question_answer

The point for the curve \[y=x{{e}^{x}}\]                                                [MNR 1990]

A)            \[x=-1\]is minimum

B)            \[x=0\]is minimum

C)            \[x=-1\]is maximum

D)             \[x=0\]is maximum

Correct Answer: A

Solution :

           Given equation (curve) \[y=x{{e}^{x}}\]                    \[\therefore \frac{dy}{dx}=x{{e}^{x}}+{{e}^{x}}={{e}^{x}}(1+x)\] and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=(x+2)\,{{e}^{x}}\]                    For maximum or minimum value of \[f(x)\],                    Þ \[\frac{dy}{dx}=0\Rightarrow x=-1\]. \[\therefore {{\left\{ {f}''(x) \right\}}_{x=-1}}=+ve\]                    Hence \[f(x)\]is minimum at \[x=-1\].