Problem 103 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(a x^{2}+b x+c=0\), then \(\lim _{x \rightarrow \frac{1}{\alpha}} \sqrt{\frac{1-\cos \left(c x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}}=\) (A) \(\left|\frac{c}{2 \alpha}\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)\right|\) (B) \(\left|\frac{c}{2 \beta}\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)\right|\) (C) \(\left|\frac{c}{\alpha \beta}\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)\right|\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is (C).

Step 1: Recall Vieta's Formulas

According to Vieta's formulas for the quadratic equation \(a x^2 + b x + c = 0\), the sum of the roots \(\alpha\) and \(\beta\) is \(\alpha + \beta = -\frac{b}{a}\), and the product of the roots is \(\alpha \beta = \frac{c}{a}\).

Step 2: Simplify the Limit Expression

The limit expression is \(\lim_{x \rightarrow \frac{1}{\alpha}} \sqrt{\frac{1-\cos(c x^2 + b x + a)}{2(1-\alpha x)^2}}\). As \(x \rightarrow \frac{1}{\alpha}\), \(1 - \alpha x \rightarrow 0\). Using the first-order Taylor expansion, \(1 - \cos(u) \approx \frac{u^2}{2}\) as \(u \rightarrow 0\).

Step 3: Substitute the Expression for cos

Let \(u = c x^2 + b x + a\). At \(x = \frac{1}{\alpha}\), \(u \approx c\left(\frac{1}{\alpha}\right)^2 + b\left(\frac{1}{\alpha}\right) + a\). Simplifying gives \(u \approx \frac{c}{\alpha^2} + \frac{b}{\alpha} + a\). Since \(\alpha\) is a root, \(\alpha^2 c + \alpha b + a = 0\). So, \(u \approx \frac{1}{\alpha} - \frac{1}{\beta}\).

Step 4: Evaluate the Limit Using Hyperbolic Angle

The expression becomes \(\lim_{x \rightarrow \frac{1}{\alpha}} \sqrt{\frac{\left(\frac{1}{2}\right)(\frac{c}{a} \left(\frac{1}{\alpha} - \frac{1}{\beta} \right)^2 )}{2(1 - \alpha x)^2}}\). This simplifies by considering the \(\sin^2\) formula approximation. Using the condition \(1 - \cos(u)\), this is effectively substituting into a complex form.

Step 5: Simplify and Select the Correct Option

Following the simplified form from Step 4, it follows the limit resolves to \(\frac{c}{\alpha \beta}\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)\) after simplifying the \(\sin^2\) approximation. This results in the choice matching option (C).

Key Concepts

Roots of the Equation

In the context of quadratic equations, the roots are simply the solutions to the equation where it equals zero. For a typical quadratic equation of the form \(ax^2 + bx + c = 0\), the roots can be found using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). These values of \(x\) make the equation true, effectively finding the points where the parabola intersects the x-axis.

In the original exercise, \(\alpha\) and \(\beta\) represent these roots. Importantly, the relationships of these roots such as their sum and product are encapsulated in Vieta's Formulas. Understanding these concepts is crucial for delving deeper into the exercise's problem-solving context, especially since knowing the algebraic manipulation of roots allows for substitutions and simplifications in further calculus operations.

Limits in Calculus

Limits are essential in calculus for understanding the behavior of functions as inputs approach a certain value. In our original problem, we tackle the limit \(\lim_{x \rightarrow \frac{1}{\alpha}}\), which examines what happens to the function as \(x\) gets incredibly close to \(\frac{1}{\alpha}\), a root of the equation.

As \(x\) nears \(\frac{1}{\alpha}\), the expression \(1 - \alpha x\) approaches zero. To deal with this, we deploy tools from calculus such as Taylor expansions. Here, the Taylor series helps approximate trigonometric terms, such as \(1 - \cos(u) \approx \frac{u^2}{2}\), simplifying the problem into a more manageable form. This approximation is key to understanding how the limit resolves, especially when plugging into and breaking down complex functions.

Vieta's Formulas

Vieta's Formulas are fundamental for relating the coefficients of a polynomial to sums and products of its roots. Specifically, in the quadratic equation \(ax^2 + bx + c = 0\), these formulas assert:

• The sum of the roots \(\alpha + \beta = -\frac{b}{a}\)
• The product of the roots \(\alpha \beta = \frac{c}{a}\)

These formulas simplify complex problems by providing a straightforward way to link root behaviors back to the original equation's coefficients.

In the exercise, Vieta's Formulas enabled the transition from the expression within the square root to a form that could be evaluated by limits. Since \(\alpha\) and \(\beta\) are roots, they satisfied the equation \(\alpha^2c + \alpha b + a = 0\), allowing substitution for further simplification in the subsequent steps of finding the limit. Understanding and applying Vieta's Formulas thus bridges algebra and calculus elegantly.