If the function \(f(x) = \begin{cases}\displaystyle\frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x}, & -1 \le x \lt 0\\ \displaystyle\frac{x + 1}{x + 2}, & 0 \le x \le 1\end{cases}\) is continuous at \(x = 0\), then the value of k is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{2}\)
From the right piece, \(f(0) = \frac{0 + 1}{0 + 2} = \frac{1}{2}\).
For the left-hand limit, rationalise the numerator:
\( \displaystyle \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x} = \frac{(1 + kx) - (1 - kx)}{x\left(\sqrt{1 + kx} + \sqrt{1 - kx}\right)} \)
\( \displaystyle = \frac{2k}{\sqrt{1 + kx} + \sqrt{1 - kx}} \)
\( \displaystyle \lim_{x \to 0^-} f(x) = \frac{2k}{1 + 1} = k \)
For continuity at 0, this must equal f(0):
\( \displaystyle k = \frac{1}{2} \)