Problem 5 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the function \(\mathrm{f}\) defined on \(\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\) by \(f(x)=\left\\{\begin{array}{c}\frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ k, & x=\frac{\pi}{4}\end{array}\right.\) is continuous, then \(\mathrm{k}\) is equal to: (a) 2 (b) \(\frac{1}{2}\) (c) 1 (d) \(\frac{1}{\sqrt{2}}\)

Step-by-Step Solution

Answer

The value of \(k\) is \(\frac{1}{2}\).

Step 1: Understand Continuity Requirement

For a function to be continuous at a point, the limit of the function as it approaches the point should equal the function's value at that point. Here, we need to find the limit of the function \(f(x)\) as \(x\) approaches \(\frac{\pi}{4}\) and ensure it equals \(k\) to keep the function continuous at \(x = \frac{\pi}{4}\).

Step 2: Set Up the Limit

We need to find \(\lim_{x \to \frac{\pi}{4}} f(x)\) for \(x eq \frac{\pi}{4}\). This requires examining the expression \(\frac{\sqrt{2} \cos x - 1}{\cot x - 1}\) because it's defined for \(x eq \frac{\pi}{4}\).

Step 3: Rewrite Cotangent Function

Recall that the cotangent function is \(\cot x = \frac{\cos x}{\sin x}\). Substitute this into the expression to get \(\frac{\sqrt{2} \cos x - 1}{\frac{\cos x}{\sin x} - 1}\). Simplify this as \(\frac{\sqrt{2} \cos x - 1}{\frac{\cos x - \sin x}{\sin x}}\).

Step 4: Simplify the Expression

Multiply numerator and denominator by \(\sin x\) to eliminate the fraction: \(\frac{\sqrt{2} \cos x - 1}{\frac{\cos x - \sin x}{\sin x}} \times \frac{\sin x}{\sin x} = \frac{\sin x (\sqrt{2} \cos x - 1)}{\cos x - \sin x}\). This becomes \(\frac{\sqrt{2} \cos x \sin x - \sin x}{\cos x - \sin x}\).

Step 5: Apply L'Hôpital's Rule

The limit's numerator and denominator both approach 0 as \(x \to \frac{\pi}{4}\). Therefore, apply L'Hôpital's rule: differentiate the numerator and denominator separately and find the limit again.

Step 6: Differentiate Numerator and Denominator

Differentiate the numerator \(\sqrt{2} \cos x \sin x - \sin x\) to get \(\sqrt{2}(-\sin x \sin x + \cos x \cos x) - \cos x = \sqrt{2} (-\sin^2 x + \cos^2 x) - \cos x\). Differentiate the denominator \(\cos x - \sin x\) to get \(-\sin x - \cos x\).

Step 7: Evaluate the Limit

Evaluate \(\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} (-\sin^2 x + \cos^2 x) - \cos x}{-\sin x - \cos x}\). Substitute \(x = \frac{\pi}{4}\) where \(\sin x = \cos x = \frac{1}{\sqrt{2}}\). Compute the derivatives' values and the limit.

Step 8: Substituting Values and Finding Limit

Substitute \(\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\) into the expression. Calculate \(\sqrt{2}(0) - \frac{1}{\sqrt{2}})\) which divides over \, \(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}\) giving \(-\frac{1}{2}\).

Step 9: Equating Limit to \(k\) for Continuity

For \(f(x)\) to be continuous at \(x = \frac{\pi}{4}\), \(k\) must equal the limit value found in the previous step, so \(k = \frac{1}{2}\).

Step 10: Verify with Given Options

Verify that \(k = \frac{1}{2}\) corresponds to option (b) in the problem, confirming the continuity of the function.

Key Concepts

limits

When examining the continuity of a function, limits play a vital role. Limits help us understand the behavior of the function as it approaches a specific point. Consider the problem where we need the function to be continuous at a specific point, say \(x = \frac{\pi}{4}\). For this, the limit of the function as \(x\) approaches \(\frac{\pi}{4}\) should equal the function's value at that point.
To solve this, we set up the expression for \( f(x) \), where \( x eq \frac{\pi}{4} \), and find \( \lim_{x \to \frac{\pi}{4}} f(x) \). This involves simplifying and examining the behavior of the given trigonometric and rational expressions.
Ultimately, this limit will determine the value that the function should take at \( x = \frac{\pi}{4}\), ensuring the function's continuity.

L'Hôpital's rule

L'Hôpital's rule is a powerful tool for evaluating limits that initially present as indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When faced with a difficult limit where both the numerator and the denominator approach 0 as \(x\) goes to a certain value, we can apply L'Hôpital’s rule.
This rule simplifies finding the limit by allowing us to differentiate the numerator and the denominator separately. In our problem, as \( x \to \frac{\pi}{4} \), both the numerator \( \sqrt{2} \cos{x} \sin{x} - \sin{x} \) and the denominator \( \cos{x} - \sin{x} \) approach zero. Differentiating these once gives us simpler functions to evaluate at \( x = \frac{\pi}{4} \).
L'Hôpital's rule thus transforms an indeterminate expression into one that's more manageable, and in this case, leads us to find \(k = \frac{1}{2}\) by evaluating the limit of the derivative expression.

trigonometric functions

Trigonometric functions like \( \cos\), \( \sin\), and \( \cot \) appear frequently in mathematical problems, including those involving limits and continuity. Understanding how these functions behave plays a crucial role in solving such problems.
In the given exercise, we manipulate these trigonometric functions to express \( f(x) \) in a form suitable for applying L'Hôpital's rule. Specifically, transforming the cotangent into a ratio of \( \cos \) and \( \sin \) helps us rewrite and simplify the expression.
It is also essential to remember certain values of these functions for standard angles, like \( \sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \), which we use during simplification and substitution after differentiation. Mastery of these functions and their properties can significantly ease the problem-solving process, providing deeper insights into why \( k = \frac{1}{2} \) ensures the continuity of the function at \( x = \frac{\pi}{4} \).