Question

On the interval $\left[0,1\right]$, the function $x^{25}(1-x^{75})$ takes its maximum value at the point

• A. $0$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B

$\frac{1}{4}$

Explanation for the correct answer:

Finding the maximum value of the function:

Given,
$f\left(x\right)=x^{25}{\left(1-x\right)}^{75}$
Differentiating,

$$\begin{gathered} f\text{'}\left(x\right)=25x^{24}+{\left(1-x\right)}^{75}+75x^{25}{\left(1-x\right)}^{74}\left[\because \frac{d{x}^a}{dx}=a{x}^{a-1}\right] \\ =x^{24}{\left[1-x\right]}^{74}\left[25(1-x)-75x\right] \\ =25x^{24}{\left[1-x\right]}^{74}\left[1-x-3x\right] \\ =25x^{24}{\left[1-x\right]}^{74}\left[1-4x\right] \end{gathered}$$

So values of $x$ are,

$$\begin{gathered} 25x^{24}=0 \\ \Rightarrow x=24 \\ {\left[1-x\right]}^{74}=0 \\ \Rightarrow 1-x=0 \\ \Rightarrow x=1 \\ \left[1-4x\right]=0 \\ \Rightarrow -4x=-1 \\ \Rightarrow x=\frac{1}{4} \end{gathered}$$

Therefore,

$x=0,1,\frac{1}{4}$

Hence, $x=\frac{1}{4}$is the maximum value.

Therefore, the correct answer is option (B).