1.A relation R is defined on the set Z of integers as follows:

(x,y)∈R iff x ^{2} + y ^{2 } = 25.

Express R and R-1 as the set of odered pairsand hence find the respective domains.

2.Let f : **R** → **R** . be defined as { 2x+1, x<= 4 Show that f is not a function

x+1 , x>=4 }

We have,

(x,y) ∈ R ⇔ x^{2} + y^{2} = 25 ⇔ y = ±√(25-x^{2})

We observe that x = 0 ⇒ y = ± 5

∴ (0,5) ∈ R and (0,-5) ∈ R

x = ± 3 ⇒y = ±√(25-9) = ± 4

∴ (3,4) ∈ R, (-3,4) ∈ R, (3,-4) ∈ R and (-3,-4) ∈ R

x = ± 4 ⇒y = ±√(25-16) = ± 3

∴ (4,3) ∈ R, (-4,3) ∈ R, (4,-3) ∈ R and (-4,-3) ∈ R

x = ± 5 ⇒y = ±√(25-25) = 0

∴ (5,0) ∈ R and (-5,0) ∈ R

We also notice that for any other integral value of x,y is not an integer.

∴ R = {(0,5), (0,-5),(3,4),(-3,4),(3,-4),(-3,-4),(4,3),(-4,3),(4,-3),(-4,-3),(5,0),(-5,0)}

and R^{-1} = {(5,0), (-5,0),(4,3),(4,-3),(-4,3),(-4,-3),(3,4),(3,-4),(-3,4),(-3,-4),(0,5),(0,-5)}

Clearly domain [R] = {0,3,-3,4,-4,5,-5} = domain (R^{-1})

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