Relations and Functions

Given,  $f:R\to R$ defined as $f(x)=3x$

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

$\Rightarrow 3x_1^{}=3x_2^{}$

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

So,  $f$ is one-one

And for $y\in R$ , there exist $\frac{y}{3}\in R$ such that

$f\left(\frac{y}{3}\right)=\frac{3*y}{3}=y$

$\therefore f$ is onto

Hence, option (A) is correct.

Given,  $f:R\to R$ defined by $f(x)=x_{}^4$

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

$\Rightarrow x_1^4=x_2^4$

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

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

So,  $f$ is not one-one

The range of $f(x)$ is a set of all positive real numbers which is not equal to co-domain $R$

So,  $f$ in not onto

$\therefore$ Option (D) is correct

Given, $f:A\to B$ defined by $f(x)=\left(\frac{x-2}{x-3}\right)$

Let $x_1,x_2\in A=R-\left\{3\right\}$ such that

$f(x_1)=f(x_2)$

$\Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}$

$\Rightarrow (x_1-2)(x_2-3)=(x_2-2)(x_1-3)$

$\Rightarrow x_1{x}_2-3x_1-2x_2+6=x_2{x}_1-3x_2-2x_1+6$

$\Rightarrow 2x_1-3x_1=2x_2-3x_2$

$\Rightarrow -x_1=-x_2$

$\Rightarrow x_1=x_{}$

So, $f$ is one-one

For $y\in B=R-\left\{1\right\}$ there exist $f(x)=y$ such that

$\frac{x_{}-2}{x_{}-3}=y$

$\Rightarrow x-2=xy-3y$

$\Rightarrow x-xy=2-3y$

$\Rightarrow x(1-y)=2-3y$

$\Rightarrow x=\frac{2-3y}{1-y}$ where $y\ne 1.\in A$

Thus, $f\left(\frac{2-3y}{1-y}\right)=\frac{\left(\frac{2-3y}{1-y}\right)-2}{\left(\frac{2-3y}{1-y}\right)-3}=\frac{2-3y-2+2y}{2-3y-3+3y}$

$\Rightarrow \frac{-y}{-1}=y$

$\therefore f$ is onto

Given, $f:N\to N$ defined $f(x)=\left(\begin{array}{cc}\hfill \frac{x+1}{2},if\hfill & \hfill xisodd\hfill \\ \hfill \frac{x}{2},if\hfill & \hfill xiseven\hfill \end{array}\right)$ $\cup x\in N$

Let $x_1=1$ and $x_2=2\in N,$

$f(x_1)=f(x_2)\Rightarrow f(1)=f(2)$

$\Rightarrow \frac{1+1}{2}=\frac{2}{2}$

$\Rightarrow 1=1$ but $1\ne 2$

So, $f$ is not one-one

For $x=$ odd and $x\in N$ , say $x=2C+1$ where $C\in N$

There exist $(4C+1)$ $\in N$ such that

$(4C+1)=\frac{4C+1+1}{2}=2C+1.\in N$

And for $x=$ even $\in N$ , say $x=2C$ where $C\in N$

There exist $(4C)\in N$ such that

$f(4C)=\frac{4C}{2}=2C.\in N$

So, $f$ is onto

But, $f$ is not bijective

Given, $f:A*B\to B*A$ defined as $f(a,b)=(b,a)$

Let $(a_1,b_1),(a_2,b_2)\in A*B$ such that

$f(a_1,b_1)=f(a_2,b_2)$

$\Rightarrow (b_1,a_1)=(b_2,a_2)$

So, $b_1=b_2$ and $a_1=a_2$

$\Rightarrow (a_1,b_1)=(a_2,b_2)$

$\therefore f$ is one-one

For $(a,b)\in B*A$

There exist $(a,b)\in A*B$ such that $f(a,b)=(b,a)$

$\therefore f$ is onto

Hence, $f$ is bijective

(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 number which is not equal to co-domain $R$

So, $f$ is not onto

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