question_answer

On the interval [0, 1] the function \[{{x}^{25}}{{(1-x)}^{75}}\]takes its maximum value at the point

A)  0                                

B)  ¼

C)  1/2                             

D)  1/3

Correct Answer: B

Solution :

Let\[f(x)={{x}^{25}}{{(1-x)}^{75}},x\in [0,1]\] \[\Rightarrow \]\[f'(x)=25{{x}^{24}}{{(1-x)}^{75}}-75{{x}^{25}}{{(1-x)}^{74}}\] \[=25{{x}^{24}}{{(1-x)}^{74}}\{(1-x)-3x\}\] \[=25{{x}^{24}}{{(1-x)}^{74}}(1-4x)\] which shows f' (x) is positive for \[x<\frac{1}{4}\] and f? (x) is negative for \[x>\frac{1}{4}.\] \[\therefore \]f(x) attains maximum at \[x=\frac{1}{4}.\]