CUET Preparation Today
On the interval [0, 1], $f(x)=x^{25}(1-x)^{75}$ takes maximum value at the point: |
0 $\frac{1}{4}$ $\frac{1}{2}$ $\frac{1}{3}$ |
$\frac{1}{4}$ |
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, $f'(c)=0$ $⇒25(1-c)-75c=0$ $⇒c=\frac{25}{100}=\frac{1}{4}$ |
