Ans: B
$\begin{array}{rcl}
\dfrac{5}{2-i} + ki & = & \dfrac{5}{2-i} \times \dfrac{2+i}{2+i} + ki \\
& = & \dfrac{10 + 5i}{(2)^2 – (i)^2} +ki \\
& = & \dfrac{10 + 5i}{5} + ki \\
& = & 2 + i + ki \\
& = & 2 +(k+1) i
\end{array}$
$\begin{array}{rcl}
\dfrac{5}{2-i} + ki & = & \dfrac{5}{2-i} \times \dfrac{2+i}{2+i} + ki \\
& = & \dfrac{10 + 5i}{(2)^2 – (i)^2} +ki \\
& = & \dfrac{10 + 5i}{5} + ki \\
& = & 2 + i + ki \\
& = & 2 +(k+1) i
\end{array}$
Since $\dfrac{5}{2-i} + ki$ is a real number, then its imaginary part is $0$. Hence, we have
$\begin{array}{rcl}
k + 1 & = & 0 \\
k & = & -1
\end{array}$
