On the interval [0, 1] the function $x^{25}(1-x)^{75}$ takes its maximum value at the point
Answer & explanation
Correct answer: option 2
Let $f(x)=x^{25}(1-x)^{75}$. Then,
$f^{\prime}(x)=x^{24}(1-x)^{74}(1-4 x)$
Now,
$f^{\prime}(x)=0 \Rightarrow x=0,1,1 / 4$
Clearly, f'(x) > 0 in the left neighborhood of 1/4 and f'(x) < 0 in the right neighbourhood of 1/4. So, f'(x) changes its sign from positive to negative in the neighbourhood of 1/4.
Hence, it attains maximum at x = 1/4.