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The switching function for a Boolean expression is the number of times the output changes from 0 to 1 or from 1 to 0 as the input variables change from one combination to another.

The Boolean expression for f(A, B, C, D) = BD + B'D' can be simplified using Boolean algebra as follows:

f(A, B, C, D) = BD + B'D' = BD + B'(B + D) = BD + B'B + B'D = BD + B'D

Now, let's create a truth table for f(A, B, C, D):

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

From the truth table, we can see that the output of the function changes only twice, from 0 to 1 and from 1 to 0. Therefore, the switching function for f(A, B, C, D) is 2.

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f(a,b,c,d)=bd(a+a')(c+c')+b'd'(a+a')(c+c')

=bdac+bdac'+bda"c+bda'c+bda'c'+b'd'ac+b'd'ac'+b'd'a'c+b'd'a'c'

=8
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1     1
  1 1  
  1 1  
1     1

this f(a,b,c,d) is filled in k map it will give 8 min terms∑(0,2,5,7,8,10,13,15) for these value cicuit will give output as high so no of switching function will be eight

to approach this type of question@radha fill the k map and find no of minterms in k map

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f(a,b,c,d)=bd(a+a')(c+c')+b'd'(a+a')(c+c')

               =bdac+bdac'+bda'c+bda'c'+b'd'ac+b'd'ac'+b'd'a'c+b'd'a'c'

so f has 8 minterms. if we use k map for this function we will get 2 sub cubes which are essential prime implicants. These 2 covers all minterms which gives only a unique minimal expression for  function so answer is 1  (if both essential prime implicants not cover all minterms then we will choose the prime implicants which are there remaing to cover remaining minterms which will give more than one minimal expression ).  correct me if am wrong

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