If $f(x)= \begin{cases}\frac{A+3 \cos x}{x^2}, & x<0 \\ B \tan \frac{\pi}{[x+3]}, & x \geq 0\end{cases}$, where [.] represents the greatest integer function, is continuous at x = 0. Then,

$A=-3, B=-\frac{\sqrt{3}}{2}$

We have,

$\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0} \frac{A+3 \cos x}{x^2}$

$\Rightarrow \lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0} \frac{A+3\left(1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !}+...\right)}{x^2}$

$\Rightarrow \lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0}\left(\frac{A+3}{x^2}-\frac{3}{2 !}+\frac{3}{4 !} x^2-\frac{3}{6 !} x^4+...\right)$

For this limit to exist, we must have A + 3 = 0 and in that case, we have

$\lim\limits_{x \rightarrow 0^{-}} f(x)=-\frac{3}{2}$

Now,

$\lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0} B \tan \left(\frac{\pi}{[x+3]}\right)=B \tan \frac{\pi}{3}=B \sqrt{3}$

Also, f(0) = $B \tan \frac{\pi}{3}=B \sqrt{3}$

Thus, if f(x) is continuous at x = 0, then

A + 3 = 0 and, $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{+}} f(x)=f(0)$

$\Rightarrow A+3=0$ and, $-\frac{3}{2}=B \sqrt{3} \Rightarrow A=-3$ and $B=-\frac{\sqrt{3}}{2}$