Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:
  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists.
  3. $\lim_{x \to c} f(x) = f(c)$.
• Left-Hand Limit (LHL): The limit of a function as $x$ approaches a point from the left side: $\lim_{x \to c^-} f(x)$.
• Right-Hand Limit (RHL): The limit of a function as $x$ approaches a point from the right side: $\lim_{x \to c^+} f(x)$.
• Standard Limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1$.
• Algebraic Manipulation for Limits: Techniques like rationalization or factoring can be used to evaluate limits.

Step-by-Step Solution

For the function $f(x)$ to be continuous for all $x$ in $\mathbb{R}$, it must be continuous at $x=0$. This requires the left-hand limit (LHL), the right-hand limit (RHL), and the function value at $x=0$ to be equal.

Step 1: Evaluate the Left-Hand Limit (LHL) at $x=0$. The LHL is the limit as $x$ approaches 0 from the left ($x < 0$). $LHL = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin (p + 1)x + \sin x}{x}$ We can split this into two separate limits: $LHL = \lim_{x \to 0^-} \frac{\sin (p + 1)x}{x} + \lim_{x \to 0^-} \frac{\sin x}{x}$ Using the standard limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$, we can rewrite the first term: $\lim_{x \to 0^-} \frac{\sin (p + 1)x}{x} = \lim_{x \to 0^-} (p + 1) \frac{\sin (p + 1)x}{(p + 1)x}$ As $x \to 0^-$, $(p+1)x \to 0$. So, this limit becomes $(p+1) \times 1 = p+1$. The second term is $\lim_{x \to 0^-} \frac{\sin x}{x} = 1$. Therefore, the LHL is: $LHL = (p + 1) + 1 = p + 2$

Step 2: Evaluate the Right-Hand Limit (RHL) at $x=0$. The RHL is the limit as $x$ approaches 0 from the right ($x > 0$). $RHL = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}}$ To simplify this expression, we can factor out $\sqrt{x}$ from the numerator: $RHL = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1 + x} - 1)}{x^{3/2}}$ $RHL = \lim_{x \to 0^+} \frac{\sqrt{1 + x} - 1}{x^{3/2} / \sqrt{x}} = \lim_{x \to 0^+} \frac{\sqrt{1 + x} - 1}{x}$ This limit is in the indeterminate form $\frac{0}{0}$. We can evaluate it using rationalization by multiplying the numerator and denominator by the conjugate of the numerator: $RHL = \lim_{x \to 0^+} \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{x(\sqrt{1 + x} + 1)}$ $RHL = \lim_{x \to 0^+} \frac{(1 + x) - 1}{x(\sqrt{1 + x} + 1)}$ $RHL = \lim_{x \to 0^+} \frac{x}{x(\sqrt{1 + x} + 1)}$ For $x \neq 0$, we can cancel out $x$: $RHL = \lim_{x \to 0^+} \frac{1}{\sqrt{1 + x} + 1}$ Now, substitute $x=0$: $RHL = \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$

Step 3: Evaluate the function value at $x=0$. From the definition of the function, $f(0) = q$.

Step 4: Apply the condition for continuity. For $f(x)$ to be continuous at $x=0$, we must have $LHL = RHL = f(0)$. From Step 1, $LHL = p + 2$. From Step 2, $RHL = \frac{1}{2}$. From Step 3, $f(0) = q$. Therefore, we have the equations:

1. $p + 2 = \frac{1}{2}$
2. $q = \frac{1}{2}$

Step 5: Solve for $p$ and $q$. From the first equation: $p = \frac{1}{2} - 2$ $p = \frac{1 - 4}{2}$ $p = -\frac{3}{2}$

From the second equation, we already have $q$: $q = \frac{1}{2}$

Thus, the values for $p$ and $q$ that make the function continuous are $p = -\frac{3}{2}$ and $q = \frac{1}{2}$.

Common Mistakes & Tips

• Incorrect application of standard limits: Ensure that the argument of the sine function goes to zero when using $\lim_{y \to 0} \frac{\sin y}{y} = 1$. In the LHL calculation, we had $\sin((p+1)x)$, and since $x \to 0$, $(p+1)x \to 0$, allowing us to use the standard limit.
• Errors in algebraic manipulation for RHL: Pay close attention to signs and exponents when rationalizing or simplifying expressions. For the RHL, ensure the denominator simplification $x^{3/2} / \sqrt{x} = x^{3/2} / x^{1/2} = x^{(3/2 - 1/2)} = x^1 = x$ is done correctly.
• Confusing LHL and RHL definitions: Always remember that LHL involves $x \to c^-$ (approaching from the left) and RHL involves $x \to c^+$ (approaching from the right).

Summary

To ensure the function $f(x)$ is continuous at $x=0$, we equated the left-hand limit, the right-hand limit, and the function's value at $x=0$. We calculated the LHL of the given function for $x<0$ to be $p+2$. We calculated the RHL for $x>0$ by simplifying the expression and using rationalization, which resulted in $\frac{1}{2}$. The function's value at $x=0$ is given as $q$. By setting these three equal, we derived the values $p = -\frac{3}{2}$ and $q = \frac{1}{2}$.

The final answer is $\boxed{p = -{3 \over 2}, q = {1 \over 2}}$, which corresponds to option (B).