Solution:

To implement the given Boolean function f(A, B, C, D) using NAND gate, we can follow the below steps:

Step 1: Obtain the truth table of the given Boolean function f(A, B, C, D)

A | B | C | D | f(A, B, C, D)
--|---|---|---|----------------
0 | 0 | 0 | 0 | 1
0 | 0 | 0 | 1 | 1
0 | 0 | 1 | 0 | 1
0 | 0 | 1 | 1 | 1
0 | 1 | 0 | 0 | 1
0 | 1 | 0 | 1 | 0
0 | 1 | 1 | 0 | 1
0 | 1 | 1 | 1 | 0
1 | 0 | 0 | 0 | 1
1 | 0 | 0 | 1 | 0
1 | 0 | 1 | 0 | 0
1 | 0 | 1 | 1 | 0
1 | 1 | 0 | 0 | 0
1 | 1 | 0 | 1 | 0
1 | 1 | 1 | 0 | 0
1 | 1 | 1 | 1 | 0

Step 2: Obtain the SOP (Sum of Products) form of the Boolean function f(A, B, C, D) from the truth table.

f(A, B, C, D) = A'B'C'D' + A'B'C'D + A'B'CD' + A'B'CD + A'BC'D' + A'BCD'

Step 3: Implement the SOP form of f(A, B, C, D) using NAND gate.

The NAND gate is a universal gate, which means that we can implement any Boolean function using only NAND gates.

We can use the De Morgan's theorem to implement the SOP form of the Boolean function f(A, B, C, D) using NAND gates.

De Morgan's theorem: (A.B)' = A' + B' and (A + B)' = A'.B'

Using De Morgan's theorem, we can represent the SOP form of the Boolean function f(A, B, C, D) as:

f(A, B, C, D) = [(A'B'C'D')' . (A'B'C'D)' . (A'B'CD')' . (A'B'CD)' . (A'BC'D')' . (A'BCD')']'

We can implement the above expression using NAND gate as follows:

- For each variable A, B, C, D, and its complement, we need one NAND gate.
- For each product term in the SOP form, we need one NAND gate.
- For the final OR operation, we need one more NAND gate.

Therefore, the minimum number of 2-input NAND gates required to implement the Boolean function f(A, B, C, D) using NAND gate is 3.

So, the correct answer is '3'.