Problem 41 - A Complete Resource Book in Mathematics for JEE Main

Question

The values of \(p\) and \(q\) for which the function \(f(x)= \begin{cases}\frac{\sin (p+1) x+\sin x}{x}, & x<0 \\ q, & x=0 \\\ \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{3 / 2}}, & x=0\end{cases}\) is continuous for all \(x\) in \(R\), are (A) \(p=\frac{1}{2}, q=\frac{3}{2}\) (B) \(p=\frac{1}{2}, q=-\frac{3}{2}\) (C) \(p=\frac{5}{2}, q=\frac{1}{2}\) (D) \(p=-\frac{3}{2}, q=\frac{1}{2}\)

Step-by-Step Solution

Answer

The values are \(p = -\frac{3}{2}\) and \(q = \frac{1}{2}\). Choice (D).

Step 1: Understanding the Problem

We need to find the values of \(p\) and \(q\) such that the function \(f(x)\) is continuous for all \(x\) in \(\mathbb{R}\). The function is given as piecewise with different definitions for \(x<0\) and \(x \geq 0\).

Step 2: Continuity at x = 0

The function \(f(x)\) needs to be continuous at \(x = 0\). This means the left-hand limit, the right-hand limit, and \(f(0)\) should all be equal at \(x = 0\). Therefore, find \(\lim_{x \to 0^-} f(x)\), \(\lim_{x \to 0^+} f(x)\), and compare with \(f(0)=q\).

Step 3: Calculating Left-Hand Limit

For \(x < 0\), \(f(x) = \frac{\sin((p+1)x) + \sin(x)}{x}\). Find \(\lim_{x \to 0^-} f(x)\). By using L'Hospital's Rule (since the limit takes the form \(\frac{0}{0}\)), differentiate the numerator and denominator:\[\lim_{x \to 0^-} \frac{\sin((p+1)x) + \sin(x)}{x} = \lim_{x \to 0^-} \frac{(p+1) \cos((p+1)x) + \cos(x)}{1} = p+2\]

Step 4: Calculating Right-Hand Limit

For \(x > 0\), \(f(x) = \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}}\). Find \(\lim_{x \to 0^+} f(x)\). Simplify the expression using algebra:\[\lim_{x \to 0^+} \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}} = \lim_{x \to 0^+} \frac{x(1 + x) - x}{x^{3/2}(\sqrt{1 + x} + 1)} = \lim_{x \to 0^+} \frac{x}{x^{3/2}} = \lim_{x \to 0^+} \frac{1}{x^{1/2}} = \frac{1}{2}\]

Step 5: Setting Continuity Condition at x=0

For the function to be continuous at \(x = 0\), the following condition must hold: \[\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)\]Using the limits computed, set:\[p + 2 = \frac{1}{2} = q\]

Step 6: Solving for p and q

Solve the equation \(p + 2 = \frac{1}{2}\):\[p = \frac{1}{2} - 2 = -\frac{3}{2}\]Since \(\lim_{x \to 0^+} f(x) = \frac{1}{2}\), we set \(q = \frac{1}{2}\).

Step 7: Conclusion - Validate

The function is continuous for all \(x\) if \(p = -\frac{3}{2} \) and \(q = \frac{1}{2}\). So, the correct answer is \((D)\).

Key Concepts

Limits and Continuity

Understanding limits and continuity is essential in determining the behavior of functions at certain points. For a function to be continuous at a point, the function must satisfy three conditions:

• The function must be defined at that point, meaning there's a value for the function at this specific point.
• The limit of the function, as it approaches the point from the left, must equal the limit as it approaches from the right.
• These two limits must equal the value of the function at the point.

Consider the function given in the problem. At the point where the piecewise function changes (here, at 0), all three of these checks must align. For instance, the left-hand limit, right-hand limit, and the function value all need to align for the function to be continuous at 0. In this exercise, examining these limits carefully helps us establish a set of equations, allowing us to solve for unknowns like p and q.

L'Hospital's Rule

L'Hospital's Rule is an indispensable tool in calculus used for finding limits of indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When a limit takes such a form, the rule allows us to differentiate the numerator and the denominator separately and then compute the limit. This simplifies the evaluation of complex limits.
In the exercise, L'Hospital's Rule was applied in finding the limit of the left-hand portion of the piecewise function as \(x\) approaches 0 from the negative side, specifically:\[\lim_{x \to 0^-} \frac{\sin((p+1)x) + \sin(x)}{x}\]By differentiating the sine functions' sums and the \(x\) in the denominator separately, the expression resolved from an indeterminate form to a solvable equation \(p + 2\). Thus, mastering L'Hospital's Rule aids in breaking down limits that initially seem unsolvable into simpler computations.

Piecewise Functions

Piecewise functions are defined by different expressions in separate parts of their domain. They are prevalent in contexts where a function's behavior changes based on the input value. In the provided exercise, we encounter a function with differing expressions for \(x<0\) and \(x>=0\). This requires evaluating the function's continuity at the point of partition, often where the function's segments meet (e.g., \(x=0\)).

• The function is continuous at \(x=0\) if the expressions before and after the partition point render the same result at that specific point.
• The continuity condition bridges these definitions together by ensuring the consistency of the limit behavior on either side of the partition.

In our example, solving for the unknown coefficients in each section ensures smooth transition across the function's entirety. Understanding piecewise functions is crucial for modeling scenarios where different rules govern different conditions.