We need the piecewise definition at $x=0$ to involve the constants $\alpha,\beta$. Interpreting the intended question in standard form:

Since only the value at $x=0$ matters for continuity, effectively we need

$f(0)=\beta=\lim_{x\to 0}\frac{1-\cos 2x}{x^2}.$

However, to make sense of both $\alpha,\beta$ appearing in the answer, the usual intended form of such questions is

with continuity and differentiability at $x=0$. Then we determine both $\alpha$ and $\beta$.

1. Continuity at $x=0$

Using the identity

$1-\cos 2x=2\sin^2 x,$

we get

$\frac{1-\cos 2x}{x^2}=2\left(\frac{\sin x}{x}\right)^2.$

Hence,

$\lim_{x\to 0}\frac{1-\cos 2x}{x^2}=2.$

For continuity at $x=0$,

$f(0)=\beta=2.$

2. Differentiability at $x=0$ to find $\alpha$

For $x\ne 0$,

$f(x)=2\left(\frac{\sin x}{x}\right)^2.$

To find the derivative at $0$, use series:

$\cos 2x=1-2x^2+\frac{(2x)^4}{24}+\cdots=1-2x^2+\frac{2}{3}x^4+\cdots$

So,

$1-\cos 2x=2x^2-\frac{2}{3}x^4+\cdots$

and therefore

$\frac{1-\cos 2x}{x^2}=2-\frac{2}{3}x^2+\cdots$

Thus near $x=0$,

$f(x)=2-\frac{2}{3}x^2+\cdots$

Hence,

$f'(0)=0.$

If the value branch is linear with slope $\alpha$, then differentiability at $0$ gives

$\alpha=f'(0)=0.$

So,

$\alpha=0,\qquad \beta=2.$

3. Compute $\alpha^2+\beta^2$

$\alpha^2+\beta^2=0^2+2^2=4.$

This does not appear in the given options. Therefore the printed question is likely missing/mis-scanned. A common corrected version is with

$f(0)=\alpha x+\beta \text{ for } x\le 0$

and matching both continuity and differentiability from both sides, which can lead to option $12$. But from the question exactly as visible, continuity alone determines only the value at $0$, not both $\alpha,\beta$.

So the stored answer $12$ cannot be justified from the displayed question.