The interval in which \(y=x^2e^{-x}\) is increasing is
- (-∞,∞)
- (−2, 0)
- (2,∞)
- (0,2)
The Correct Option is D
Solution and Explanation
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.
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