Logic functions and Minimization

By Definition of XNOR, So Option-A is correct. Also by Definition of XOR, Option-B is So Option-B is also correct. Option-C is Option-C is also correct. Option-D x'⊕y' = x''y' + x'y'' = xy' + x'y = x⊕y ≠ x⊙y Therefore option (D) is false. This explanation is provided by Chirag Manwani.

There are two prime implicants in the following K-Map- Prime Implicant higlighted in Green = Prime Implicant higlighted in Orange = So the Boolean expression is- Therefore option (B) is correct. This explanation is provided by Chirag Manwani.

All options except D produce XOR as described below :

See following : (P+Q'+R').(P+Q'+R).(P+Q+R') = From the K-map, POS form is : P + Q'.R'

K-map,

= PQ + QR’ + PR’ 
= PQ(R+R’) + (P+P’)QR’ + P(Q+Q’)R’
= PQR + PQR’ +PQR’ +P’QR’ + PQR’ + PQ’R’ 
= PQR(m7) + PQR'(m6)+P’QR'(m2) +PQ’R'(m4) 
= m2 + m4 + m6 + m7 

Option (A) is correct.

AB+C = (A+C)(B+C) = ((A+C)' + (B+C)')' So, '3' 2-input NOR gates are required.

One group consists of (0000, 0010, 1000, 1010) which gives b’d’ Other group is (0000, 0001, 1000, 1001) which gives a’d’ So, solution is b’d’ + a’d’

From logic diagram  we have f=f1.f2+f3 f=m(4,5,6,7,8).f2+m(1,6,15)----(1) from eq(1) we need to find such f2 so that we can get  f=m(1,6,8,15) eq(1)  says we can get m(1,6,15) from f3 ,so only  8 left now from option (a,b,d) we get (4,6), (4,8) ,(4,6,8) respectively for m(4,5,6,7,8)f2 which is not required as m4 is undesired. But option (C) m(4,5,6,7,8)(6,8)+(1,6,15)

• m(6,8)+m(1,6,15)
• m(1,6,8,15)an

Option (C) is correct.

Step by step explanation :

= (P + Q’)(PQ’ + PR)(P’R’ + Q’)
= (PPQ’ + PPR + PQ’Q’ + PQ’R) (P’R’ + Q’) 
= (PQ’ + PR + PQ’ + PQ’R) (P’R’ + Q’)
= (PP’Q’R’ + PP’R’R + PP’Q’R’ + PP’Q’RR’ + PQ’Q’ + PQ’R + PQ’Q’ + PQ’Q’R)
= (0 + 0 + 0 + 0 + PQ’ + PQ’R + PQ’ + PQ’R)
= PQ’ + PQ’R
= PQ'(1 + R)
= PQ’ 

So, option (A) is correct.

On solving K-MAP we get ZX’+XZ’ so  it is independent of w,y Ans (B) part.