Ncert Solutions Maths class 12th

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

The $fx$ n is $f(x)=\frac{1}{x}$ , which is a $f:R_{*}$ $\to$ $R_{*}$ and $R_{*}$ is set of all non-zero real numbers

For, $x_1,x_2\in R_{*},f(x_1)=f(x_2)$

$\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

$\Rightarrow x_1=x_2$ So, $f$ is one-one

For, $y\in R_{*},$ $x=\frac{1}{f(x)}=\frac{1}{y}$ such that

So, $f(x)=y$

So, every element in the co-domain has a pre-image in $f$

So, $f$ is onto

If $f:N\to R_{*}$ such th

The given relation in set N defined by

$R=\left\{\left(a,b\right):a=b-2,b>6\right\}$

For (2,4),        4>6 is not true

For (3,8),     8>6  but  3= 8-2 ⇒3=6 is not true

For (6,8),      8>6 and 6= 8-2 ⇒6=6 is true

And for (8,7), 7>6 but 8= 7-2 ⇒8=5 is not true

Hence, option (C) is correct

The set in $A=\left\{1,2,3,4\right\}$

The relation in this set $A$ is given by

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

$R$ is reflexive as $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right)\in R$

As, $\left(1,2\right)\in R$ but $\left(2,1\right)\notin R$

$R$ is not symmetric

For $\left(1,2\right)\in R$ and $\left(2,2\right)\in R;\left(1,2\right)\in R$

And for $\left(1,3\right)\in R$ and $\left(3,2\right)\in R;\left(1,3\right)\in R$

∴ $R$ is transitive

Hence, option (B) is correct

The given relation in the set $L=$ all lines in $XY-$ plane is defined as

$R=\{\left(L_1,L_2\right):L_1$ is parallel to $L_2\}$

Let $L_1\in A$ then as $L_1$ is parallel to $L_1$ ,

$\left(L_1,L_1\right)\in R$

So, $R$ is reflexive

Let $L_1,L_2\in A$ and $\left(L_1,L_1\right)\in R$

Then, $L_1$ is parallel to $L_2$

$L_2$ is parallel to $L_1$

So, $\left(L_2,L_1\right)\in R$

i.e.,&

The given relation in set A of all polygons is defined as

R= $\{\left(P_1,P_2\right):P_1$ and $P_2$ have same number of sides $\}$

Let $P_1\in A$ ,

As number of sides $\left(P_1\right)$ = number of sides $\left(P_1\right)$

$\left(P_1,P_1\right)\in R$

So, R is reflexive.

Let $P_1,P_2\in A$ and $\left(P_1,P_2\right)\in R$

Then, number of sides of $P_1$ = number of sides of $P_2$

Number of