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GATE 2014 [Set-2]
December 23, 2024NTA UGC NET JUNE-2023 Paper-2
December 23, 2024GATE 2014 [Set-2]
| Question 16 |
The dual of a Boolean function F(x1, x2, …, xn, +, ⋅, ‘), written as FD, is the same expression as that of F with + and ⋅ swapped. F is said to be self-dual if F = FD. The number of self-dual functions with n Boolean variables is
| 2n | |
| 2(n-1) | |
| 2(2n ) | |
| 2(2(n-1) ) |
Question 16 Explanation:
Number of possible minterms = 2n.
Number of mutually exclusive pairs of minterms = 2n-1.
There are 2 choices for each pair i.e., we can choose one of the two minterms from each pair of minterms for the function.
Therefore number of functions = 2 x 2 x …. 2n-1 times.
= 2(2(n-1))
Number of mutually exclusive pairs of minterms = 2n-1.
There are 2 choices for each pair i.e., we can choose one of the two minterms from each pair of minterms for the function.
Therefore number of functions = 2 x 2 x …. 2n-1 times.
= 2(2(n-1))
Correct Answer: D
Question 16 Explanation:
Number of possible minterms = 2n.
Number of mutually exclusive pairs of minterms = 2n-1.
There are 2 choices for each pair i.e., we can choose one of the two minterms from each pair of minterms for the function.
Therefore number of functions = 2 x 2 x …. 2n-1 times.
= 2(2(n-1))
Number of mutually exclusive pairs of minterms = 2n-1.
There are 2 choices for each pair i.e., we can choose one of the two minterms from each pair of minterms for the function.
Therefore number of functions = 2 x 2 x …. 2n-1 times.
= 2(2(n-1))