Let $f(x) =\left\{\begin{matrix}\frac{x(1+a\cos x)-b\sin x}{x^3}&;x≠0\\1&;x=0\end{matrix}\right.$; If f(x) is continuous at x = 0 then a & b are given by

$-\frac{5}{2},-\frac{3}{2}$

f(x) is continuous at x = 0

$∴\underset{x→0}{\lim}f(x) = f(0)$

$\underset{x→0}{\lim}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$

$\underset{x→0}{\lim}\frac{x\left\{1+a\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+...\right)\right\}-b\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}+...\right)}{x^3}=1$

For existing of above $1 + a – b = 0$ .....(1)

$\frac{-a}{2}+\frac{b}{6}=1$   ....(2)

on solving (1) & (2) we get $a = –5/2, b = –3/2$