Which one of the following circuits implements the Boolean function given below? f(x, y,z) = m0 + m1 + m3 +m4 + m5 + m6, where mi is the ith minterm.
Which one of the following circuits implements the Boolean function given below? f(x, y,z) = m0 + m1 + m3 +m4 + m5 + m6, where mi is the ith minterm. Correct Answer <img alt="F1 Raju.S 24-02-21 Savita D11" src="//storage.googleapis.com/tb-img/production/21/02/F1_Raju.S_24-02-21_Savita_D11.png" style="width: 177px; height: 156px;">
Answer: Option 1
[ alt="F1 Shraddha Raju 17.03.2021 D4" src="//storage.googleapis.com/tb-img/production/21/03/F1_Shraddha_Raju_17.03.2021_D4.png" style="width: 389px; height: 168px;">
Concept :
- Boolean expression : S̅ 1S̅ 0I0 + S̅ 1 S0I1 + S1S̅ 0I2 + S1S0I3
Explanation :
Given :
F( x, y, z ) = m0 + m1 + m3 + m4 + m5 + m6
F( x, y, z ) = x’y’z’ + x’y’z + x’yz + xy’z’ + xy’z + xyz’
Now if we observe Given Question is a MCQ i.e. There will be only one Correct Option.
In given all options, Multiplexers have y, z as Select lines.
So we will try classify the input based on select line.(possible inputs)
And other variables on input side.
[ alt="F1 Shraddha Raju 17.03.2021 D5" src="//storage.googleapis.com/tb-img/production/21/03/F1_Shraddha_Raju_17.03.2021_D5.png" style="width: 333px; height: 267px;">
So Correct options is 1.
Alternatively
Alternatively,
Option 4 :
Boolean expression : y’z’x’ + y’z(1) + yz’(x) + yz(1)
= x’y’z’ + xyz’ + y’z + yz
= x’y’z’ + xyz’ + (x+x’)y’z + (x+x’)yz
= x’y’z’ + x’y’z + x’yz + xy’z + xyz’ + xyz
= m0 + m1 + m3 + m5 + m6 + m7
This is not equivalent to given expression in the Question.
Option 3 :
Boolean Expression : y’z’(1) + y’z(1) + yz’(x’) + yz(x)
= x’yz’ + xyz + (x+x’)y’z’ + (x+x’)y’z
= x’y’z’ + x’y’z + x’yz’ + xy’z’ + xy’z + xyz
= m0 + m1 + m2 + m4 + m5 + m7
Again, This is not equivalent to given expression in the Question.
Option 2 :
Boolean Expression : y’z’(x) + y’z(1) + yz’(x’) + yz(1)
= x’yz’ + xy’z’ + y’z + yz
= x’yz’ + xy’z’ + (x+x’)y’z + (x+x’)yz
= x’y’z + x’yz’ +x’yz + xy’z’ + xy’z + xyz
= m1 + m2 + m3 + m4 + m5 + m7
Again, This is not equivalent to given expression in the Question.
Option 1 :
Boolean Expression : y’z’(1) + y’z(1) + yz’(x) + yz(x’)
= xyz’ + x’yz + y’z’ + y’z
= xyz’ + x’yz + (x+x’)y’z’ + (x+x’)y’z
= x’y’z’ + x’y’z + x’yz + xy’z’ +xy’z + xyz’
= m0 + m1 + m3 + m4 + m5 + m6
Hence this is the Correct Option.

