#### Answer

For the derivative to be continuous at $x=1$ we have to choose $a=4$.

#### Work Step by Step

By definition of continuity of the derivative at $x=a$, $f'$ is continuous if
$$\lim_{x\to a^+}f'(x)=\lim_{x\to a^-}f'(x)=f'(a).$$
We will first find the derivative on the both sides of $x=1$.
If $x\leq1$ then
$$f'(x)=\lim_{h\to0}\frac{2(x+h)^2-2x^2}{h}=\lim_{h\to0}\frac{2(x^2+h^2+2xh)-2x^2}{h}=\lim_{h\to0}\frac{2x^2+2h^2+4xh-2x^2}{h}=\lim_{h\to0}\frac{2h^2+4xh}{h}=\lim_{h\to0}(2h+4x)=2\cdot0+4x=4x.$$
If $x>1$ then
$$f'(x)=\lim_{h\to0}\frac{a(x+h)-2-(ax-2)}{h}=\lim_{h\to0}\frac{ax+ah-2-ax+2}{h}=\lim_{h\to0}\frac{ah}{h}=\lim_{h\to0}a=a.$$
Now we will find the one-sided limits.
Because when $x\leq1$ $f'(x)=4x$ we have
$$\lim_{x\to 1^{+}}4x=4\cdot1=4.$$
When $x>1$ then $f'(x)=a$ so we have
$$\lim_{x\to1^-}a=a$$. For those one-sided limits to be equal we see that $a=4$. Then also $f'(1)=4$ so when $a=4$ the derivative is continuous.