Problem 134 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(\alpha\) and \(\beta\) be the distinct roots of \(a x^{2}+b x+c=0\), then \(\lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^{2}+b x+c\right)}{(x-\alpha)^{2}}\) is equal to (A) \(\frac{a^{2}}{2}(\alpha-\beta)^{2}\) (B) 0 (C) \(-\frac{a^{2}}{2}(\alpha-\beta)^{2}\) (D) \(\frac{1}{2}(\alpha-\beta)^{2}\)

Step-by-Step Solution

Answer

(A) \( \frac{a^2}{2}(\alpha-\beta)^2 \)."

Step 1: Understanding the Limit Expression

We need to evaluate the limit \( \lim_{x \to \alpha} \frac{1 - \cos(f(x))}{(x-\alpha)^2} \), where \( f(x) = ax^2 + bx + c \). We'll use the standard limit \( \lim_{u \to 0} \frac{1 - \cos(u)}{u^2} = \frac{1}{2} \).

Step 2: Taylor Expansion of Cosine

For small values of \( u \), \( 1 - \cos(u) \approx \frac{u^2}{2} \). So, \( 1 - \cos(f(x)) \approx \frac{(f(x))^2}{2} \) as \( x \to \alpha \).

Step 3: Expressing the Quadratic in Factored Form

Since \( \alpha \) and \( \beta \) are roots, the quadratic can be written as \( a(x-\alpha)(x-\beta) \). So, \( f(x) = a(x-\alpha)(x-\beta) \).

Step 4: Substitute and Simplify the Limit Expression

Replace \( f(x) \) in the limit: \( \lim_{x \to \alpha} \frac{1 - \cos(a(x-\alpha)(x-\beta))}{(x-\alpha)^2} \). This becomes \( \lim_{x \to \alpha} \frac{\frac{(a(x-\alpha)(x-\beta))^2}{2}}{(x-\alpha)^2} \).

Step 5: Simplifying the Fraction

The numerator becomes \( \frac{a^2 (x-\alpha)^2 (x-\beta)^2}{2} \). Dividing by \( (x-\alpha)^2 \) results in \( \frac{a^2 (x-\beta)^2}{2} \).

Step 6: Evaluate the Limit as x approaches α

As \( x \to \alpha \), the expression \( (x-\beta)^2 \to (\alpha-\beta)^2 \). Therefore, the limit is \( \frac{a^2}{2} (\alpha-\beta)^2 \).

Key Concepts

Quadratic Equations

Quadratic equations are polynomials of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). These equations have at most two solutions, known as roots, which can be found using various methods like factoring, the quadratic formula, or completing the square.

• Factoring involves expressing the quadratic as a product of its linear factors: \( a(x - \alpha)(x - \beta) \), where \( \alpha \) and \( \beta \) are the roots of the equation.
• The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is a general method for finding the roots of any quadratic equation.
• The discriminant \( b^2 - 4ac \) determines the nature of the roots:
  • Positive: Two distinct real roots,
  • Zero: One real root, repeated,
  • Negative: Two complex roots.

For the given exercise, factored form is particularly useful for evaluating limits by simplifying expressions that stem from the factorized equation.

Limit Evaluation

Limit evaluation is a fundamental concept in calculus that involves determining the value that a function approaches as the input approaches a certain point. In the context of the given problem, we need to understand how to evaluate a specific form of limit:
\[ \lim_{x \to \alpha} \frac{1 - \cos(f(x))}{(x-\alpha)^2} \]
Here’s how it's typically evaluated:

• Recognize the standard trigonometric limit \( \lim_{u \to 0} \frac{1 - \cos(u)}{u^2} = \frac{1}{2} \), which shows how the cosine function behaves when \( u \) is close to zero.
• Expand the expression using Taylor series for \( \cos(u)\). For small \( u \), \( 1 - \cos(u) \approx \frac{u^2}{2} \). This approximation simplifies evaluating the limit.
• Link the problem to this standard result by substituting \( f(x) = a(x-\alpha)(x-\beta) \). This facilitates simplification by directly applying the Taylor approximation.

Limit evaluation is crucial because it helps understand the behavior of functions near specific points, which is fundamental to calculus and continuous functions.

Trigonometric Limits

Trigonometric limits deal with finding the limit of functions that involve trigonometric expressions. Specifically, in this scenario, we are focusing on limits that involve the cosine function.
Understanding these limits requires both a solid grasp of trigonometric identities and the ability to apply Taylor series approximations:

• The approximation \( 1 - \cos(u) \approx \frac{u^2}{2} \) when \( u \) is small allows us to simplify many trigonometric limits.
• For the exercise, we replaced \( f(x) \) into the trigonometric limit, simplifying the expression using the approximation. This enabled the evaluation of the limit at \( x = \alpha \).
• It is this trigonometric identity and approximation that converts the problem into a manageable form that can be directly solved using standard limit rules.

Trigonometric limits are key in calculus, allowing mathematicians to tackle difficult questions involving periodic functions and oscillatory behaviors by using expansions like the Taylor series.