Question

Find the value of x for which $x{{\left( 1-x \right)}^{2}},0\le x\le 2$ is max,
A. 1
B. 2
C.\[\dfrac{1}{3}\]
D. 3

Answer 1

Hint: We will be using the basic concepts of function to solve the question. Also, other concepts like that of graphing is required. For drawing graph concepts of calculus specifically differential calculus must be strong.
Complete step by step answer:
Now, we have been given a function $x{{\left( 1-x \right)}^{2}}$and we have to find its maximum value in domain $0\le x\le 2$.
We will write the given functions as $f\left( x \right)=x{{\left( 1-x \right)}^{2}},0\le x\le 2$. To solve this question we will be using a calculus and analytical approach to find the maximum value.
We know that,
$\begin{align}
  & f\left( x \right)=x{{\left( 1-x \right)}^{2}} \\
 & =x\left( 1+{{x}^{2}}-2x \right) \\
\end{align}$
Using ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-ab$.
Now, further opening the bracket,
$\begin{align}
  & f\left( x \right)=x+{{x}^{3}}-2{{x}^{2}} \\
 & f\left( x \right)={{x}^{3}}-2{{x}^{2}}+x \\
\end{align}$
We can see that the given function is cubic. So, we will be drawing its graph with the help of calculus. To draw a graph we will first find its point of maxima and minima by differentiating f(x).
$\begin{align}
  & \dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( {{x}^{3}}-2{{x}^{2}}+x \right) \\
 & =3{{x}^{2}}-4x+1............\left( 1 \right) \\
\end{align}$
To find point of maxima and minima we will equate (1) to zero,
$\begin{align}
  & 3{{x}^{2}}-4x+1=0 \\
 & 3{{x}^{2}}-3x-x+1=0 \\
 & 3x\left( x-1 \right)-1\left( x-1 \right)=0 \\
 & \left( 3x-1 \right)\left( x-1 \right)=0 \\
 & 3x=1\ or\ x=1 \\
\end{align}$
$x=\dfrac{1}{3},1$ are the points of maxima or minima.
Now, substituting these values in f(x) we will get the value of function corresponding to it.
$\begin{align}
  & f\left( 1 \right)=1{{\left( 1-1 \right)}^{2}}=0...........\left( 2 \right) \\
 & f\left( \dfrac{1}{3} \right)=\dfrac{1}{3}{{\left( \dfrac{2}{3} \right)}^{2}}=\dfrac{4}{27}...........\left( 3 \right) \\
\end{align}$
Clearly we can see that 1 is point of minima and $\dfrac{1}{3}$is point of maxima as $f\left( 1 \right)So,
$\begin{align}
  & f\left( 0 \right)=0{{\left( 1-0 \right)}^{2}}=0............\left( 4 \right) \\
 & f\left( 2 \right)=2{{\left( 1-2 \right)}^{2}}=2...........\left( 5 \right) \\
\end{align}$
Since 1 is a point of minima and $\dfrac{1}{3}$ point of maxima.
We can also see from the graph the given situation more clearly.
On comparing (2), (3), (4) & (5) we can see that f(2) is maximum in $0\le x\le 2$.
Hence, the value of function is maximum at 2.
So, (B) is the correct option.

Note: We can also solve the question by option elimination easily. Substituting the value of x in option (a), (b), (c), (d) we get option (d) that is x = 3 at the point where $x{{\left( 1-x \right)}^{2}}$ is maximum but since, $0\le x\le 2$ we choose option (b) x = 2 as the answer because it is the point at which the function is highest and also x lies in the domain.