A) \[-\frac{1}{e}\]
B) \[\frac{1}{e}\]
C) \[-e\]
D) e
Correct Answer: A
Solution :
[a] Let \[y=x{{e}^{x}}\]. Differentiate both side w.r.t. ?x?. \[\Rightarrow \frac{dy}{dx}={{e}^{x}}+x{{e}^{x}}={{e}^{x}}(1+x)\] Put \[\frac{dy}{dx}=0\] \[\Rightarrow {{e}^{x}}(1+x)=0\Rightarrow x=-1\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}+{{e}^{x}}(1+x)={{e}^{x}}(x+2)\] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{(x=-1)}}=\frac{1}{e}+0>0\] Hence, \[y=x{{e}^{x}}\] is minimum function and \[{{y}_{\min }}=-\frac{1}{e}\].
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