Relations and Functions

(i) We have, $R=\{\left(x,y\right):3x-y=0\}$ a relation in set A= $\{1,2,3..........14\}$

For $x\in A,y=3x$ or $y\ne x$ i.e.,

$\left(x,x\right)$ does not exist in R

$\therefore$ R is not reflexive.

For $\left(x,y\right)\in R,y=3x$

Then $\left(y,x\right)x\ne 3y$

So $\left(y,x\right)\notin R$

$\therefore$ R is not symmetric

For $\left(x,y\right)\in R$ and $\left(y,z\right)\in R$ . We have

$y=3x$ and $z=3y$

Then $z=3\left(3x\right)=9x$

i.e., $\left(x,z\right)\notin R$

$\therefore$ R is not Transitive

(ii) We have,

R= $\{\left(x,y\right):y=x+5\&x<4\}$ is a relation in N

= $\{\left(1,1+5\right),\left(2,2+5\right),\left(3,3+5\right)\}$

= $\{\left(1,6\right),\left(2,7\right),\left(3,8\right)\}$

Clearly, R is not reflexive as $\left(x,x\right)\notin R$ and $x<4\&x\in N$

Also, R is not symmetric as $\left(1,6\right)\in R$ but $\left(6,1\right)\notin R$

And for $\left(x,y\right)\in R\left(y,z\right)\notin R$ . Hence, R is not Transitive.

(iii) R= $\{\left(x,y\right);y$ is divisible by x $\}$ is a relation in set

A= $\{1,2,3,4,5,6\}$

So, R= $\{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,2\right),\left(2,4\right),\left(2,6\right),\left(3,3\right),\left(3,6\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\}$

Hence, R is reflexive because $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\in R$ i.e., $\left(x,x\right)\in R$

R is not symmetric as $\left(1,2\right)\in R$ but $\left(2,1\right)\notin R$

And for $x,y,z\in A$ and $\left(x,y\right)\in R\&\left(y,z\right)\in R$

$\frac{y}{z}=n$ and $\frac{z}{y}=m$ where $n\&m\in N$

Then, $z=my=m(nx)=mn.x$

$\frac{z}{x}=m.n,m,mn\in N$

Hence, $\left(x,z\right)\in R$

$\therefore$ R is transitive

(iv) R= $\{\left(x,y\right):x-y$ is an integer $\}$ is a relation in set Z

For $x\in Z$

$x-x=0$ is an integer

So, $\left(x,x\right)\in R$ i.e., R is reflexive

For $x,y\in Z$ and $\left(x,y\right)\in R$

$x-y$ is an integer

$-\left(y-x\right)$ is an integer

$\left(y-x\right)$ is an integer

So, $\left(y,x\right)\in R$ i.e., R is symmetric

For $x,y,z\in R$ and $\left(x,y\right)\in R\&\left(y,z\right)\in R$ We have,

$x-y$ is an integer

$y-z$ is an integer

So, $\left(x-y\right)+\left(y-z\right)$ is also an integer

$x-z$ is an integer

So, $\left(x,z\right)\in R$ i.e., R is transitive.

(v) (a) R= $\{\left(x,y\right):x$ and $y$ work at same place $\}$ in set A of human being.

For $x\in A$ we get

$x\&x$ work at same place

$\left(x,x\right)\in R$ So, R is reflexive.

For $x,y\in A$ and $\left(x,y\right)\in R$ We get,

$x\&y$ work at same place

$y\&x$ work at same place

$\left(y,x\right)\in R$ . So, R is symmetric.

For $x,y,z\in A$ and $\left(x,y\right)$ and $\left(y,z\right)\in R$ . We have,

$x\&y$ work at same place

$y\&z$ work at same place

$x\&z$ work at same place

i.e., $\left(x,z\right)\in R$ . So, R is Transitive.

(b) R= $\{\left(x,y\right):x\&y$ live in same locality $\}$

For $x\in A$ , we get,

$x\&x$ live in same locality

$\left(x,x\right)\in R$ So, R is reflexive.

For $x,y\in A$ and $\left(x,y\right)\in R$ we get,

$x\&y$ live in same locality

$y\&x$ live in same locality

i.e., $\left(y,x\right)\in R$ . So, R is symmetric.

For $x,y,z\in A$ and $\left(x,y\right)$ & $\left(y,z\right)\in R$ . Then

$x,y$ live in same locality

$y,z$ live in same locality

$x,z$ live in same locality

So, $\left(x,z\right)\in R$ i.e., R is transitive.

(c) R= $\{\left(x,y\right):x$ is exactly 7 cm taller than y $\}$

For $x\in A$ ,

Height $\left(x\right)\ne 7+$ height of $\left(x\right)$

So, $\left(x,x\right)\notin R$ . i.e., R is not reflexive.

For, $x,y\in A$ and $\left(x,y\right)\in R$ we have,

Height $\left(x\right)$ $=7+height\left(y\right)$

But $height\left(y\right)\ne 7+height\left(x\right)$

So, $\left(y,x\right)\in R$ i.e., R is not symmetric.

For $x,y,z\in A$ and $\left(x,y\right)\in R$ and $\left(y,z\right)\in R$ we have,

Height $\left(x\right)=7+height\left(y\right)$

And $height\left(y\right)=7+height\left(z\right)$

So, $height\left(x\right)=7+7+height\left(z\right)=14+height\left(z\right)$

i.e., $\left(x,z\right)\notin R$

So, R is not transitive.

(d) R= $\{\left(x,y\right):x$ is wife of y $\}$

For $x\in A,$

X is not wife of x.

So, $\left(x,x\right)\notin R$ i.e., R is not reflexive.

For $x,y\in A$ and $\left(x,y\right)\in R$ ,

X is wife of y but y is not wife of x

So, $\left(y,x\right)\notin R$ . i.e., R is not symmetric.

For $x,y,z\in A$ and $\left(x,y\right)$ and $\left(y,z\right)\in R$

X is wife of y

Y is wife of z

But y can never be husband & wife simultaneously

So, R is not transitive.

(e) R= $\{\left(x,y\right):x$ is father of y $\}$

For $x\in A$ ,

X is not a father of x

So, $\left(x,x\right)\notin R$ i.e., R is not reflexive.

For $x,y\in A$ and $\left(x,y\right)\in R$

X is father of y

Y is father of z

But x is not father of z

So, $\left(x,z\right)\notin R$ i.e., R is not transitive.

An inverse of a function does just opposite work as the original function does in other words it essentially "undoes" what the original function does. If a function f maps an input x to an output y (written as f (x) = y),

 then its inverse function is denoted as  f? ¹ (y) and  f? ¹ (y) = x

However, Students must be aware that not all functions have inverses. For an inverse function, the function must be one-one (injective) and onto (surjective): So that every possible output value is covered.

Together, these two properties make the function bijective, and only bijective functions have well-defined inverses. If a function isn't one-one (like f (x) = x²), it can't have a true inverse unless its domain is restricted.

A relation deals with, how elements from any set  (A) are related to elements of another set (B). there are various form of relations, such as Reflexive, Symatric, Transitive and Equivalance. On the other hand, Functions are special case of Relations where every element in Set A is related to exactly one element in Set B. Functions works like mathematics formulas, for a given set of input the out put will always remain the same. Students will be abel to check the type of relations, and application in class 11 Math Relation and Function chapter. We have provided Class 11 Math Relations and Functions NCERT Solutions on shiskha