Find the value of $k$ so that the function $f(x) = \begin{cases} \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x + 1}{x - 1}, & \text{if } 0 \leq x \leq 1 \end{cases}$ at $x = 0$ is continuous at $x=0$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → -1 ##
We have,
$f(x) = \begin{cases} \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x + 1}{x - 1}, & \text{if } 0 \leq x \leq 1 \end{cases} \text{ at } x = 0$
$∴\text{LHL} = \lim\limits_{x \to 0^-} \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x}$
$= \lim\limits_{x \to 0^-} \left( \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x} \right) \cdot \left( \frac{\sqrt{1 + kx} + \sqrt{1 - kx}}{\sqrt{1 + kx} + \sqrt{1 - kx}} \right)$
$[∵\text{multiply and divided by } \sqrt{1 + kx} + \sqrt{1 - kx}]$
$= \lim\limits_{x \to 0^-} \frac{1 + kx - (1 - kx)}{x[\sqrt{1 + kx} + \sqrt{1 - kx}]} \quad [∵a^2 - b^2 = (a + b)(a - b)]$
$= \lim\limits_{x \to 0^-} \frac{2kx}{x[\sqrt{1 + kx} + \sqrt{1 - kx}]}$
Put $x = 0 - h$,
$= \lim\limits_{h \to 0} \frac{2k}{\sqrt{1 + k(0 - h)} + \sqrt{1 - k(0 - h)}}$
$= \lim\limits_{h \to 0} \frac{2k}{\sqrt{1 - kh} + \sqrt{1 + kh}} = \frac{2k}{2} = k$
and $f(0) = \frac{2 \times 0 + 1}{0 - 1} = -1$
$\Rightarrow k = -1 \quad [∵\text{LHL} = \text{RHL} = f(0)]$