If $\alpha, \beta$ are the roots of the equation $ax^{2}+bx+c=0$, then : $lim_{x\rightarrow\alpha}\frac{1-cos(ax^{2}+bx+c)}{(x-\alpha)^{2}}=$

Question

Accepted Answer

To solve the problem, we need to evaluate the limit:

$$
\lim_{x \rightarrow \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}
$$

Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0$, we know:

$$ a\alpha^2 + b\alpha + c = 0 $$

This implies that at $x = \alpha$, the expression $ax^2 + bx + c$ becomes zero. Therefore, near $x = \alpha$, we can approximate:

$$ ax^2 + bx + c \approx a(x - \alpha)(x - \beta) $$

For small values of $\theta$, we know that $1 - \cos(\theta) \approx \frac{\theta^2}{2}$. Applying this approximation, we have:

$$ 1 - \cos(ax^2 + bx + c) \approx \frac{(ax^2 + bx + c)^2}{2} $$

Substituting the approximation for $ax^2 + bx + c$:

$$ 1 - \cos(ax^2 + bx + c) \approx \frac{(a(x - \alpha)(x - \beta))^2}{2} $$

Now, substituting this into the limit expression:

$$
\lim_{x \rightarrow \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2} = \lim_{x \rightarrow \alpha} \frac{\frac{(a(x - \alpha)(x - \beta))^2}{2}}{(x - \alpha)^2}
$$

Simplifying the expression:

$$
= \lim_{x \rightarrow \alpha} \frac{a^2(x - \alpha)^2(x - \beta)^2}{2(x - \alpha)^2}
$$

Canceling $(x - \alpha)^2$ from the numerator and the denominator:

$$
= \lim_{x \rightarrow \alpha} \frac{a^2(x - \beta)^2}{2}
$$

Substituting $x = \alpha$ into the remaining expression:

$$
= \frac{a^2(\alpha - \beta)^2}{2}
$$

Thus, the correct answer is Option 1: $\frac{a^2}{2}(\alpha - \beta)^2$.

**Explanation for Incorrect Options:**

- **Option 2: $\frac{a}{2}(\alpha - \beta)^2$**: This option incorrectly assumes that the factor of $a$ is not squared, which does not match the derived expression where $a^2$ is present.

- **Option 3: $\frac{-a^2}{2}(\alpha - \beta)^2$**: This option introduces an incorrect negative sign. The expression derived from the limit does not have a negative sign.

- **Option 4: $\frac{-a}{2}(\alpha - \beta)^2$**: Similar to Option 3, this option incorrectly includes a negative sign and also fails to square the factor of $a$.

Therefore, the correct answer is indeed Option 1, as it accurately reflects the derived expression from the limit evaluation.