#### Please Solve R.D. Sharma class 12 Chapter relations Exercise 1.1 Question 10  Maths textbook Solution.

Domain of $R$ is x ∈ N where x∈$\left \{ 1,2,3............20 \right \}$and

range of $R$ is y ∈ N where $\left \{ 39,37,35,......3,1 \right \}$

The relation having properties of being neither symmetric nor transitive nor reflexive.

Hint:

A relation R on set A is

Reflexive relation:

If $(a, a) \in R$ for every $a \in A$

Symmetric relation:

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

Transitive relation:

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

Given:

$R=\{(x, y): x, y \in N, 2 x+y=41\}$

Solution:

The domain of $R \text { is } x \in N, \text { such that } 2 x+y=41$

$x=(41-y) / 2$

Since $y \in N$, largest value that $x$ can take corresponds to the smallest value that $y$ can take.

$\therefore \quad x=\{1,2,3 \ldots .20\}$

Range $R$ of is  $y \in N$such that

Since

$\begin{gathered} 2 x+y=41 \\ y=41-2 x \\ x=\{1,2,3 \ldots 20\} \end{gathered}$

$y=\{39,37,35, \ldots . .3,1\}$

Since $(2,2) \notin R, R$,is not reflexive.

Also, since $(1,39) \in R(39,1) \notin R, R$ is not symmetric.

Finally, since $(15,11) \in R \text { and }(11,19) \in R \text { but }(15,19) \notin R, R$ is not transitive.

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