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Question 13 (OR 2
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Check whether the relation R in the set R of real numbers, defined by R = {(a, b) : 1 + ab > 0}, is reflexive, symmetric or transitive.

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1 Important

Question 2

Question 3

Question 4 (Or 1st) Important

Question 4 (Or 2nd)

Question 5

Question 6

Question 7 Important

Question 8 (Or 1st) Important

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st) Important

Question 10 (Or 2nd)

Question 11 Important

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st) Important

Question 13 (Or 2nd) You are here

Question 14 Important

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd) Important

Question 17

Question 18

Question 19 Important

Question 20 Important

Question 21 (Or 1st)

Question 21 (Or 2nd) Important

Question 22

Question 23 Important

Question 24 (Or 1st)

Question 24 (Or 2nd) Important

Question 25

Question 26 (Or 1st) Important

Question 26 (Or 2nd)

Question 27 (Or 1st) Important

Question 27 (Or 2nd) Important

Question 28

Question 29 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Oct. 1, 2019 by Teachoo

**
Question 13 (OR 2
**
**
nd
**
**
question)
**

Check whether the relation R in the set R of real numbers, defined by R = {(a, b) : 1 + ab > 0}, is reflexive, symmetric or transitive.

Question 13 (OR 2nd question) Check whether the relation R in the set R of real numbers, defined by R = {(a, b) : 1 + ab > 0}, is reflexive, symmetric or transitive. R = {(a, b) : 1 + ab > 0}, Checking for reflexive If the relation is reflexive, then (a ,a) ∈ R i.e. 1 + a2 > 0 Since square numbers are always positive Hence, 1 + a2 > 0 is true for all values of a. So, the given relation it is reflexive. R = {(a, b) : 1 + ab > 0}, Checking for symmetric, To check whether symmetric or not, If (a, b) ∈ R, then (b,a) ∈ R i.e., if 1+ ab < 0, then 1 + ba > 0 Since if 1 + ab > 0, then 1 + ba > 0 is always true for all value of a & b Hence, the given relation is symmetric R = {(a, b) : 1 + ab > 0}, Checking transitive To check whether transitive or not, If (a, b) ∈ R & (b, c) ∈ R , then (a, c) ∈ R i.e., if 1 + ab > 0, & 1 + bc > 0, then 1 + ac > 0 Let’s take an example a = –8, b = –2, c = 1/4 1 + ab = 1 + (–8) × (–2)= 1 + 16= 17> 0 1 + bc = 1 + (–2) × 1/4= 1 – 1/2 = 1/2> 0 1 + ac = 1 + (–8) × 1/4= 1 – 2= –1≯ 0 Since 1 + ac ≯ 0 when 1 + ab > 0 and 1 + bc > 0 ∴ The condition is not true for all values of a, b, c. Hence, the given relation it is not transitive Therefore, the given relation is reflexive and symmetric, but not transitive