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

• A

  $\frac{3}{2}$
• Correct answer
  B

  $\frac{9}{2}$
• C

  $\frac{81}{4}$
• D

  $\frac{9}{4}$

Hence, the correct answer is option: 'B'.