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: Understand continuity here.
The function changes its rule at $x = 0$. For it to be continuous there, the left-hand limit, right-hand value, and the value at 0 must all be equal.
Step 2: Find the value from the right.
For $x \ge 0$, $f(x) = \dfrac{2x+1}{x-2}$. At $x = 0$: \[ f(0) = \frac{2(0)+1}{0-2} = -\frac{1}{2} \]
Step 3: Set up the left-hand limit.
For $x < 0$, $f(x) = \dfrac{\sqrt{1+px} - \sqrt{1-px}}{x}$. Direct substitution gives $\tfrac00$, so we rationalise.
Step 4: Rationalise the numerator.
Multiply top and bottom by $\sqrt{1+px} + \sqrt{1-px}$: \[ \frac{(1+px) - (1-px)}{x(\sqrt{1+px} + \sqrt{1-px})} = \frac{2px}{x(\sqrt{1+px} + \sqrt{1-px})} \]
Step 5: Cancel and take the limit.
Cancel $x$ and let $x \to 0$: \[ \frac{2p}{\sqrt{1} + \sqrt{1}} = \frac{2p}{2} = p \]
Step 6: Match the two sides.
For continuity the left limit equals $f(0)$: \[ p = -\frac{1}{2} \] \[ \boxed{p = -\dfrac{1}{2}} \]