Let R be a relation in N defined by R = {(x, y) : x + 2y = 8}. The range of R is

{1, 2, 3}

R = {(x, y) : x + 2y = 8, x, y ∈ N}

$x+2 y=8 \Rightarrow y=\frac{8-x}{2}$

$x=1 \Rightarrow y=\frac{7}{2} \notin N ; x=2 \Rightarrow y=3 \in N$

$x=3 \Rightarrow y=\frac{5}{2} \notin N ; x=4 \Rightarrow y=2 \in N$

$x=5 \Leftrightarrow y=\frac{3}{2} \notin N ; x=6 \Rightarrow y=1 \in N$

$x=7 \Rightarrow y=\frac{1}{2} \notin N ; x=8 \Rightarrow y=0 \notin N$

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∴ R = {(2, 3), (4, 2), (6, 1)

∴ Range of R = {y : (x, y) ∈ R) = {1, 2, 3}.

Hence (2) is the correct answer.