If a function f(x) = begin{cases} frac{sqrt[3 | vedclass

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(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = 5$. First,consider the left-hand limit:

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$\frac{9}{2}$

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(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = 5$. First,consider the left-hand limit: $\lim_{x 	o 0^-} \frac{(1+ax^2+bx^3)^{1/3} - (1-ax^2-bx^3)^{1/3}}{x^2}$. Using the binomial expansion $(1+u)^{1/3} \approx 1 + \frac{1}{3}u - \frac{1}{9}u^2$,we get: $\lim_{x 	o 0^-} \frac{(1 + \frac{1}{3}(ax^2+bx^3)) - (1 + \frac{1}{3}(-ax^2-bx^3))}{x^2} = \lim_{x 	o 0^-} \frac{\frac{2}{3}ax^2 + \frac{2}{3}bx^3}{x^2} = \frac{2}{3}a$. Since the limit is $5$,we have $\frac{2}{3}a = 5 \implies a = \frac{15}{2}$. Next,consider the right-hand limit: $\lim_{x 	o 0^+} \frac{	an 3x - \sin 3x}{bx^3}$. Using Taylor series $	an 3x \approx 3x + \frac{(3x)^3}{3}$ and $\sin 3x \approx 3x - \frac{(3x)^3}{6}$: $\lim_{x 	o 0^+} \frac{(3x + 9x^3) - (3x - \frac{27}{6}x^3)}{bx^3} = \lim_{x 	o 0^+} \frac{9x^3 + 4.5x^3}{bx^3} = \frac{13.5}{b} = \frac{27}{2b}$. Since the limit is $5$,we have $\frac{27}{2b} = 5 \implies b = \frac{27}{10}$. Wait,re-evaluating the expansion: $	an 3x - \sin 3x = \sin 3x (\sec 3x - 1) \approx (3x)(1 + \frac{(3x)^2}{2} - 1) = \frac{27x^3}{2}$. So $\frac{27/2}{b} = 5 \implies b = \frac{27}{10}$. Geometric mean of $a$ and $b$ is $\sqrt{ab} = \sqrt{\frac{15}{2} 	imes \frac{27}{10}} = \sqrt{\frac{3 	imes 27}{2 	imes 2}} = \sqrt{\frac{81}{4}} = \frac{9}{2}$.

If a function $f(x) = \begin{cases} \frac{\sqrt[3]{1+ax^2+bx^3}-\sqrt[3]{1-ax^2-bx^3}}{x^2}, & x < 0 \\ 5, & x=0 \\ \frac{\tan 3x - \sin 3x}{bx^3}, & x > 0 \end{cases}$ is continuous at $x=0$,then the geometric mean of $a$ and $b$ is

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