Let $\text{⨍ (x)=}\frac{\mathrm{x}(1+\mathrm{acos} \mathrm{x}-\mathrm{bsin} \mathrm{x})}{{\mathrm{x}}^3}\text{, x≠0}$ and $\text{⨍(0)=1,}$ then values if $ʻ\mathrm{a}ʼ$ and $ʻ\mathrm{b}ʼ$ so that $\text{ʻ⨍ ʼ}$ is continuous are

$\begin{array}{l}f\left(x\right)=\underset{x\to 0}{lim} \frac{x\left(1+a\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\dots \dots \right)\right)-b\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\right)}{x^3}\\ \underset{x\to 0}{lim} \frac{(1+a-b)+x^2\left(\frac{-a}{2!}+\frac{b}{3!}\right)}{x^2}\\ \Rightarrow 1+a-b=0\dots .\left(\mathrm{i}\right)\\ \text{ and }\frac{-a}{2}+\frac{b}{6}=1\dots ..\text{ (ii) }\\ \Rightarrow \mathrm{a}=\frac{-5}{2};\mathrm{b}=\frac{-3}{2}\end{array}$