Applications Of Derivatives Question 158
Question: Maximum value of $ x{{(1-x)}^{2}} $ when $ 0\le x\le 2 $ , is
[MP PET 1997]
Options:
A) $ \frac{2}{27} $
B) $ \frac{4}{27} $
C) 5
D) 0
Show Answer
Answer:
Correct Answer: B
Solution:
Given $ f(x)=x{{(1-x)}^{2}} $ , $ f(x)=x^{3}-2x^{2}+x $ Now $ f’(x)=3x^{2}-4x+1 $ Put $ f’(x)=0 $ i.e., $ 3x^{2}-4x+1=0 $
$ 3x^{2}-3x-x+1=0 $
therefore $ x=1,1/3 $
$ {f}’’(x)=6x-4 $
$ \therefore {f}’’,(1)=2= $ positive and $ f’’(1/3)=-2= $ -ve
Hence maximum value will be at $ x=\frac{1}{3} $
Maximum value $ f,( \frac{1}{3} )=\frac{4}{27} $ .
BETA
