Applications Of Derivatives Question 233
Question: On the interval [0, 1], the function $ x^{25}{{(1-x)}^{75}} $ takes its maximum value at the point
[IIT 1995]
Options:
A) 0
B) 1/2
C) 1/3
D) ¼
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)=x^{25}{{(1-x)}^{75}} $
$ f’(x)=x^{25}(75){{(1-x)}^{74}}(-1)+25x^{24}{{(1-x)}^{75}} $
For maxima and minima,
$ -75x^{25}{{(1-x)}^{74}}+25x^{24}{{(1-x)}^{75}}=0 $
therefore $ 25x^{24}{{(1-x)}^{74}}[(1-x)-3x]=0 $
therefore Either $ x=0 $ or $ x=1 $ or $ x=\frac{1}{4} $
At $ x=\frac{1}{4},\ \ f’,( \frac{1}{4}-h )>0 $ and $ f’( \frac{1}{4}+h )<0 $
$ \therefore f(x) $ is maximum at $ x=\frac{1}{4} $ .
Trick: Check with the options.
BETA
