For the given Boolean theorem
AB + A′C + BC = AB + A′C
Apply dual property, we get:
(A + B) (A′ + C) (B + C) = (A + B) (A′ + C)
Hence alternative (A) is the correct answer.

Correct Option: A

For the given Boolean theorem
AB + A′C + BC = AB + A′C
Apply dual property, we get:
(A + B) (A′ + C) (B + C) = (A + B) (A′ + C)
Hence alternative (A) is the correct answer.

= (A′ + B′ + C′)′
= A′′. B′′. C′′
= ABC
Hence (B) is the correct alternative.

Correct Option: B

= (A′ + B′ + C′)′
= A′′. B′′. C′′
= ABC
Hence (B) is the correct alternative.

Apply hit and trial method
(A) A + AC + AB + BC = A + BC: i.e., False
(B) AC + AB + BC: i.e., False
(C) AC + AB + BC = AC + AB = BC + AA = A (A + B) + C (A + B) = (A + C) (A + B)
which is ≠ LHS i.e., False
(D) A + AC + AB + BC
= A (1 + C) + AB + BC
= A + BC
= L.H.S. True
Hence alternative (D) is the correct answer.

Correct Option: D

Apply hit and trial method
(A) A + AC + AB + BC = A + BC: i.e., False
(B) AC + AB + BC: i.e., False
(C) AC + AB + BC = AC + AB = BC + AA = A (A + B) + C (A + B) = (A + C) (A + B)
which is ≠ LHS i.e., False
(D) A + AC + AB + BC
= A (1 + C) + AB + BC
= A + BC
= L.H.S. True
Hence alternative (D) is the correct answer.

Each bit require one line so that for n lines, the size of memory in the address bus is 2ⁿ
for 16 bit, 2¹⁶ = 65, 536 or 64 K.

Correct Option: A

Each bit require one line so that for n lines, the size of memory in the address bus is 2ⁿ
for 16 bit, 2¹⁶ = 65, 536 or 64 K.

The gate period is made equal to main supply period, 1/50 = 0·02 s or 20 ms. for 60 Hz, it is 16·67 ms.

Correct Option: B

The gate period is made equal to main supply period, 1/50 = 0·02 s or 20 ms. for 60 Hz, it is 16·67 ms.