If the function defined by $f(x) =\left\{\begin{matrix}kx^2 + 1&,\text{if x ≤ 1}\\2&,\text{if x > 1}\end{matrix}\right.$ is continuous at $x = 1$, then $k$ is equal to

1

The correct answer is Option (4) → 1

Given:

$f(x) = \begin{cases} kx^2 + 1, & \text{if } x \leq 1 \\ 2, & \text{if } x > 1 \end{cases}$

For $f(x)$ to be continuous at $x = 1$,

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x) = f(1)$

From the left: $f(1^-) = k(1)^2 + 1 = k + 1$

From the right: $f(1^+) = 2$

Set equal for continuity: $k + 1 = 2 \Rightarrow k = {1}$