If $f(x)=\left\{\begin{array}{ll}e^x, & x<2 \\ a x+b, & x \geq 2\end{array}\right.$ is differentiable for all $x \in R$, then

$a=e^2, b=-e^2$

Clearly, f(x) is everywhere continuous and differentiable except possible at x = 2.

At x = 2, we have,

$\lim\limits_{x \rightarrow 2^{-}} f(x)=\lim\limits_{x \rightarrow 2^{-}} e^x=e^2$

$\lim\limits_{x \rightarrow 2^{+}} f(x)=\lim\limits_{x \rightarrow 2^{+}} a x+b=2 a+b$

$f(2)=2 a+b$

Also,

(LHD at x = 2) = $\left(\frac{d}{d x}\left(e^x\right)\right)_{x=2}=e^2$

(RHD at x = 2) = $\left(\frac{d}{d x}(a x+b)\right)_{x=2}=a$

For f(x) to be differentiable at x = 2, it should be both continuous and differentiable at x = 2

∴  $\lim\limits_{x \rightarrow 2^{-}} f(x)=\lim\limits_{x \rightarrow 2^{+}} f(x)=f(2)$

and, (LHD at x = 2) = (RHD at x = 2)

$\Rightarrow e^2=2 a+b$ and $a=e^2 \Rightarrow a=e^2$ and $b=-e^2$