Let ` alpha and beta ` be the distinct root of ` ax^(2) + bx + c=0 ` then` lim_(x to 0) (1- cos (ax^(2)+ bx+c))/((x-alpha)^(2))` is equal to

Options

A. `1/2 (alpha-beta)^(2)`
B. `- a^(2)/2 (alpha -beta)^(2)`
C. 0
D. `a^(2)/2 (alpha -beta)^(2)`

Answer

To solve the limit problem, we start with the given expression:

\[ \lim_{x \to 0} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2} \]

where \(\alpha\) and \(\beta\) are the distinct roots of the quadratic equation \(ax^2 + bx + c = 0\).

### Step 1: Rewrite the quadratic expression
Since \(\alpha\) and \(\beta\) are the roots, we can express the quadratic in factored form:

\[ ax^2 + bx + c = a(x - \alpha)(x - \beta) \]

### Step 2: Evaluate the limit
We need to evaluate the limit as \(x\) approaches \(\alpha\) (since \(\alpha\) is a root of the quadratic). Thus, we rewrite the limit:

\[ \lim_{x \to \alpha} \frac{1 - \cos(a(x - \alpha)(x - \beta))}{(x - \alpha)^2} \]

### Step 3: Use the small-angle approximation
As \(x\) approaches \(\alpha\), the expression \(a(x - \alpha)(x - \beta)\) approaches \(0\). We can use the approximation \(1 - \cos(u) \approx \frac{u^2}{2}\) for small \(u\):

\[ 1 - \cos(a(x - \alpha)(x - \beta)) \approx \frac{(a(x - \alpha)(x - \beta))^2}{2} \]

### Step 4: Substitute the approximation into the limit
Substituting this approximation into our limit gives:

\[ \lim_{x \to \alpha} \frac{\frac{(a(x - \alpha)(x - \beta))^2}{2}}{(x - \alpha)^2} \]

### Step 5: Simplify the limit
This simplifies to:

\[ \lim_{x \to \alpha} \frac{a^2(x - \alpha)^2(x - \beta)^2}{2(x - \alpha)^2} \]

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

\[ \lim_{x \to \alpha} \frac{a^2(x - \beta)^2}{2} \]

### Step 6: Evaluate the limit
Now, substituting \(x = \alpha\):

\[ \frac{a^2(\alpha - \beta)^2}{2} \]

### Final Result
Thus, the limit evaluates to:

\[ \frac{a^2(\alpha - \beta)^2}{2} \]

### Summary
The final answer is:

\[ \lim_{x \to 0} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2} = \frac{a^2(\alpha - \beta)^2}{2} \]