Q. 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) $\frac{1}{3}$
Explanation

(b)
Given, $f(x) = x^{25}(1 - x)^{75}$
$\Rightarrow \ \ \ \ \ f^{'}(x) = 25x^{24}\ (1 - x)^{75} - 75x^{25}(1 - x)^{74}$
$= 25x^{24}(1 - x)^{74}(1 - 4x)$
Put, $f^{'}(x) = 0\ \ \ \Rightarrow \ \ \ \ x = 0,\ 1,\frac{1}{4}$
If $x < \frac{1}{4}$, then
$f^{'}(x) = 25x^{24}\ (1 - x)^{74}(1 - 4x) > 0$
And if $x > \frac{1}{4}$, then
$f^{'}(x) = 25x^{24}\ (1 - x)^{24}(1 - 4x) < 0$
Thus, $f^{'}(x)$ changes its sign from positive to negative as $x$ passes through 1/4 from left to right.
Hence, $f(x)$ attains its maximum at x=1/4

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.