Let $f (x) = {\begin{cases} x e^{a x}; x \leq 0 \\ x + a x^{2} - x^{3}; x > 0 \end{cases}$ where a is constant. In which of the following intervals $f^{1} (x)$ is increasing

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$(- \infty, \frac{a}{3}), a < 0$

$(- \infty, \frac{- 2}{a}) \cup (\frac{a}{3}, \infty)$

$(- \frac{2}{a}, \frac{a}{3}), a > 0$

$(- \frac{2}{a}, \infty), a < 0$

answer is C.

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Detailed Solution

$L e t f (x) = {\begin{cases} x e^{a x} \\ x + a x^{2} - x^{3^{...............}} \end{cases} f (x) = {\begin{cases} x e^{a x; x \leq 0} \\ x + a x^{2} - x^{3} x > 0 \end{cases} [f (x) i s c o n t i n u o u s a t x = 0] \Rightarrow f (x) = {\begin{cases} (a x + 1) e^{a x}; x \leq 0 \\ 1 + 2 a x - 3 x^{2}; x > 0 \end{cases}$

[ $f^{1} (x)$ is continuous at $x = 0$ ] And $f^{11} (x) = {\begin{cases} 2 a e^{a x} + a^{2} \times e^{a x}; x \leq 0 \\ 2 a - 6 x; x > 0 \end{cases}$
$[f^{11} (x) i s c o n t i n u o u s a t x = 0]$
Now $f^{11} (x) > 0 \Rightarrow 2 a - 6 x > 0$ if $x > 0$ or $2 a e^{a x} + a^{2} \times e^{a x} > 0$ if $x \leq 0$
$\Rightarrow x < \frac{a}{3}$ if $x > 0$ or $x > - \frac{2}{a}$ if $x \leq 0 \Rightarrow f^{11} (x) > 0$ if $- \frac{2}{a} < x < \frac{a}{3}$
$∴ f^{11} (x)$ increases at $x \in (- \frac{2}{a}, \frac{a}{3})$