The correct option is(D): 0,2.
We have,
\( y=x^2e^{-x}\)
∴ \(\frac{dy}{dx}\)\(=2xe^{-x}-x^2e^{-x}=xe^{-x}(2-x)\)
Now, \(\frac{dy}{dx}=0\)
⇒ x=0, and x=2
The points x = 0 and x = 2 divide the real line into three disjoint intervals
i.e.,(-∞,0), (0,2), and(2,∞).
In intervals (-∞,0) and (2,∞), f'(x)<0 as e-x is always positive.
∴ f is decreasing on (-∞,0) and (2,∞).
In interval (0, 2), f'(x)>0.
f is strictly increasing on (0, 2).
Hence, f is strictly increasing in interval (0, 2).
The correct answer is D.

A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:
(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?