5 Marks Questions - Page 2 - MATHS STD 12 Science Questions - Vidyadip

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5 Marks Questions

Question 515 Marks

Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by $f(\text{x})=\Big(\frac{\text{x}-2}{\text{x}-3}\Big).$ Is f one-one and onto? Justify your answer.

Answer

A = R - {3}, B = R - {1}
f: A → B is defined as $f(\text{x})=\Big(\frac{\text{x}-2}{\text{x}-3}\Big)$
Let $\text{x},\text{y}\in\text{A}$ such that f(x) = f(y)
$\Rightarrow\frac{\text{x}-2}{\text{x}-3}=\frac{\text{y}-2}{\text{y}-3}$
⇒ (x - 2)(y - 3) = (y - 2)(x - 3)
⇒ xy - 3x - 2y + 6 = xy - 3y - 2x + 6
⇒ -3x - 2y = -3y - 2x
⇒ 3x - 2x = 3y - 2y
⇒ x = y
$\therefore$ f is one-one.
Let $\text{y}\in\text{B}=\text{R}-\{1\}.$ Then, $\text{y}\neq1.$
The function is onto if there exists $\text{x}\in\text{A}$ such that f(x) = y.
Now,
f(x) = y
$\Rightarrow\frac{\text{x}-2}{\text{x}-3}=\text{y}$
⇒ x - 2 = xy - 3y
⇒ x(1 - y) = -3y + 2
$\Rightarrow\text{x}=\frac{2-3\text{y}}{1-\text{y}}\in\text{A}\ \ \ \ \ [\text{y}\neq1]$
Thus, for any $\text{y}\in\text{B},$ there exists $\frac{2-3\text{y}}{1-\text{y}}\in\text{A}$ such nthat
$f\Big(\frac{2-3\text{y}}{1-\text{y}}\Big)=\frac{\Big(\frac{2-3\text{y}}{1-\text{y}}\Big)-2}{\Big(\frac{2-3\text{y}}{1-\text{y}}\Big)-3}=\frac{2-3\text{y}-2+2\text{y}}{2-3\text{y}-3+3\text{y}}=\frac{-\text{y}}{-1}=\text{y}$
$\therefore$ f is onto.
Hence, function f is one-one and onto.

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Question 525 Marks

Let f: N → N be defined by ${f(n)}=\begin{cases}\frac{\text{n}+1}{2},\text{ if n is odd}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for all}\text{ n }\in\text{ N}.\\\frac{\text{n}}{2}\ \ \ \ ,\text{if n is even}\end{cases}$ State whether the function f is bijective. Justify your answer.

Answer

${f(n)}=\begin{cases}\frac{\text{n}+1}{2},\text{ if n is odd}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for all}\text{ n }\in\text{ N}.\\\frac{\text{n}}{2}\ \ \ \ ,\text{if n is even}\end{cases}$
f: N → N is defined as
It can be observed that:
$\therefore{f}(1)={f}(2),\ \text{where }1\neq2.$
$\therefore$ f is not one-one.
Consider a natural number (n) in co-domain N.
Case I: n is odd
$\therefore$ n = 2r + 1 for some $\text{r}\in\text{N}.$ Then, there exists $4\text{r}+1\in\text{N}$ such that
$f(4\text{r}+1)=\frac{4\text{r}+1+1}{2}=2\text{r}+1$
Case II: n is even
$\therefore$ n = 2r for some $\text{r}\in\text{N}.$ Then, there exists $4\text{r}\in\text{N}$ such that $f(4\text{r})=\frac{4\text{r}}{2}=2\text{r}$
$\therefore$ f is onto.
Hence, f is not a bijective function.

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Question 535 Marks

Let $f: W \rightarrow W$ be defined as $f(n) = n – 1,$ if n is odd and $f (n) = n + 1,$ if $n$ is even. Show that $f$ is invertible. Find the inverse of $f.$ Here, $W$ is the set of all whole numbers.

