Get Answers to all your Questions

header-bg qa

Need solution for RD Sharma maths Class 12 Chapter Relation Exercise 1.1 Question 5 Subquestion (ii) maths textbook solution.

Answers (1)


Reflexive and symmetric but not transitive.


A relation R on set A is

 Reflexive relation:

If  (a,a)\in \; R for every a\in \; R

Symmetric relation:

If (a,b) is true then (b,a)  is also true for every a, b \in A

Transitive relation:

If (a,b) and (b, c) \in R, then (a, c) \in R  for every \mathrm{a}, \mathrm{b}, \mathrm{c} \in A

Given :

a R b \text { if } 1+a b>0

Solution :


  Let a be an arbitrary element of R

  Then, a\in \; R

            \begin{aligned} &1+a \times a>0 \\ &1+a^{2}>0 \end{aligned}
Since, Square of any number is positive.

So, the given relation is reflexive.


  Let  (a, b) \in R

          \begin{aligned} &1+a b>0 \\ &1+b a>0 \\ &(b, a) \in R \end{aligned}

So, the given relation is symmetric.


\begin{aligned} &\text { Let }(a, b) \in R \text { and }(b, c) \in R \\ &\text { Let } \mathrm{a}=-8, \mathrm{~b}=-2, \mathrm{c}=\frac{1}{4} \\ &\text { Then } 1+a b>0 \text { i.e; } 1+(-8)(-2)=17>0 \\ &\text { and } 1+b c>0 \text { i.e; } 1+(-2) \frac{1}{4}=\frac{1}{2}>0 \text { But } 1+a c \neq 0 \text { i.e; } 1+(-8) \frac{1}{4}=-1 \ngtr 0 \\ &(a, c) \epsilon R \end{aligned}


So, the given relation is not transitive.


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support