Applications Of Derivatives Question 350
Question: On
[1, e] the greatest value of $ x^{2}\log x $ [AMU 2002]
Options:
A) $ e^{2} $
B) $ \frac{1}{e}\log \frac{1}{\sqrt{e}} $
C) $ e^{2}\log \sqrt{e} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=x^{2}\log x $
therefore $ {f}’(x)=(2\log x+1)x $
Now $ {f}’(x)=0 $
therefore $ x={e^{-1/2}},,0 $
$ \because $ $ 0<{e^{-1/2}}<1 $
$ \because $ None of these critical points lies in the interval [1, e] So we only complete the value of $ f(x) $ at the end points 1 and e. We have $ f(1)=0,,f(e)=e^{2} $ Greatest value = $ e^{2}. $
BETA
