Maths

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

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, the given relation is a fxⁿ.

So, domain = {2, 4, 6, 8, 10, 12, 14}

range = {1,2,3,4,5,6,7}

(iii) Domain of the given relation = {1,2}

As element 1 has more than one image i.e., 3 and 5, the given relation is not a fxⁿ.

(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

49.

Let P(3, 9) and Q (-1, 2) be the point. Let M bisects PQ at M so,

Co-ordinate of M = $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$=\left(\frac{3+(-1)}{2},\frac{4+2}{2}\right)$

$=\left(\frac{3-1}{2},\frac{6}{2}\right)=\left(\frac{2}{2},\frac{6}{2}\right)=(1,3)$

Now, slope of AB, m = $\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}.$

As the bisects AB perpendicular it has slope $-\frac{1}{m}$ and it passes through M(1, 3) it has the equation of the form,

$\frac{-1}{m}=\frac{y-y{}_{\mathrm{\sigma }}}{x-x{}_0}$

$\frac{-1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{y-3}{x-1}$

$\Rightarrow -2=\frac{y-3}{x-1}$

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

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

$\Rightarrow 2x+y-3-2=0$

2x + y - 5 = 0

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

48.

Given, slope of line 1, m₁ = 2.

Let m₂ be the slope of line 2.

If θ is the angle between the two lines then we can write,

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