$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\left(x+3 x^{2}\right)^{1 / 3}-x^{1 / 3}}{x^{4 / 3}}$
$=\lim _{x \rightarrow 0} \frac{(1+3 x)^{1 / 3}-1}{x}=1$
$f(0)=b$
for continuity at $x=0$ $\lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x)$
$\Rightarrow \quad a+3=b=1$
$\therefore \quad a=-2, \quad b=1$
$\therefore \quad a+2 b=0$
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($A$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the green region below the line $\mathrm{L}_{\mathrm{h}}$
($B$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the red region below the line $\mathrm{L}_h$
($C$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the red region below the line $L_h$
($D$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the green region below the line $\mathrm{L}_{\mathrm{h}}$