Question
The function $$f\left( x \right) = \frac{{{x^2}}}{{{e^x}}}$$ monotonically increasing if :
A.
$$x < 0{\text{ only}}$$
B.
$$x > 2{\text{ only}}$$
C.
$$0 < x < 2$$
D.
$$x\, \in \left( { - \infty ,\,0} \right) \cup \left( {2,\,\infty } \right)$$
Answer :
$$0 < x < 2$$
Solution :
$$\eqalign{
& f\left( x \right) = \frac{{{x^2}}}{{{e^x}}}\,; \cr
& f'\left( x \right) = \frac{{2x.{e^x} - {e^x}.{x^2}}}{{{{\left( {{e^x}} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{2x - {x^2}}}{{{e^x}}} \cr} $$
as $${e^x}$$ is always positive and for monotonically increasing ;
$$\eqalign{
& 2x - {x^2} > 0 \cr
& \Rightarrow {x^2} - 2x < 0 \cr
& \Rightarrow x\left( {x - 2} \right) < 0 \cr
& \Rightarrow x\, \in \,\left( {0,\,2} \right)\, \cr} $$