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The interval in which $y=x^2 e^{-x}$ is increasing is
(a) $(-\infty, \infty)($ b) $(-2,0)($ c) $(2, \infty)$ (d) $(0,2)$
(a) $(-\infty, \infty)($ b) $(-2,0)($ c) $(2, \infty)$ (d) $(0,2)$
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(d) $f^{\prime}(x)=2 x e^{-x}+x^2\left(-e^{-x}\right)=x e^{-x}(2-x)=e^{-x} x(2-x)$
Now $\mathrm{e}^{-\mathrm{x}}$ is positive for all $\mathrm{x} \in \mathrm{R} \mathrm{f}^{\prime}(\mathrm{x})=0$ at $\mathrm{x}=0,2$
$x=0, x=2$ divide the number line into three disjoint intervals.
viz. $(-\infty, 0),(0,2),(2, \infty)$
(a) Interval $(-\infty, 0) \quad \mathrm{x}$ is $+\mathrm{ve}$ and $(2-\mathrm{x})$ is $+\mathrm{ve}$ $\therefore \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \mathrm{x}(2-\mathrm{x})=(+)(-)(+)=-$ ve
$\Rightarrow \mathrm{f}$ is decreasing in $(-\infty, 0)$
(b) Interval $(0,2) \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \mathrm{x}(2-\mathrm{x})$
$=(+)(+)(+)=+$ ve $\Rightarrow f$ is increasing in $(0,2)$
(c) Interval $(2, \infty) \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \mathrm{x}(2-\mathrm{x})=(+)(+)(-)$
$=-\mathrm{ve} \Rightarrow \mathrm{f}$ is decreasing in the interval $(2, \infty)$
Now $\mathrm{e}^{-\mathrm{x}}$ is positive for all $\mathrm{x} \in \mathrm{R} \mathrm{f}^{\prime}(\mathrm{x})=0$ at $\mathrm{x}=0,2$
$x=0, x=2$ divide the number line into three disjoint intervals.
viz. $(-\infty, 0),(0,2),(2, \infty)$
(a) Interval $(-\infty, 0) \quad \mathrm{x}$ is $+\mathrm{ve}$ and $(2-\mathrm{x})$ is $+\mathrm{ve}$ $\therefore \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \mathrm{x}(2-\mathrm{x})=(+)(-)(+)=-$ ve
$\Rightarrow \mathrm{f}$ is decreasing in $(-\infty, 0)$
(b) Interval $(0,2) \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \mathrm{x}(2-\mathrm{x})$
$=(+)(+)(+)=+$ ve $\Rightarrow f$ is increasing in $(0,2)$
(c) Interval $(2, \infty) \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \mathrm{x}(2-\mathrm{x})=(+)(+)(-)$
$=-\mathrm{ve} \Rightarrow \mathrm{f}$ is decreasing in the interval $(2, \infty)$
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