If the function \(f(x) = \begin{cases}\displaystyle\frac{kx}{|x|}, & x \lt 0\\ 3, & x \ge 0\end{cases}\) is continuous at \(x = 0\), then the value of k is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −3
For x < 0, |x| = −x, so the left piece is a constant:
\( \displaystyle \frac{kx}{|x|} = \frac{kx}{-x} = -k \)
\( \displaystyle \lim_{x \to 0^-} f(x) = -k \)
The right-hand limit and f(0) are both 3. For continuity at 0:
\(-k = 3\)
\(k = -3\)