Relations and Functions

20. (i) Domain of the given relation = {2,5,8,11,14,17}

Since every element of the domain has one and only one image, the given relation is a fxⁿ.

So, domain = {2,5,8,11,14,17}

range = {1}

(ii) Domain of the given relation = {2,4,6,8,10,12,14}

Since every element of the domain has one and only one image

(i) $f:R\to R$ defined as $f(x)=3-4x$

For $x_1,x_2\in R$ such that $f(x_1)=f(x_2)$

$\begin{array}{l}\Rightarrow 3-4x_1=3-4x_2\\ \Rightarrow 4x_1=4x_2\\ \Rightarrow x_1=x_2\end{array}$

So, $f$ is one-one

For $y\in R$ , there exist

$f\left(\frac{.3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right)=3-3+y=y$

Hence, $f$ is onto

$\therefore f$ is bijective

(ii) Given, $f:R\to R$ defined as $f(x)=1+x_{}^2$

For $x_1,x_2\in R$ such that $f(x_1)=f(x_2)$

$\begin{array}{l}\Rightarrow 1+x_1^2=1+x_2^2\\ \Rightarrow x_1^2=x_2^2\\ \Rightarrow x_1^{}=\pm x_2^{}\end{array}$

$\Rightarrow x_1^{}=x_2^{}$ or $x_1^{}=-x_2^{}$

$\therefore f$ is not one-one

The range of $f(x)$ is always a positive real

Given,  $f:A\to B$ and $f=\{(1,4),(2,5),(3,6)\}$

$\begin{array}{l}\therefore f(1)=4\\ f(2)=5\\ f(3)=6\end{array}$

i.e., the image elements of $A$ under the given $f_X^n$ $f$ are unique

So,  $f$ is one-one

The $f{x}_{}^n$ $f:R?R$ is given by $f(x)=\left(\begin{array}{ccc}\hfill 1_{}\hfill & \hfill if\hfill & \hfill x>0\hfill \\ \hfill 0_{}\hfill & \hfill if\hfill & \hfill x=0\hfill \\ \hfill ?1_{}\hfill & \hfill if\hfill & \hfill x<0\hfill \end{array}\right)$

For $x_1=1,x_2=2,?R$

$f(x_1)=f(1)=1$

$f(x_2)=f(2)=1$ but $1?2$

So,  $f$ is not one-one

And the range of $f(x)=\{1,0,?1\}$ hence it is not equal to the co-domain $R$

So,  $f$ is not onto

The $f{x}_{}^n$ $f:R\to R$ is given by $f(x)=\left|x\right|$

$\Rightarrow f(x)=\left(\begin{array}{cc}\hfill x_{},if\hfill & \hfill x\ge 0\hfill \\ \hfill -x_{},if\hfill & \hfill x<0\hfill \end{array}\right)$

For $x_1=-1$ and $x_2=1$

$f(x_1)=f(-1)=\left|-1\right|=1$

$f(x_2)=f(1)=\left|1\right|=1$

So, $f(x_1)=f(x_2)$ but $x_1\ne x_2$

i.e., $f$ is not one-one

For $x=-1\in R$

$f(x)=\left|x\right|$

i.e., $f(-1)=\left|-1\right|=1$

So, range of $f(x)$ is always a positive real number and is not equal to the co-domain $R$

i.e., $f$ is not onto

The $f{x}_{}^n$ $f:R\to R$ is given by $f(x)=[x]$

Let $x_1=1.5$ and $x_2=1.2\in R$ Then,

$f(x_1)=f(1.5)=[1.5]=1$

$f(x_2)=f(1.2)=[1.2]=1$

So, $f(x_1)=f(x_2)$ but $x_1\ne x_2$

i.e., $f(1.5)=f(1.2)$ but $1.5\ne 1.2$

So, $f$ is not one-one

The range of $f(x)$ is a set of all integers, $Z$ which is not a co-domain of $R$

$\therefore f$ is not onto

(i) $f:N\to N$ given by $f(x)=x_{}^2$

For, $x_1,x_2\in N$ , $f(x_1)=f(x_2)$

$\Rightarrow x_1^2=x_2^2$

$\Rightarrow x_1^{}=x_2^{}\notin N$

So, $f$ is one-one/ injective

For $x\in N$ , i.e., $x=1,2,3....$

Range of $f(x)=\{1_{}^2,2_{}^2,3_{}^2...\}=\{1,4,9...\}\ne N$

i.e., co-domain of $N$

So, $f$ is not onto/ subjective

(ii) $f:Z\to Z$ given by $f(x)=x_{}^2$

For, $x_1,x_2\in Z$ , $f(x_1)=f(x_2)$

$\Rightarrow x_1^2=x_2^2$

$\Rightarrow x_1^{}=\pm x_2^{}\notin Z$

i.e., $x_1=x_2$ and $x_1=-x_2$

So, $f$ is not one-one/ injectiv

19. Given, R= { (a, b): a, b $\in$ z and a – b is an integer}

We know that, the difference of two integers is also an integer.

R= { (a, b): a – b $\in$ z & a, b $\in$ z}

Domain of R=Z.

Range of R= Z.

18. Given, A={x, y, z}so, n(A)=3

B={1,2} so n(B)=2

? n(A × B)=n(A) ×n(B)=3 × 2=6

Hence, no. of relation from A to B=Number of subsets of A × B

=2⁶

=64.

17. GivenR= { (x, x³) : x is a prime number less than 10}

R = { (x, x³) : x = 2,3,5,7}

= { (2,2³), (3,3³), (5,5³), (7,7³)}

= { (2,8), (3,27), (5,125), (7,343)}