Problem 31 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

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 \([2005]\) (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

The limit is option (a) \(\frac{a^2}{2}(\alpha-\beta)^2\).

Step 1: Understand the Problem

We need to evaluate the limit \( \lim_{x \rightarrow \alpha} \frac{1 - \cos(f(x))}{(x-\alpha)^2} \) where \( f(x) = ax^2 + bx + c \). This involves applying the limit properties and possibly L'Hôpital's Rule since we have a zero over zero form.

Step 2: Express the Quadratic Function

Since \( \alpha \) is a root of the quadratic equation \( ax^2 + bx + c = 0 \), we can say that \( f(\alpha) = a\alpha^2 + b\alpha + c = 0 \). The function \( f(x) \) can thus be factored around \( \alpha \) as \( f(x) = a(x-\alpha)(x-\beta) \) using the roots \( \alpha \) and \( \beta \).

Step 3: Simplify \( \cos(f(x)) \) Near \( \alpha \)

Near \( \alpha \), we have \( f(x) \approx a(x-\alpha)(x-\beta) \). This implies that for small deviations around \( \alpha \), \( \cos(f(x)) \approx 1 - \frac{(f(x))^2}{2} \) because \( \cos y \approx 1 - \frac{y^2}{2} \) when \( y \to 0 \).

Step 4: Evaluate the Numerator

The numerator becomes \( 1 - \cos(f(x)) \approx \frac{(f(x))^2}{2} \). Substituting \( f(x) \approx a(x-\alpha)(x-\beta) \), we get \((f(x))^2 \approx [a(x-\alpha)(x-\beta)]^2\), leading to the numerator being \( \frac{[a(x-\alpha)(x-\beta)]^2}{2} \).

Step 5: Divide by \((x-\alpha)^2\)

Now, divide the whole expression by \((x-\alpha)^2\) to get the limit:\[ \frac{[a(x-\alpha)(x-\beta)]^2}{2(x-\alpha)^2} = \frac{a^2(x-\beta)^2}{2} \] as \( x \to \alpha \).

Step 6: Substitute \(x = \alpha\)

Finally, evaluate the limit by substituting \( x = \alpha \):\[ \frac{a^2(\alpha-\beta)^2}{2} \] Hence, the result gives us the limit expression.

Step 7: Compare to Options

Comparing \( \frac{a^2(\alpha-\beta)^2}{2} \) with the given choices, we find that option (a) \( \frac{a^2}{2}(\alpha-\beta)^2 \) matches.

Key Concepts

Limits in Calculus

Limits are a fundamental concept in calculus that help us understand the behavior of functions as they approach a specific point. When you see an expression like \( \lim_{x \rightarrow \alpha} \), it indicates what happens to a function as \( x \) gets very close to \( \alpha \). For our context, the limit \( \lim_{x \rightarrow \alpha} \frac{1 - \cos(f(x))}{(x-\alpha)^2} \) involves checking how the function behaves around the root \( \alpha \) of a quadratic equation.
To solve problems involving limits, especially when you have indeterminate forms like \( \frac{0}{0} \), one must manipulate the expression to allow simplification or apply special rules. Understanding limits allows us to transition into topics like L'Hôpital's Rule or other calculus techniques for evaluating these expressions. In simple terms, limits give us the power to make sense of function behavior in otherwise undefined situations.

L'Hôpital's Rule

L'Hôpital's Rule is incredibly useful when calculating limits that result in indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule provides a way to resolve these forms by taking derivatives of the numerator and the denominator.

• To apply L'Hôpital's Rule, first ensure your limit is in an indeterminate form.
• Then, differentiate the top and bottom of the fraction separately.
• Lastly, calculate the limit of the new fraction.

Remember, L'Hôpital's Rule can be applied repeatedly if the resulting limit still gives an indeterminate form. However, in some situations—like our function with \( 1 - \cos(f(x)) \) and \( (x-\alpha)^2 \)—using approximations and algebraic simplifications can sometimes be a more straightforward approach than resorting to derivatives. This leads smoothly into our exploration of quadratic roots and how they simplify such calculations.

Roots of Polynomials

Polynomial roots are the points where the polynomial equals zero. For a quadratic equation \( ax^2 + bx + c = 0 \), roots \( \alpha \) and \( \beta \) are solutions to this equation. Knowing these roots allows you to factor the quadratic into \( a(x-\alpha)(x-\beta) \).
In our exercise, understanding the roots is vital because it helps refactor the quadratic expression to evaluate the limit. By expressing the polynomial as \( a(x-\alpha)(x-\beta) \), calculations become simpler. You can substitute this form into your original limit problem, reducing complexity. This ultimately leads to extracting terms directly related to the distance between roots, \( (\alpha - \beta) \), as seen in the limit calculated in the solution.

• This kind of manipulation is common; it turns a complex problem into simpler algebraic forms.
• By figuring out the roots, you pave the way for transforming quadratic polynomials, enabling easier calculations and deeper insights into the function's behavior near these points.

Grasping polynomial roots plays a crucial role in calculus, as understanding them facilitates tackling broader mathematical challenges.