A) x < 0 only
B) x > 2 only
C) 0 < x < 2
D) \[x\in (-\infty ,0)\cup (2,\infty )\]
Correct Answer: C
Solution :
[c] \[f(x)=\frac{{{x}^{2}}}{{{e}^{x}}};f'(x)=\frac{2x.{{e}^{x}}-{{e}^{x}}.{{x}^{2}}}{{{\left( {{e}^{x}} \right)}^{2}}}\] \[f'(x)=\frac{2x-{{x}^{2}}}{{{e}^{x}}}\] As \[{{e}^{x}}\] is always positive and for monotonically increasing; \[2x-{{x}^{2}}>0\] \[\Rightarrow {{x}^{2}}-2x<0\Rightarrow x(x-2)<0\Rightarrow x\in (0,2)\]
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