Explanation
Step -1: Find values of x at which the given function changes its nature.
The given function is y(x)=x2e−x
dydx=x2ddxe−x+e−xddxx2
=−x2e−x+2xe−x
Put dydx=0.
⇒−x2e−x+2xe−x=0
⇒xe−x(−x+2)=0
∵
\therefore\text{Either }x=0\text{ or }2-x=0
\Rightarrow x=0\text{ or }2
\textbf{Step -2: Find the interval in which the given function is increasing.}
\text{The points }x=0\text{ and }x=2\text{ divide the real line into three disjoint intervals,}
\text{i.e., }(-\infty,0),(0,2)\text{ and }(2,\infty)
\because\text{In the interval }(0,2),\dfrac{dy}{dx}>0.
\therefore \text{In }(0,2),\text{ the given function is increasing.}
\textbf{Hence, The correct option is D.}
\left| dy \right| >\left| dx \right|
\left| \dfrac { dy }{ dx } \right| >1
{ x }^{ 3 }=12y
diff wrto y
3{ x }^{ 2 }\dfrac { dx }{ dy } =12
\dfrac { dx }{ dy } =\dfrac { 12 }{ { 3x }^{ 2 } } =\dfrac { 4 }{ { x }^{ 2 } }
\left| \dfrac { y }{ { x }^{ 2 } } \right| >1
\left| { x }^{ 2 } \right| >4
x>\pm 2
\left( 2,-2 \right)
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