We have,
$$
\begin{aligned}
f(x) & =\left\{\begin{array}{cc}
a x^2+b ; & x < -1 \\
b x^2+a x+4 ; & x \geq-1
\end{array}\right. \\
\therefore \quad f^{\prime}(x) & =\left\{\begin{array}{cc}
2 a x, & x < -1 \\
2 b x+a, & x \geq-1
\end{array}\right.
\end{aligned}
$$
$\because f(x)$ is differentiable at $x=-1$
$\therefore$ It is continuous at $x=-1$ and hence
$$
\begin{array}{ll}
& \lim _{\left(x \rightarrow-1^{-}\right)} f(x)=\lim _{x \rightarrow-1^{+}} f(x) \\
\Rightarrow \quad & a+b=b-a+4 \\
\Rightarrow \quad & a=2 \\
\text { and also, } & \lim _{\left(x \rightarrow-1^{-}\right)} f^{\prime}(x)=\lim _{x \rightarrow-1^{+}} f^{\prime}(x) \\
\Rightarrow \quad & -2 a=-2 b+a \\
\Rightarrow \quad & 3 a=2 b \Rightarrow b=3 \quad(\because a=2)
\end{array}
$$
Hence, $a=2, b=3$