Applications Of Derivatives Question 234
Question: The function $ f(x)=\int\limits_{-1}^{x}{t(e^{t}-1)(t-1){{(t-2)}^{3}}{{(t-3)}^{5}}}dt $ has a local minimum at x =
[IIT 1999]
Options:
A) 0
B) 1
C) 2
D) 3
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=\int_{-1}^{x}{t(e^{t}-1)(t-1){{(t-2)}^{3}}{{(t-3)}^{5}}}dt $
$ \therefore f’(x)=x(e^{x}-1)(x-1){{(x-2)}^{3}}{{(x-3)}^{5}} $
For local minima, slope i.e. $ f’(x) $ should change sign from $ -ve $ to $ +ve $ .
$ f’(x)=0\Rightarrow x=0,,1,,2,,3 $
If $ x=0-h, $ where h is a very small number, then
$ f’(x)=(-)(-)(-1)(-1)(-1)=-ve $
If $ x=0+h $ ; $ f’(x)=(+)(+)(-)(-1)(-1)=-ve $
Hence at $ x=0 $ neither maxima nor minima.
If $ x=1-h $
$ f’(x)=(+)(+)(-)(-1)(-1)=-ve $
If $ x=1+h $ ; $ f’(x)=(+)(+)(+)(-1)(-1)=+ve $
Hence at $ x=1 $ there is a local minima.
If $ x=2-h $ ; $ f’(x)=(+)(+1)(+)(-)(-)=+ve $
If $ x=2+h $ ; $ f’(x)=(+)(+)(+)(+)(-1)=-ve $
Hence at $ x=2 $ there is a local maxima.
If $ x=3-h $ ; $ f’(x)=(+)(+)(+)(+)(-)=-ve $
If $ x=3+h $ ; $ f’(x)=(+)(+)(+)(+)(+)=+ve $
Hence at $ x=3 $ there is a local minima.
BETA
