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The figure gives a system of logic gates...

The figure gives a system of logic gates. From the study of the truth tabe, it can be found that to produce a high output (1) at R, we must have

A

X= 0, Y = 1

B

X=1, Y=1

C

X=1, Y=0

D

X=0, Y=0

Text Solution

Verified by Experts

The correct Answer is:
C


`[bar((bar(X)+Y)+Xbar(Y))]=R`
`R=(bar(bar(X)+Y))*(bar(Xbar(Y)))`
`R=(barX barY) (barX+barY)`
`R=X barY(barX+Y)`
`R=X barX barY+X barY`
`R=XbarY`
i.e., `{:(X=1),(Y=0):}] rArr R=1`
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