If $\text{⨍}\left(\mathrm{x}\right)\text{=}\frac{\mathrm{a} \cos \mathrm{x}-\cos \mathrm{bx}}{{\mathrm{x}}^2}\text{, x≠0 and ⨍(0)=4}$ is continues at x = 0 . Then the ordered pair (a, b) is

$$\begin{gathered} \text{⨍(0)=}\underset{\mathrm{x}\to 0}{\lim }\text{⨍(x)=}\underset{\mathrm{x}\to 0}{\lim }\left(\frac{\mathrm{a} \mathrm{cosx}-\mathrm{cosbx}}{{\mathrm{x}}^2}\right) \\ \Rightarrow \text{⨍(0)=}\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{a}\left(1-{\displaystyle \frac{{\mathrm{x}}^2}{2!}}\right)-\left(1-{\displaystyle \frac{{\mathrm{b}}^2{\mathrm{x}}^2}{2!}}\right)}{{\mathrm{x}}^2} \\ \Rightarrow \underset{\mathrm{x}\to 0}{\lim }\frac{\left(\mathrm{a}-1\right)}{{\mathrm{x}}^2}\text{-}\left(\frac{\mathrm{a}}{2}\text{-}\frac{{\mathrm{b}}^2}{2}\right)\text{=4} \\ \Rightarrow \text{a-1=0, }\frac{\mathrm{a}}{2}\text{-}\frac{{\mathrm{b}}^2}{2}\text{=-4} \\ \Rightarrow \mathrm{a}=1, \frac{\mathrm{a}}{2}-\frac{{\mathrm{b}}^2}{2}=-4 \\ \Rightarrow \mathrm{a}-{\mathrm{b}}^2=-8\Rightarrow 1+8={\mathrm{b}}^2\Rightarrow \mathrm{b}=\pm 3 \\ \therefore \left(\mathrm{a}, \mathrm{b}\right)=\left(1,\pm 3\right) \end{gathered}$$