Answer

It is given that: $f: W \rightarrow W$ is defined as
$f(\text{n})=\begin{cases}\text{n}-1,\ \text{if n is odd}\\\text{n}+1,\ \text{if n is even}\end{cases}$ One$-$one:
Let $f(n) = f(m).$ It can be observed that if $n$ is odd and $m$ is even,
then we will have $n - 1 = m + 1.$
$ \Rightarrow n - m = 2$
However, this is impossible. Similarly, the possibility of $n$ being even and $m$ being odd can also be ignored under a similar argument.
$\therefore$ Both $n$ and $m$ must be either odd or even.
Now, if both $n$ and $m$ are odd, then we have: $f(n) = f(m)$
$ \Rightarrow n - 1 = m - 1$
$ \Rightarrow n = m$ Again, if both $n$ and $m$ are even,
then we have: $f(n) = f(m) $
$\Rightarrow n + 1 = m + 1$
$ \Rightarrow n = m$
$\therefore f $is one$-$one. It is clear that any odd number $2r + 1$ in co$-$domain $N$ is the image of $2r$ in domain $N$ and any even number $2r$ in co$-$domain $N$ is the image of $2r + 1$ in domain$ N.$
$\therefore f$ is onto.
Hence$, f$ is an invertible function.
Let us define $g: W \rightarrow W$ as:
$\text{g(m)}=\begin{cases}\text{m}+1,\ \text{if m is odd}\\\text{m}-1,\ \text{if m is even}\end{cases}$
Now, when $n$ is odd: $gof(n) = g(f(n)) = g(n - 1) = n - 1 + 1 = n$ And,
when n is even: $gof(n) = g(f(n)) = g(n + 1) = n + 1 - 1 = n$ Similarly,
when $m$ is odd: $fog(m)=f(g(m)) = f(m - 1) = m - 1 + 1 = m$
When m is even: $fog(m) = f(g(m)) = f(m + 1) = m + 1 - 1 = m$
$\therefore gof = I_W$ and $fog = I_W$
Thus, $f$ is an invertible and the inverse of $f$ is given by $f^{-1} = g,$
which is the same as $f.$ Hence, the inverse of $f$ is fitself.

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Question 545 Marks

Consider $f: R_{+ }\rightarrow [– 5, \infty )$ given by $f(x) = 9x^2 + 6x – 5.$ Show that $f$ is invertible with $f^{-1}(\mathrm{y})=\left(\frac{(\sqrt{\mathrm{y}+6})^{-1}}{3}\right)$

Answer

SOLUTION Consider $f: \mathrm{R}_{+} \rightarrow[-5, \infty]$ and $\mathrm{f}(\mathrm{x})=9 \mathrm{x}^2+6 \mathrm{x}-5$.
Let $\mathrm{x}_1, \mathrm{x}_2 \in \mathrm{R} \rightarrow[-5, \infty]$,
then $f\left(\mathrm{x}_1\right)=9 \mathrm{x}_1^2+6 \mathrm{x}_1-5$ and $f\left(\mathrm{x}_2\right)=9 \mathrm{x}_2^2+6 \mathrm{x}_2-5$
Now, $f\left(x_1\right)=f\left(x_2\right)$
then $9 x_1^2+6 x_1-5=9 x_2^2+6 x_2-5$
$\Rightarrow 9 \mathrm{x}_1^2+6 \mathrm{x}_1=9 \mathrm{x}_2^2+6 \mathrm{x}_2 $
$\Rightarrow 9\left(\mathrm{x}_1^2-\mathrm{x}_2^2\right)+6\left(\mathrm{x}_1-\mathrm{x}_2\right)=0$
$\Rightarrow\left(x_1-x_2\right)\left[9\left(x_1+x_2\right)+6\right]=0$
$ \Rightarrow x_1-x_2=0 $
$\Rightarrow x_1=x_2 $
$\therefore f$ is one$-$one.
Now, again $y=9 x^2+6 x-5 $
$\Rightarrow 9 x^2+6 x-(5+y)=0$
$\Rightarrow \mathrm{x}=\frac{-6 \pm \sqrt{(6)^2+4 \times 9(5+\mathrm{y})}}{18}$
$=\frac{-6 \pm 6 \sqrt{1+5+\mathrm{y}}}{18}$
$=\frac{-6 \pm 6 \sqrt{\mathrm{y}+6}}{18}$
$=\frac{\sqrt{\mathrm{y}+6}-1}{3}$
$ \therefore f(\mathrm{x})=f\left(\frac{\sqrt{\mathrm{y}+6}-1}{3}\right)$
$=9\left(\frac{\sqrt{\mathrm{y}+6}-1}{3}\right)^2+6\left(\frac{\sqrt{\mathrm{y}+6}-1}{3}\right)-5$
$ =9\left(\frac{\mathrm{y}+6+1-2 \sqrt{\mathrm{y}+6}-1}{9}\right)+2(\sqrt{\mathrm{y}+6}-1)-5$
$ =\mathrm{y}+7-2 \sqrt{\mathrm{y}+6}+2 \sqrt{\mathrm{y}+6}-2-5=\mathrm{y}$
Therefore, $f(x)$ is invertible and $f^{-1}(x)=\frac{\sqrt{y+6}-1}{3}$.

