On the interval [0, 1], $f(x)=x^{25}(1-x)^{75}$ takes maximum value at the point:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{4}$
$f(x)=x^{25}(1-x)^{25}$
$f'(x)=24x^{24}(1-x)^{75}-75x^{25}(1-x)^{74}$
for critical point,
$f'(c)=0$
$⇒25(1-c)-75c=0$
$⇒c=\frac{25}{100}=\frac{1}{4}$