Correct Option: A
There are two control line one is $K$ and another is $C_{0}$ .
- When $K=1$ , $C_{0}= 1$ we can perform $A-B$
- When $K=0$, $C_{0}=0$ we can perform $A+B$
But without manipulating $B(B_{0},B_{1},\ldots)$ we cannot perform A+1. But here we have only two control lines which is $K,C_{0}$. Therefore the answer is A.
Note:
For A+B : $C_{0}= 0$ , $K=0$, and $0 \oplus x = x$
$\frac{\begin{matrix} &A3 &A2 &A1 &A0 \\ +& B3 & B2 & B1 & B0 \end{matrix}}{\begin{matrix} &S3&S2&S1&S0\end{matrix}}$
$\frac{\begin{matrix} &C3&C2&C1&C0=0\\ &A3 &A2 &A1 &A0 \\ \qquad\qquad\qquad\qquad+& B3 & B2 & B1 &B0 \end{matrix}}{\begin{matrix} Sum \enspace Output:&S3&S2&S1&S0\\Carry \enspace Output:\qquad&C4&C3&C2&C1\end{matrix}}$
For A-B : $C_{0} = 1$ , $K = 1$, and $1\oplus x=\bar x$
given that numbers are in 2's complement representations, So A - B = A + 2's complement of B
How to get 2's complement of B ?
2's complement of B = 1's complement of B + 1
So, by keeping K=1, we get 1's complement of B and By keeping C$_{0}$ = 1, we are adding 1 to 1's complement of B.