Let f(x)=\ array cc 1+(x)/(2 k), & 0 < x < 2 k x, & 2 ≤ x < 4 array...

$f(x)=$\begin{cases}1+\frac{x}{2 k}, & 0 < x < 2 \\ k x, & 2 \leq x < 4\end{cases}Hence, $\lim _{x \rightarrow 2} f(x)$ exists so we can write that\begin{aligned} & \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x) \\ & \lim _{x \rightarrow 2^{-}}\left(1+\frac{x}{2 k}\right)=\lim _{x \rightarrow 2^{+}}(k x) \\ & \Rightarrow 1+\frac{2}{2 k}=2 k \\ & \Rightarrow 1+\frac{1}{k}=2 k \\ & \Rightarrow k+1=2 k^2 \\ & \Rightarrow 2 k^2-k-1=0 \\ & \Rightarrow 2 k^2-2 k+k-1=0 \\ & \Rightarrow 2 k(k-1)+1(k-1)=0 \\ & \Rightarrow(2 k+1)(k-1)=0 \\ & \therefore k=1 \text{\text{ or } } k=-\frac{1}{2} \end{aligned}$$