1. Since $\alpha$ and $\beta$ are the distinct roots of $ax^2+bx+c=0,$ we can factor the quadratic as $ax^2+bx+c=a(x-\alpha)(x-\beta).$

2. We need to evaluate $\lim_{x\to \alpha}\frac{1-\cos(ax^2+bx+c)}{(x-\alpha)^2}.$ Using the factorization, $ax^2+bx+c=a(x-\alpha)(x-\beta).$ So the limit becomes $\lim_{x\to \alpha}\frac{1-\cos\big(a(x-\alpha)(x-\beta)\big)}{(x-\alpha)^2}.$

3. Let $t=a(x-\alpha)(x-\beta).$ As $x\to \alpha$, we have $t\to 0$. Use the standard limit $1-\cos t \sim \frac{t^2}{2} \quad (t\to 0).$ Hence,

Substitute $t=a(x-\alpha)(x-\beta)$:

4. Now take the limit $x\to \alpha$:

5. Therefore, $\boxed{\lim_{x\to \alpha}\frac{1-\cos(ax^2+bx+c)}{(x-\alpha)^2}=\frac{a^2(\alpha-\beta)^2}{2}}.$

6. Checking options:

• A: $\dfrac{a^2(\alpha-\beta)^2}{2}$ ✅
• B: $0$ ❌
• C: $-\dfrac{a^2(\alpha-\beta)^2}{2}$ ❌
• D: $\dfrac{(\alpha-\beta)^2}{2}$ ❌

So the correct option is A.