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, $f\left(x\right)=x^{25}(1-x{)}^{75}$
$\Rightarrow f^{\prime }\left(x\right)=25x^{24}(1-x{)}^{74}(1-4x).$
Now, $f^{\prime }\left(x\right)=0$
$\Rightarrow x=0,1,\frac{1}{4}$Clearly $f^{\prime }\left(x\right)>0$ in the left neighbourhood of $\frac{1}{4}$ and $f^{\prime }\left(x\right)<0$ in the right neighbourhood of $\frac{1}{4}$.
So, $f^{\prime }\left(x\right)$ changes its sign from positive to negative in the neighbourhood of $\frac{1}{4}.$
Hence, it attains maximum at $x=\frac{1}{4}$