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Solution


Q.

Let A = {–3, –2, –1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if 0<x2+2y≤4. Let l be the number of elements in R and m be the minimum number of elements required to be added in R to make it reflexive relation. Then l + m is equal to          [2025]
1 18  
2 20  
3 17  
4 19  

Ans.

(1)

A = {–3, –2, –1, 0, 1, 2, 3}

xRy if and only if –2y≤x2≤4–2y

y = –3          6≤x2≤10               ⇒ x∈{–3,3}

y = –2          4≤x2≤8                 ⇒ x∈{–2,2}

y = –1          2≤x2≤6                 ⇒ x∈{–2,2}

y = 0            0≤x2≤4                 ⇒ x∈{–2,–1,0,1,2}

y = 1            –2≤x2≤2              ⇒ x∈{–1,0,1}

y = 2            –4≤x2≤0              ⇒ x∈{0}

y = 3            –6≤x2≤–2           ⇒ Value of x does not exist

R = {(–3, –3), (3, –3), (–2, –2),(2, –2), (–2, –1), (2, –1), (–2, 0), (–1, 0), (0,0), (1, 0), (2, 0), (–1, 1), (0, 1), (1, 1), (0, 2)}

∴   l = 15

To make it reflexive we will add (–1, –1), (2, 2), (3, 3) in R

∴   l + m = 15 + 3 = 18.