Question

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

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

Answer 1

The correct option is D

$\frac{1}{4}$

Explanation for the correct option.

Finding maximum value of the given function

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

Differentiating with respect to .$x$
$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=25x^{24}{(1-x)}^{75}-75x^{25}{(1-x)}^{74} \\ =25x^{24}{(1-x)}^{74}(1-4x) \\ \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow x=0,1,\frac{1}{4} \end{gathered}$$

If$x<\frac{1}{4},$ then
$f\prime \left(x\right)=25x^{24}{(1-x)}^{74}(1-4x)>0$
and if$x>\frac{1}{4}$​,then
$f\prime \left(x\right)=25x^{24}{(1-x)}^{74}(1-4x)<0$

Thus $f\prime \left(x\right)$changes its sign from positive to negative as $x$ passes through $\frac{1}{4}$ from left to right.
Hence $f\left(x\right)$ attains its maximum at $x=\frac{1}{4}$

Hence the correct answer is option (D).