Problem 1 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\alpha\) is the positive root of the equation, \(p(x)=x^{2}-x-2=0\), then \(\lim _{x \rightarrow \alpha^{+}} \frac{\sqrt{1-\cos (p(x))}}{x+\alpha-4}\) is equal to: [Sep. \(\left.\mathbf{0 5}, \mathbf{2 0 2 0}(\mathbf{I})\right]\) (a) \(\frac{3}{2}\) (b) \(\frac{3}{\sqrt{2}}\) (c) \(\frac{1}{\sqrt{2}}\) (d) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

The limit is \( \frac{3}{\sqrt{2}} \), option (b).

Step 1: Solve the Quadratic Equation

We begin by finding the positive root of the equation \( p(x) = x^2 - x - 2 = 0 \). Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -2 \).Substituting these values we have:\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \]The roots are \( x = 2 \) and \( x = -1 \). The positive root is \( \alpha = 2 \).

Step 2: Simplify the Function

We calculate \( \lim_{x \to \alpha^+} \frac{\sqrt{1-\cos(p(x))}}{x+\alpha-4} \). First, note that since \( \alpha = 2 \) satisfies \( p(\alpha) = 0 \), then near \( \alpha \) we have that \( p(x) \approx 0 \).Therefore, \( \cos(p(x)) \approx 1 \) for \( x \approx \alpha \), which leads \( 1 - \cos(p(x)) \approx 0 \).

Step 3: Apply the Trigonometric Limit

As \( p(x) \to 0 \), it is well-known that \( 1 - \cos(p(x)) \approx \frac{p(x)^2}{2} \) (using the standard approximation for small angles). Thus \( \sqrt{1-\cos(p(x))} \approx \sqrt{\frac{p(x)^2}{2}} = \frac{|p(x)|}{\sqrt{2}} \).

Step 4: Evaluate the Limit

Substituting the approximation from Step 3 into the limit, we have:\[ \lim_{x \to 2^+} \frac{\frac{|p(x)|}{\sqrt{2}}}{x + 2 - 4} = \lim_{x \to 2^+} \frac{|p(x)|}{\sqrt{2} \cdot (x - 2)} \]Since \( p(x) = x^2 - x - 2 \), expand this:\[ p(x) = (x - 2)(x + 1) \]Near \( x = 2 \), \( p(x) \) becomes positively small which means \(([x - 2])\) is positive, thus:\[ \frac{|p(x)|}{\sqrt{2} \cdot (x - 2)} = \frac{(x - 2)(x + 1)}{\sqrt{2} \cdot (x - 2)} = \frac{x + 1}{\sqrt{2}} \]This simplifies to \( \frac{3}{\sqrt{2}} \) as \( x \to 2^+ \).

Step 5: Choose the Correct Option

The final result of the limit is \( \frac{3}{\sqrt{2}} \). Thus, the correct choice is option (b) \( \frac{3}{\sqrt{2}} \).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and make an important appearance in many mathematical contexts. They are polynomials of degree two and take the general form of \( ax^2 + bx + c = 0 \). The solutions, or roots, of quadratic equations can be found using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In this formula, \( a \), \( b \), and \( c \) are coefficients of the equation. This formula is derived from completing the square method and allows us to find both real and complex roots.
In our exercise, we find the positive root of the specific quadratic equation \( x^2 - x - 2 = 0 \). Substituting the values into the quadratic formula, we have:

• \( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2} \)

This results in roots 2 and -1. Since we're interested in the positive root, \( \alpha = 2 \). Understanding these roots is critical when evaluating limits or analyzing the behavior of quadratic functions.

Trigonometric Limits

Trigonometric limits are often encountered in calculus, particularly when dealing with small angle approximations. These are limits that involve trigonometric functions like sine, cosine, and tangent. One of the classic results when finding such limits involves the function of cosine.
As seen in the problem, when a function \( p(x) \) is close to zero, \( 1 - \cos(p(x)) \) can be approximated using the formula:

• \( 1 - \cos(p(x)) \approx \frac{p(x)^2}{2} \)

This formula is applicable for small angles due to the Taylor series expansion of the cosine function: \( \cos(x) \approx 1 - \frac{x^2}{2} \) for small \( x \). In the exercise, this approximation helps transform the limit of \( \frac{\sqrt{1-\cos(p(x))}}{x+\alpha-4} \) into a simpler form.
It's essential when simplifying expressions in calculus to recognize and utilize these approximations effectively.

Limit Evaluation

Limit evaluation is key in understanding the behavior of functions as they approach specific values. Limits can often alleviate the complexities of directly substituting into expressions, especially when indeterminate forms are present.
In our specific problem, we're working with the limit:

• \( \lim_{x \to \alpha^+} \frac{\sqrt{1-\cos(p(x))}}{x+\alpha-4} \)

Here, \( \alpha = 2 \). To evaluate this limit, we first simplify the expression using known approximations for small angles:

• \( 1 - \cos(p(x)) \approx \frac{p(x)^2}{2} \)
• \( \sqrt{1-\cos(p(x))} \approx \frac{|p(x)|}{\sqrt{2}} \)

Substituting these into the limit expression provides a more approachable form:

• \( \lim_{x \to 2^+} \frac{|p(x)|}{\sqrt{2} \cdot (x - 2)} \)

Given \( p(x) = (x - 2)(x + 1) \), we can simplify to \( \frac{x + 1}{\sqrt{2}} \) when \( x - 2 \) is positive. Evaluating the limit gives \( \frac{3}{\sqrt{2}} \), guiding us to option (b) for the solution.
Understanding this step-by-step process allows a deeper comprehension of how to handle limits involving trigonometric identities and rational functions.