Question

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

• A. \((\alpha - \beta)^2\)
• B. \(\frac{1}{2}(\alpha - \beta)^2\)
• C. \(\frac{a^2}{4}(\alpha - \beta)^2\)
• D. \(\frac{a^2}{2}(\alpha - \beta)^2\)

Hint:

When dealing with limits involving trigonometric functions and polynomials, use Taylor series expansions around the point of interest to simplify the expressions and solve the limit.

Answer 1

The Correct Option is D

Step 1: The quadratic equation is \(ax^2 + bx + c = 0\) with roots \(\alpha\) and \(\beta\). Vieta's formulas provide:

\[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}. \]

Step 2: The limit's expression contains \(1 - \cos(ax^2 + bx + c)\), which becomes \(1 - \cos(0)\) at the roots, necessitating a series expansion.

Step 3: Employ the Taylor series for \(\cos z\) around \(z = 0\): \[ \cos(z) \approx 1 - \frac{z^2}{2}. \] Apply this approximation to \(ax^2 + bx + c\), specifically near \(x = \beta\): \[ ax^2 + bx + c \approx a(x - \beta)^2. \]

Step 4: Substitute into the limit: \[ \lim_{x \to \beta} \frac{1 - \cos(ax^2 + bx + c)}{(x - \beta)^2} \approx \lim_{x \to \beta} \frac{1 - \left(1 - \frac{a^2(x - \beta)^4}{2}\right)}{(x - \beta)^2}. \] Simplify: \[ \lim_{x \to \beta} \frac{\frac{a^2(x - \beta)^4}{2}}{(x - \beta)^2} = \frac{a^2}{2}(x - \beta)^2. \]

Step 5: Simplify further to obtain: \[ \frac{a^2}{2}(\alpha - \beta)^2. \] The solution is: \[ \boxed{\frac{a^2}{2}(\alpha - \beta)^2}. \]