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.

| S. No. | Particulars | Amount (in ₹ crore) |
|---|---|---|
| (i) | Operating Surplus | 3,740 |
| (ii) | Increase in unsold stock | 600 |
| (iii) | Sales | 10,625 |
| (iv) | Purchase of raw materials | 2,625 |
| (v) | Consumption of fixed capital | 500 |
| (vi) | Subsidies | 400 |
| (vii) | Indirect taxes | 1,200 |