Long Answer Type NCERT Exemplar Problems and Solution Part 1

Question – 16: If $A = (1, 2, 3, 4)$, define relations on $A$ which have properties of being

(a) Reflexive, transitive but not symmetric

(b)  Symmetric but neither reflexive nor transitive.

(c)  Reflexive, symmetric and transitive.

Solution: Let $R_1$= { $(1,1),(2,2),(1,2)$}

Thus, it is clear that $(1,1)\in R_1$ and $(2,2)\in R_1$

Thus $R_1$ is reflexive.

Again, $(1,2)\in R_1$ $but\hspace{5pt} (2,1)\notin R_1$

Thus it is not symmetric.Now,  $(2,1) \in R_1,$ and $(1,1)\in R_1$ $\Rightarrow (2,1)\in R_1$ Thus $R_1$ is reflexive and transitive but not symmetric.

Remark-1: Some students get confused while checking Transitive property. So, remember, if we are unable to find pairs like $(a,b), (b,c)\in R$. Then $R$ is transitive. This is just because of the property of $p\Rightarrow q$ Recall the truth table of $p\Rightarrow q$. If $p$ is false then $p\Rightarrow q$ is always true.

(b) Symmetric but neither reflexive nor transitive

Solution: Let $R_2$={$(1,2),(2,1)$} Thus, it is clear that $(1,2)\in R_2$ and $(2,1)\in R_2$ Therefore, $R_2$ is symmetric. But $(1,1)\notin R_2$ Thus $R_2$ is neither reflexive nor transitive.

(c) Reflexive, symmetric and transitive.

Solution: Let $R_3$={$(1,1),(2,2),(3,3),(4,4)$}

It is clear that it is reflexive and symmetric and it is also transitive. See the Remark-1 and use it to show that it is Transitive.

Question- 17:  Let $R$ be relation defined on the set of natural number $N$ as follows: $R$={$(x,y):x\in N, 2x+y=41$}. Find the domain and range of the relation $R$. Also, verify whether $R$ is reflexive, symmetric and transitive.

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

Thus, Domain of $R$={$1,2,3,...20$} It's just because $x\geq 0$ So solving $\frac{y-41}{2} \geq 0$ We can find Domain of $R$. Range of $R$ will be all such $y=41-2x$ $x\in Domain$. Which is equal to :{$1,3,5,7,9...39$} It is not Symmetric since $(2,2)\notin R$ Since $2\times2+2=6\neq 41$ Also $(1,39)\in R$ as $2\times 1+39=41$ but $(39,1)\in R$ as $39\times 2+1 \neq 41$. $(11,19),(19,3)\in R$

Since $2\times 11+19=41$ and $19\times 2 +3=41$ but $(11,3)\notin R$. Since, $2\times 11+3\neq 41$

Thus, R is neither reflexive, nor symmetric and nor transitive.♦

Question – 18 : Given $A$= {$2, 3, 4$}, $B$ = {$2, 5, 6, 7$}. Construct an example of each of the following:

(a) an injective mapping from A to B

Solution: Let $f:A\rightarrow B$ denote a mapping such that

$f$={$(x,y):y=x+3$}

But this is an injective mapping.

(b) a mapping from $A$ to $B$ which is not injective

Let $g:A\rightarrow B$ denotes a mapping given by $g$={$(2,2),(3,5),(4,5)$}

Here it is clear that it is not an injective mapping. Since for $3\neq 4$ $\Rightarrow g(3)=g(4)=5$.

(c) a mapping from B to A

Solution: Any Subset of $B\times A$ For example {$(2,2)$} ♦