Problem 124 - A Complete Resource Book in Mathematics for JEE Main

Question

The real values of \(p\) and \(q\) for which the function \(f(x)=\left\\{\begin{array}{l}\frac{\sin (p+1) x+\sin x}{x}, x<0 \\ q, x=0 \quad \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{3 / 2}}, x>0\end{array} \quad\right.\) is continuous for all \(x\) in \(R\), is (A) \(p=\frac{5}{2}, q=\frac{1}{2}\) (B) \(p=-\frac{3}{2}, q=\frac{1}{2}\) (C) \(p=\frac{1}{2}, q=\frac{3}{2}\) (D) \(p=\frac{1}{2}, q=-\frac{3}{2}\)

Step-by-Step Solution

Answer

The correct values are \(p = -\frac{3}{2}\) and \(q = \frac{1}{2}\); option (B).

Step 1: Analyze the Continuity Condition

To ensure the function \(f(x)\) is continuous at \(x=0\), we need \(\lim_{{x \to 0^-}} f(x) = f(0) = \lim_{{x \to 0^+}} f(x) = q\).

Step 2: Evaluate the Left-Hand Limit

Compute \(\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^-}} \frac{\sin((p+1)x) + \sin x}{x}\). Using L'Hôpital's Rule:\[\lim_{{x \to 0^-}} \frac{\sin((p+1)x) + \sin x}{x} = (p+1) + 1 = p+2.\]

Step 3: Set Left-Hand Limit Equal to q

For continuity, the left-hand limit must be equal to \(q\). Therefore, \(p + 2 = q\).

Step 4: Evaluate the Right-Hand Limit

For \(x > 0\), determine \(\lim_{{x \to 0^+}} \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}}\). Both the numerator and denominator approach zero:\[\lim_{{x \to 0^+}} \frac{\sqrt{x(1+x)} - \sqrt{x}}{x^{3/2}} = \lim_{{x \to 0^+}} \frac{x(1/2+1/(4\sqrt{1+x}))}{3/2 \cdot x^{1/2}} = \lim_{{x \to 0^+}} \frac{1+x}{2\sqrt{1+x}} = \frac{1}{2}.\]

Step 5: Set Right-Hand Limit Equal to q

The right-hand limit also equals \(q\). Hence, \(q = \frac{1}{2}\).

Step 6: Solve for p

From the equation \(p+2 = q\) and substituting \(q = \frac{1}{2}\), we find:\[p + 2 = \frac{1}{2} \Rightarrow p = \frac{1}{2} - 2 = -\frac{3}{2}.\]

Step 7: Verify Solutions

Check to ensure both limits are consistent with continuity conditions. The correct values are \(p = -\frac{3}{2}\) and \(q = \frac{1}{2}\).

Key Concepts

Limits of Functions

Understanding the concept of limits is crucial when dealing with the continuity of functions. Limits help us analyze the behavior of a function as the input value approaches a particular point, in this case, zero. Simply put, the limit of a function as it approaches a certain point tells us what value the function is getting infinitely close to. This idea is particularly important when we deal with piecewise functions, like in the problem above.

For example, consider the left-hand limit of the given function when \(x < 0\). We focus on how the function behaves as \(x\) gets close to zero but remains negative. Similarly, we determine the right-hand limit for when \(x > 0\). Both of these must equal \(f(0)\) for the function to be continuous at \(x = 0\). Understanding and calculating these limits accurately allow us to investigate continuity properly and deduce that the function can meet the criteria to be continuous across its domain.

In mathematical analysis, limits are a fundamental tool, providing a foundation for concepts like continuity, derivatives, and integrals. Developing a strong understanding of limits is key in progressing through calculus and understanding more complex mathematical theories.

L'Hôpital's Rule

L'Hôpital's Rule comes in handy when we deal with limits that result in indeterminate forms, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). In the original exercise, L'Hôpital's Rule helps evaluate the limit \(\lim_{{x \to 0^-}} \frac{\sin((p+1)x) + \sin x}{x}\).

To apply L'Hôpital's Rule, we differentiate the numerator and denominator separately and then compute the result as \(x\) approaches the point \(0^-\). This results in the limit simplifying to \((p+1) + 1 = p+2\). For continuous function arguments, this outcome must equal \(q\), establishing a crucial relationship between \(p\) and \(q\).

• Why use L'Hôpital's Rule? It provides a straightforward approach to solving limits when direct substitution leads to indeterminate forms.
• When does it apply? Only when both the numerator and denominator approach zero or both approach infinity.

Understanding this elegant rule not only simplifies the solution process significantly but also provides deeper insight into limits beyond basic numerical substitution.

Continuity at a Point

Continuity at a point is a fundamental concept in calculus that signifies a function does not have breaks, jumps, or holes at that point. For a function to be continuous at a point \(c\), three conditions must be satisfied:

• The function \(f(x)\) is defined at \(c\), i.e., \(f(c)\) exists.
• The limit \(\lim_{{x \to c}} f(x)\) exists.
• \(\lim_{{x \to c}} f(x) = f(c)\).

The presented exercise revolves around ensuring continuity at \(x = 0\). We achieve this by matching the left-hand limit (as \(x\) approaches 0 from the left) and the right-hand limit (as \(x\) approaches 0 from the right) with the function value at 0 (i.e., \(q\)).

By solving the limits and equating them to \(q\), which the problem states must equal \(f(0) \), we determine the parameters \(p\) and \(q\). This ensures the function is seamless at the transition between different behaviors around \(x = 0\).

Understanding the conditions for continuity allows one to evaluate complex functions’ behaviors and ensure they act predictably across their domains.