Correct : a

To minimize this function, we use a 4-variable K-Map with variables b₃, b₂, b₁, b₀. The function is given as F = Σ(0, 2, 4, 8, 10, 11, 12), so we place 1s at these minterm positions and 0s everywhere else. Since there are no don't care terms, every minterm must be covered exactly.
Let's first convert the minterms to their binary equivalents (b₃b₂b₁b₀):
0 → 0000, 2 → 0010, 4 → 0100, 8 → 1000, 10 → 1010, 11 → 1011, 12 → 1100
Now we identify the groups on the K-Map:
Group 1 — {0, 2, 8, 10}: These four minterms all have b₁=0 and b₀=0 in common, regardless of b₃ and b₂. So this group gives the term b̄₁b̄₀.
Group 2 — {0, 4, 8, 12}: These four minterms all have b₂=0 and b₀=0 in common. This group gives the term b̄₂b̄₀.
Group 3 — {10, 11}: These two minterms have b₃=1, b₂=0, b₁=1 in common, with b₀ varying. This pair gives the term b₁b̄₂b₃.
Now, the critical thing to check — is minterm 11 covered by any of the first two groups? Minterm 11 is 1011, which has b₀=1. Group 1 requires b̄₀ and Group 2 also requires b̄₀, so neither covers minterm 11. This means the third term b₁b̄₂b₃ is absolutely essential and cannot be dropped.
This is exactly why Option B (which gives only b̄₁b̄₀ + b̄₂b̄₀) is wrong — it covers only 6 out of the 7 required minterms and misses minterm 11 completely.
The final minimized SOP expression is b̄₁b̄₀ + b̄₂b̄₀ + b₁b̄₂b₃, which covers all 7 minterms with no redundancy. That's Option A.
A useful habit during GATE — always verify your minimized expression by checking that every minterm in the original sum is satisfied by at least one term in your answer. It takes 30 seconds and saves you from picking an incomplete expression like Option B.