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

Answer: (B)

$\frac{1}{4}$

Given $f\left(x\right)=x^{25}{(1-x)}^{75}$
$\Rightarrow$ $f^{\prime }\left(x\right)=25x^{24}{(1-x)}^{75}-75x^{25}{(1-x)}^{74}$
$=25x^{24}{(1-x)}^{74}(1-4x)$
$\therefore$ $f^{\prime }\left(x\right)=0$
$\Rightarrow$ $x=0,1,\frac{1}{4}$
If $x<\frac{1}{4}$, them
$f^{\prime }\left(x\right)=25x^{24}{(1-x)}^{74}(1-4x)>0$
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 throough $\frac{1}{4}$ from left to right.
Hence $f\left(x\right)$ attains its maximum at $x=\frac{1}{4}$