If $g(x) =\left\{\begin{matrix}\frac{ax}{|x|},&\text{if x<0}\\5,&\text{if x ≥ 0}\end{matrix}\right.$ is continuous at $x = 0$, then the value of $a$ is

-5

The correct answer is Option (3) → -5

Given: \( g(x) = \begin{cases} \frac{ax}{|x|}, & \text{if } x < 0 \\ 5, & \text{if } x \geq 0 \end{cases} \) is continuous at x = 0

To ensure continuity at x = 0, the left-hand limit (LHL), right-hand limit (RHL), and value at the point must be equal:

Right-hand limit (x → 0⁺): Since \( x \geq 0 \Rightarrow g(x) = 5 \), ⇒ RHL = 5

Left-hand limit (x → 0⁻): For \( x < 0 \), \( |x| = -x \), so \( g(x) = \frac{ax}{-x} = -a \) ⇒ LHL = −a

g(0) = 5

For continuity at x = 0: LHL = RHL = g(0) ⇒ −a = 5 ⇒ a = −5