MCQ
The function $f(x) = x^2e^{-x}$ is monotonic increasing when :
- A$\text{x}\in\text{R}-[0,2]$
- ✓$0 < \text{x} < 2$
- C$2 < \text{x} < \infty$
- D$\text{x} < 0$
✓
Answer
Correct option: B.
$0 < \text{x} < 2$
$f(x) = x^2e^{-x}$
$\Rightarrow f\ '(x) = -x^2e^{-x} + 2x\ e^{-x}$
$\Rightarrow f\ '(x) = -e^{-x}x(x - 2)$
Given that function is monotonically increasing.
$-e^{-x}x(x - 2) > 0$
$x(x - 2) < 0$
$0 < x < 2$
$\Rightarrow f\ '(x) = -x^2e^{-x} + 2x\ e^{-x}$
$\Rightarrow f\ '(x) = -e^{-x}x(x - 2)$
Given that function is monotonically increasing.
$-e^{-x}x(x - 2) > 0$
$x(x - 2) < 0$
$0 < x < 2$
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