If [x] denotes the greatest integer not exceeding x and if the function f defind by $f\left(x\right)=\left\{\begin{array}{l}\frac{a+2\cos x}{x^2} x<0\\ b \tan \frac{\mathrm{\pi }}{\left[x+4\right]} x\ge 0\end{array}\right.$ is continuous at x = 0 then (a, b) =

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}\frac{\mathrm{a}+2 \cos \mathrm{x}}{{\mathrm{x}}^2} :& \mathrm{x}<0\\ \mathrm{b} \tan \frac{\mathrm{\pi }}{[\mathrm{x}+4]} :& \mathrm{x}>=0 \mathrm{is} \mathrm{continuous} \mathrm{at} \mathrm{x}=0\end{array}\right. \\ \underset{\mathrm{x}\to 0^{-}}{\lim }\frac{\mathrm{a}+2 \cos \mathrm{x}}{{\mathrm{x}}^2}=\underset{\mathrm{x}\to 0^{+}}{\lim }\mathrm{b} \tan \frac{\mathrm{\pi }}{[\mathrm{x}+4]} \\ \mathrm{Limit} \mathrm{exists} \mathrm{if} \mathrm{a}+2=0\Rightarrow \mathrm{a}=-2 \\ \underset{\mathrm{x}\to 0}{\lim }\frac{-2+2 \cos \mathrm{x}}{{\mathrm{x}}^2}=-2 \underset{\mathrm{x}\to 0}{\lim }\frac{1-\cos \mathrm{x}}{{\mathrm{x}}^2}=-2\left(\frac{1}{2}\right)=-1 \\ -1=\mathrm{b} \tan \frac{\mathrm{\pi }}{4}\Rightarrow \mathrm{b}=-1 \\ \left(\mathrm{a},\mathrm{b}\right)=\left(-2,-1\right) \end{gathered}$$