Problem 71 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{x \rightarrow 1 / \alpha} \frac{1-\cos \left(c x^{2}+b x+a\right)}{(1-x \alpha)^{2}}\), where \(\alpha\) is a root of \(a x^{2}+b x+c=0\), is equal to (A) \(\frac{b^{2}-4 a c}{2 \alpha^{2}}\) (B) \(\frac{b^{2}-4 a c}{\alpha^{2}}\) (C) \(\frac{4 a c-b^{2}}{2 \alpha^{2}}\) (D) None of these

Step-by-Step Solution

Answer

The answer is (B) \( \frac{b^{2}-4ac}{\alpha^{2}} \).

Step 1: Identify the Root

The problem states that \( \alpha \) is a root of the quadratic equation \( ax^2 + bx + c = 0 \). To find the roots of the equation, we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Thus, the roots are \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 2: Substitute \( x = \frac{1}{\alpha} \) into the Expression

The expression to evaluate is \( \lim_{x \to \frac{1}{\alpha}} \frac{1 - \cos(cx^2 + bx + a)}{(1 - x \alpha)^2} \). Substitute \( x = \frac{1}{\alpha} \) into \( cx^2 + bx + a \). Given that \( \alpha \) is a root, \( a\alpha^2 + b\alpha + c = 0 \), so \( c \left(\frac{1}{\alpha^2}\right) + b \left(\frac{1}{\alpha}\right) + a = \frac{c}{\alpha^2} + \frac{b}{\alpha} + a \) simplifies via the root property to a related equivalent expression.

Step 3: Simplify \( 1 - \cos(0) \) near the Root

Using the fact that \( \alpha \) is a root and the expressions simplifies to zero, note that \( cx^2 + bx + a = 0 \) for \( x = \frac{1}{\alpha} \). Therefore the expression \( 1 - \cos(0) \) expands to \( 1 \), since \( \cos(0) = 1 \).

Step 4: Evaluate Limit Using L'Hôpital's Rule

Notice that \( \frac{0}{0} \) is an indeterminate form. Apply L'Hôpital's Rule, which allows us to differentiate the numerator and the denominator separately to find the limit. The derivative of \( 1 - \cos(u) \) with respect to \( u \) is \( \sin(u) \cdot u' \), and the derivative of \( (1 - x\alpha)^2 \) is \(-2\alpha(1-x\alpha) \).

Step 5: Apply Simplified Expressions and Evaluate the Limit

Apply the derivatives from the previous step: \( \sin(cx^2 + bx + a) \times (2cx + b) \) over \(-2\alpha(1-x\alpha)\). Evaluate these at the point where \( x \to \frac{1}{\alpha} \). Use Taylor expansion for sine \( \sin(u) \approx u \) for small \( u \); finally, plug in the value and find the limit using simplified algebraic manipulation.

Step 6: Simplify Further to Match Answer Choices

The steps give \( \frac{(cx^2 + bx + a)(2cx + b)}{\alpha^2} \); since \( cx^2 + bx + a \to 0 \), simplify to \( \frac{b^2 - 4ac}{\alpha^2} \). Check which answer option matches this expression, confirming it matches option \( B \).

Key Concepts

Quadratic Equation Roots

In the given exercise, the quadratic equation takes center stage. A quadratic equation typically looks like this: \( ax^2 + bx + c = 0 \). To find the roots of this equation, one can apply the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). These roots are essentially the values of \( x \) that make the equation true, meaning when you substitute these values into the equation, the left-hand side equals zero.
Understanding how to derive these roots involves a few key ideas:

• Discriminant \( (b^2 - 4ac) \): This part of the formula helps determine the number and nature of the roots. If the discriminant is positive, you have two distinct real roots. If zero, there's one real root (a repeated root), and if negative, you'll end up with complex roots.
• Completing Square: An alternate method to derive the roots by rewriting the quadratic in a squared form. While the quadratic formula is often faster, completing the square provides a deeper understanding of the quadratic expressions.

In the exercise, \( \alpha \) is one of these roots. This is crucial, as the expression \( cx^2 + bx + a \) simplifies when \( x = \frac{1}{\alpha} \), leading us to easier manipulation of the complex limit problem at hand.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus to solve limits that arise in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In our exercise, as \( x \to \frac{1}{\alpha} \), both the numerator \( 1 - \cos(cx^2 + bx + a) \) and the denominator \( (1 - x \alpha)^2 \) approach zero, forming a \( \frac{0}{0} \) situation.
Here's how to apply the rule correctly:

• Derivatives Needed: Differentiate both the numerator and denominator independently. This means you need to know derivative formulas for functions, like \( \cos(u) \), which derives into \( -\sin(u) \times u' \), and polynomial terms like \( ax \), which simply become \( a \).
• Evaluate New Limit: Once all derivatives are calculated, re-evaluate the limit with these derivatives. Often, it'll simplify out of the indeterminate form, allowing computation of the actual limit value.

L'Hôpital's Rule is used in step 4 of the exercise solution to handle the expression at \( x = \frac{1}{\alpha} \), helping us navigate through complex algebraic forms into a manageable computation.

Indeterminate Forms in Calculus

Indeterminate forms in calculus often arise in limit problems when straightforward substitution results in undefined or ambiguous expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or even more complex forms such as \( 0 \cdot \infty \). During the evaluation of our limit, specifically at \( x = \frac{1}{\alpha} \), we encounter the form \( \frac{0}{0} \), naturally prompting the use of L'Hôpital's Rule.

• Recognizing Indeterminate Forms: Knowing when an expression takes one of these forms is crucial for applying techniques like L'Hôpital's Rule or algebraic manipulation to find definitive answers.
• Other Techniques: Besides L'Hôpital's Rule, other methods include algebraic simplification, multiplying by conjugates, or using Taylor series expansions to rewrite complex expressions.

Indeterminate forms challenge our basic operations but also pave the way for deeper manipulation of mathematical expressions. These forms are not merely about undefined edges, they invite exploration into the behavior of functions around these interesting values.