}}\times 25{{x}^{24}} \\
 & \Rightarrow f'\left( x \right)=25{{x}^{24}}{{\left( 1-x \right)}^{74}}\left[ -3x+\left( 1-x \right) \right] \\
 & \Rightarrow f'\left( x \right)=25{{x}^{24}}{{\left( 1-x \right)}^{74}}\left( 1-4x \right) \\
\end{align}$
Substituting $f'\left( x \right)=0$ we get,
\[\Rightarrow {{x}^{24}}=0\], \[{{\left( 1-x \right)}^{74}}=0\] or \[\left( 1-4x \right)=0\]
\[\Rightarrow x=0\], \[x=1\] or \[x=\dfrac{1}{4}\]
Now, again differentiating the function with respect to x and using the product rule of derivative for the product of three function (u, v and w) given as $\dfrac{d\left( u\times v\times w \right)}{dx}=\left( uv\times \dfrac{dw}{dx} \right)+\left( uw\times \dfrac{dv}{dx} \right)+\left( vw\times \dfrac{du}{dx} \right)$ we get,
\[\begin{align}
  & \Rightarrow f''\left( x \right)=25\left[ {{x}^{24}}{{\left( 1-x \right)}^{74}}\times \dfrac{d\left( 1-4x \right)}{dx}+{{x}^{24}}\left( 1-4x \right)\times \dfrac{d\left( {{\left( 1-x \right)}^{74}} \right)}{dx}+{{\left( 1-x \right)}^{74}}\left( 1-4x \right)\times \dfrac{d\left( {{x}^{24}} \right)}{dx} \right] \\
 & \Rightarrow f''\left( x \right)=25\left[ {{x}^{24}}{{\left( 1-x \right)}^{74}}\times \left( -4 \right)+{{x}^{24}}\left( 1-4x \right)\times \left( -73{{\left( 1-x \right)}^{73}} \right)+{{\left( 1-x \right)}^{74}}\left( 1-4x \right)\times \left( 24{{x}^{23}} \right) \right] \\
 & \Rightarrow f''\left( x \right)=25\left[ -4{{x}^{24}}{{\left( 1-x \right)}^{74}}-73{{\left( 1-x \right)}^{73}}{{x}^{24}}\left( 1-4x \right)+24{{x}^{23}}{{\left( 1-x \right)}^{74}}\left( 1-4x \right) \right] \\
\end{align}\]
We know that if at x = a $f''\left( a \right)>0$ then that point is the point of minima, if $f''\left( a \right)<0$ then that point is the point of maxima and if $f''\left( a \right)=0$ then x = a is an inflection point and there may not be a minimum or maximum at that point. Let us check the sign of $f''\left( x \right)$ for the values of x obtained in the substitution $f'\left( x \right)=0$.
(i) At $x=0$ we have,
\[\begin{align}
  & \Rightarrow f''\left( 0 \right)=25\left[ -0-0+0 \right] \\
 & \Rightarrow f''\left( 0 \right)=0 \\
\end{align}\]
Clearly $f''\left( 0 \right)=0$ so the point $x=0$ is an inflection point and therefore there may not be any maximum or minimum at this point.
(ii) At $x=1$ we have,
\[\begin{align}
  & \Rightarrow f''\left( 1 \right)=25\left[ -0-0+0 \right] \\
 & \Rightarrow f''\left( 1 \right)=0 \\
\end{align}\]
Here also we have $f''\left( 1 \right)=0$ so the point $x=1$ is an inflection point and therefore there may not be any maximum or minimum at this point also.
(ii) At $x=\dfrac{1}{4}$ we have,
\[\begin{align}
  & \Rightarrow f''\left( \dfrac{1}{4} \right)=25\left[ -{{\left( \dfrac{1}{4} \right)}^{24}}{{\left( 1-\dfrac{1}{4} \right)}^{74}}-0+0 \right] \\
 & \Rightarrow f''\left( \dfrac{1}{4} \right)=-25{{\left( \dfrac{1}{4} \right)}^{24}}{{\left( \dfrac{3}{4} \right)}^{74}} \\
\end{align}\]
Clearly we have $f''\left( \dfrac{1}{4} \right)<0$ so the point $x=\dfrac{1}{4}$ is a point of maxima and ${{x}^{25}}{{\left( 1-x \right)}^{75}}$ will take its maximum value at this point.
Hence, option (b) is the correct answer.


Note: The second – order derivative test fails in the case where $f''\left( a \right)=0$ and higher – order derivative test is used to check if the point x = a is a local maxima, minima or saddle point. In the above situation the exponent is too large so we cannot keep on differentiating the function until we get ${{f}^{\left( n \right)}}\left( a \right)\ne 0$, where $\left( n \right)$ is the ${{n}^{th}}$ derivative of $f\left( x \right)$, therefore we don’t have to worry about these points here.