Question

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

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

Answer 1

The correct option is B $\frac{1}{4}$

Let $f\left(x\right)=x^{25}{(1-x)}^{75}$
On differentiating with respect to $x$, we get
$f^{\prime }\left(x\right)=25x^{24}{(1-x)}^{75}-75x^{25}{(1-x)}^{74}$
$=25x^{24}\left[{(1-x)}^{74}(1-x-3x)\right]$
$=25x^{24}{(1-x)}^{74}(1-4x)$

$f$ takes maximum value $⟹f$ is increasing

$⟹f^{\prime }\left(x\right)>0⟹(1-4x)>0⟹x<\frac{1}{4}$
$\therefore$ $f^{\prime }$ changes sign about $x=\frac{1}{4}$ only.

Hence, $f$ takes its maximum value at the point $\frac{1}{4}$.