The given function f(x) will be continuous at x = 2, if $\lim\limits_{x\rightarrow2^{-}}$ f (x) = $\lim\limits_{x\rightarrow2^{+}}$ f (x) = f (2) $\lim\limits_{x\rightarrow2^{-}}$ = $\lim\limits_{x\rightarrow2^{-}}$ 2x + 1 = 2 × 2 + 1 = 5 $\lim\limits_{x\rightarrow2^{-}}$ = $\lim\limits_{x\rightarrow2^{-}}$ 3x + 1 = 3 × 2 - 1 = - 5 ∴ f (2) = k ⇒ k = 5 Thus, for k = 5, the given function is continuous at x = 2.