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The interval in which f(x) = $x e^{2 - x}$ increases is

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$(0, \infty)$

$(1, \infty)$

$(- \infty, 1)$

$(- \infty, 2)$

answer is A.

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Detailed Solution

$f (x) = {xe}^{2 - x} f^{I} (x) = e^{2 - x} + {xe}^{2 - x} (- 1) = e^{2 - x} (1 - x)$

$f (x)$ is increasing

$\Rightarrow f^{I} (x) > 0 e^{2 - x} (1 - x) > 0 \Rightarrow 1 - x > 0 \Rightarrow x < 1 \Rightarrow x \in (- \infty, 1)$

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