The given function f(x) will be continuous at x = 2, if
$\underset{\mathit{x}\to 2^{-}}{\lim }$ f (x) = $\underset{\mathit{x}\to 2^{+}}{\lim }$ f (x) = f (2)
$\underset{\mathit{x}\to 2^{-}}{\lim }$ = $\underset{\mathit{x}\to 2^{-}}{\lim }$ 2x + 1 = 2 × 2 + 1 = 5
$\underset{\mathit{x}\to 2^{-}}{\lim }$ = $\underset{\mathit{x}\to 2^{-}}{\lim }$ 3x + 1 = 3 × 2 - 1 = - 5
∴ f (2) = k
⇒ k = 5
Thus, for k = 5, the given function is continuous at x = 2.