Question

Let $f(x)$ be a function defined as: $$f(x) = \begin{cases} \frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x + 1}{x - 2}, & \text{if } 0 \leq x \leq 1 \end{cases}$$ If $f(x)$ is continuous in the interval $[-1, 1]$, then $p =$

• A. $1$
• B. $-1$
• C. $-\frac{1}{2}$
• D. $\frac{1}{2}$

Hint:

For limits of the form $\lim_{x \to 0} \frac{\sqrt{1+ax} - \sqrt{1-bx}}{x}$, you can use a standard limit shortcut template which evaluates directly to $\frac{a - (-b)}{2} = \frac{a+b}{2}$. Here, it immediately yields $\frac{p - (-p)}{2} = p$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a piecewise function $f(x)$ with a change in definition at the boundary point $x = 0$.
The problem states that the function is continuous throughout the interval $[-1, 1]$, which strictly implies it must also be continuous at the critical point $x = 0$.

Step 2: Key Formula or Approach:
For a function to be continuous at a point $x = c$, the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and value of the function at that point must all be equal:
$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$$ We will evaluate the LHL using rationalization and equate it to the direct value from the second piece of the function.

Step 3: Detailed Explanation:
First, let's find the Right-Hand Limit and functional value at $x = 0$ using the second sub-function:
$$f(0) = \lim_{x \to 0^+} \frac{2x + 1}{x - 2} = \frac{2(0) + 1}{0 - 2} = -\frac{1}{2}$$ Next, evaluate the Left-Hand Limit at $x = 0$:
$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}$$ Since direct substitution yields an indeterminate $\frac{0}{0}$ form, we rationalize the numerator by multiplying the top and bottom by the conjugate expression $(\sqrt{1 + px} + \sqrt{1 - px})$:
$$\lim_{x \to 0^-} \frac{(\sqrt{1 + px} - \sqrt{1 - px})(\sqrt{1 + px} + \sqrt{1 - px})}{x(\sqrt{1 + px} + \sqrt{1 - px})}$$ Apply the algebraic identity $(a-b)(a+b) = a^2 - b^2$:
$$= \lim_{x \to 0^-} \frac{(1 + px) - (1 - px)}{x(\sqrt{1 + px} + \sqrt{1 - px})} = \lim_{x \to 0^-} \frac{2px}{x(\sqrt{1 + px} + \sqrt{1 - px})}$$ Cancel out the common variable $x$ from the numerator and denominator:
$$= \lim_{x \to 0^-} \frac{2p}{\sqrt{1 + px} + \sqrt{1 - px}}$$ Now, perform direct substitution for $x = 0$:
$$= \frac{2p}{\sqrt{1 + 0} + \sqrt{1 - 0}} = \frac{2p}{2} = p$$ Since the function is continuous at $x = 0$, equate the LHL to the value of the function:
$$p = -\frac{1}{2}$$

Step 4: Final Answer:
The value of the parameter $p$ is $-\frac{1}{2}$, which corresponds to option (C).