Problem 15 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the function \(f(x)=\left\\{\begin{array}{l}\frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}}, x \neq \pi \\ k \quad, x=\pi\end{array}\right.\) is continuous at \(x=\pi\), then \(\mathrm{k}\) equals: (a) 0 (b) \(\frac{1}{2}\) (c) 2 (d) \(\frac{1}{4}\)

Step-by-Step Solution

Answer

k = \(\frac{1}{4}\); option (d) is correct.

Step 1: Understand Continuity at a Point

For a function to be continuous at a point, the limit of the function as it approaches that point must equal the function's value at that point. Here, we need \(\lim_{{x \to \pi}} f(x) = f(\pi)\). This means \(\lim_{{x \to \pi}} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}} = k\).

Step 2: Substitute Limit Expression

Substitute \(f(x)\) with the given expression: \(\lim_{{x \to \pi}} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}}\). Investigate any potential indeterminate form, which in this case arises because direct substitution leads to \(\frac{0}{0}\).

Step 3: Simplify the Limit Expression

Use the limit properties and trigonometric identities. As \(x\) approaches \(\pi\), apply L'Hopital's Rule or algebraic manipulation to resolve the indeterminate form. Recognizing that near \(\pi\), \(\cos x \approx -1\), rewrite: \(\sqrt{2+(-1)} = 1\). Factor differently or multiply by the conjugate to simplify further.

Step 4: Apply L'Hopital's Rule

Since the limit evaluates to the indeterminate form \(\frac{0}{0}\), apply L'Hopital's Rule (differentiating the numerator and denominator separately). After differentiation, evaluate the new expression to find the limit.

Step 5: Calculate the Derivatives

The derivative of the numerator, \((\sqrt{2+\cos x})\), is \(-\frac{\sin{x}}{2\sqrt{2+\cos{x}}}\). The derivative of the denominator, \((\pi-x)^{2}\), is \(-2(\pi-x)\). Substitute these into the L'Hopital's Rule expression.

Step 6: Evaluate the Limit After Differentiation

Substitute back into the L'Hopital's formula: \(\lim_{{x \to \pi}} \frac{-\frac{\sin{x}}{2\sqrt{2+\cos{x}}}}{-2(\pi-x)}\). Simplify and compute as \(x \to \pi\). With \(\sin \pi = 0\), the limit simplifies further, ultimately yielding \(\frac{1}{4}\).

Step 7: Conclude k Value

Since \(\lim_{{x \to \pi}} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}} = k\), and from the computation, we found the limit is \(\frac{1}{4}\). Therefore, \(k = \frac{1}{4}\).

Key Concepts

L'Hopital's Rule

L'Hopital's Rule is a powerful tool used to evaluate limits that take the indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule requires differentiating both the numerator and the denominator of a limit separately. If after substituting a limit value into a function, the result is an indeterminate form, L'Hopital's Rule can simplify the expression and find the limit.
To use L'Hopital's Rule, you must:

• Ensure the original limit yields an indeterminate form like \( \frac{0}{0} \).
• Differently differentiate the numerator and the denominator.
• Re-evaluate the limit with these new derivatives. Repeat the process if necessary.

This method is particularly useful in the original exercise, as the direct substitution into the function led to an indeterminate form \( \frac{0}{0} \). By differentiating the numerator \( (\sqrt{2+\cos x}) \) and the denominator \( (\pi-x)^{2} \), we can successfully find the limit and establish the continuity of the function at \( x=\pi \).

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values within their domains. In the realm of limit evaluation, these identities are helpful for simplifying expressions before or after applying techniques like L'Hopital's Rule.
In the original exercise, identifying that near \( x=\pi \), \( \cos x \approx -1 \) becomes vital. This approximation helps simplify the expression \( \sqrt{2+\cos x} \), as it can be rewritten more conveniently when evaluating the limit. Recognizing identities and knowing approximations for different angles allows for easier manipulations of trigonometric expressions, making it essential in solving complex limits.
These identities, such as \( \sin^2x + \cos^2x = 1 \), often assist in rewriting expressions into simpler forms. Familiarity with them not only aids in limit evaluation but also in solving a wide range of mathematical problems.

Limit Evaluation

Limit evaluation is a critical concept in calculus, particularly when determining continuity at a point. For a function to be continuous at \( x=a \), the following must be true:

• The limit \( \lim_{{x \to a}} f(x) \) exists.
• This limit equals \( f(a) \).

In the original problem, we need to find \( k \) for the function to be continuous at \( x=\pi \). This means evaluating the given limit \( \lim_{{x \to \pi}} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}} \) is key.
The evaluation process involves first attempting direct substitution. If the result is indeterminate, such as \( \frac{0}{0} \), further techniques like L'Hopital's Rule must be applied. By deriving and substituting appropriately, we find the limit value fulfills the condition for continuity, and in this case, we calculate \( k = \frac{1}{4} \).
Understanding limit evaluation is fundamental to solving problems that assess the behavior of functions as they approach specific points. It connects principles from calculus to broader concepts like continuity and differentiability.