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 \[x>0\]
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, \[{f}'(x)>0\] Þ \[x{{e}^{-{{x}^{2}}/2}}>0\] Þ \[x>0\] and \[f(x)\] to be decreasing for \[x<0\].
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