Applications Of Derivatives Question 251
Question: The function $ f(x)=1-{e^{-x^{2}/2}} $ is
[AMU 1999]
Options:
A) Decreasing for all x
B) Increasing for all x
C) Decreasing for $ x<0 $ and increasing for $ x>0 $
D) Increasing for $ x<0 $ and decreasing for $ x>0 $
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Answer:
Correct Answer: C
Solution:
$ f(x)=1-{e^{-x^{2}/2}} $
$ {f}’(x)=-{e^{-x^{2}/2}}(-x)=x{e^{-x^{2}/2}} $
For $ f(x) $ to be increasing, $ {f}’(x)>0 $
therefore $ x{e^{-x^{2}/2}}>0 $
therefore $ x>0 $ and $ f(x) $ to be decreasing for $ x<0 $ .
BETA
