Let $f\left(x\right)=\left\{\begin{array}{cc}ax,& x<2\\ a{x}^2+bx+3,& x\ge 2\end{array}\right.$ If $f$ is differentiable for all $x$ then the value of $(a, b)$ is equal to

$\underset{x\to 2-}{lim} f\left(x\right)=\underset{x\to 2-}{lim} ax=2a$ and $\underset{x\to 2+}{lim} f\left(x\right)=\underset{x\to 2+}{lim}$

$\left(a{x}^2+bx+3\right)=4a+2b+3$

Since$f$ is differentiable so continuous, hence 

$2a+2b+3=0$

Also $\underset{h\to 0+}{lim} \frac{f(2+h)-f\left(2\right)}{h}$

$=\underset{h\to 0+}{lim} \frac{a(2+h{)}^2+b(2+h)+3-(4a+2b+3)}{h}$

$=\underset{h\to 0+}{lim} \frac{a{h}^2+4ah+bh}{h}$

$=4a+b$

$\underset{h\to 0-}{lim} \frac{f(2+h)-f\left(2\right)}{h}$

$=\underset{h\to 0-}{lim} \frac{a(2+h)-(4a+2b+3)}{h}$

$=a(\text{ using (i)) }$

So $a=4a+b\Rightarrow -3a=b$. Putting this in (i),

we get $a=\frac{3}{4},b=-\frac{9}{4}$