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Question 555 Marks

Let $A = \{– 1, 0, 1, 2\}, B = \{– 4, – 2, 0, 2\}$ and $f, g: A \rightarrow B$ be functions defined by $f(\text{x})=\text{x}^2-\text{x},\ \text{x}\in\text{A}$ and $\text{g(x)}=2\Big|\text{x}-\frac{1}{2}\Big|,\ \text{x}\in\text{A}.$ Are $f$ and $g$ equal? Justify your answer. $($Hint: One may note that two functions $f: A \rightarrow B$ and $g: A \rightarrow B$ such that $f(\text{a}) = \text{g(a)}\ \forall \text{a} \in \text{A},$ are called equal functions$)$.

Answer

It is given that $A = \{-1, 0, 1, 2\}, B =\{-4, -2, 0, 2\}.$
Also, it is given that $f, g: A \rightarrow B$ are defined by
$f(\text{x})=\text{x}^2-\text{x},\ \text{x}\in\text{A}$ and $\text{g(x)}=2\Big|\text{x}-\frac{1}{2}\Big|,\ \text{x}\in\text{A}.$
It is observed that:
$f(-1) = (-1)^2 - (-1) = 1 + 1 = 2$
$\text{g}(-1)=2\Big|(-1)-\frac{1}{2}\Big|-1=2\Big(\frac{3}{2}\Big)-1=3-1=2$
$\Rightarrow f(-1) = g(-1)$
$f(0) = (0)^2 - 0 = 0$
$\text{g}(0)=2\Big|0-\frac{1}{2}\Big|-1=2\Big(\frac{1}{2}\Big)-1=1-1=0$
$\Rightarrow f(0) = g(0)$
$f(1) = (1)^2 - 1 = 1 - 1 = 0$
$\text{g}(1)=2\Big|1-\frac{1}{2}\Big|-1=2\Big(\frac{1}{2}\Big)-1=1-1=0$
$\Rightarrow f(1) = g(1)$
$f(2) = (2)^2 - 2 = 4 - 2 = 2$
$\text{g}(2)=2\Big|2-\frac{1}{2}\Big|-1=2\Big(\frac{3}{2}\Big)-1=3-1=2$
$\therefore f(\text{a})=\text{g(a)}\ \forall\text{a}\in\text{A}$
Hence, the functions $f$ and $g$ are equal.

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Question 565 Marks

Show that the relation R defined by R = {(a, b): a - b is divisible by 3; a, b ∈ Z} is an equivalence relation.

Answer

We observe the following relations of relation R.
Reflexivity: Let a be an arbitrary element of R. Then,
a - a = 0 = 0 × 3
⇒ a - a is divisible by 3
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{Z}$
So, R is reflexive on Z.
Symmetry: Let $(\text{a, b})\in\text{R}$
⇒ a - b is divisible by 3
⇒ a - b = 3p for some $\text{p}\in\text{Z}$
⇒ b - a = 3(-p)
Here, $-\text{p}\in\text{Z}$
⇒ b - a is divisible by 3
⇒ (b, a)∈R for all a, $\text{b}\in\text{Z}$
So, R is symmetric on Z.
Transitivity: Let a, b and b, c $\in\text{R}$
⇒ a - b and b - c are divisible by 3
⇒ a - b = 3p for some $\text{p}\in\text{Z}$
and b - c = 3q for some $\text{q}\in\text{Z}$
Adding the above two, we get
a - b + b - c = 3p + 3q
⇒ a - c = 3(p + q)
Here, p + q $\in\text{Z}$
⇒ a - c is divisible by 3
$ \Rightarrow \text{a, c} \in \text{R}$ for all a, c $\in\text{Z}$
So, R is transitive on Z.
Hence, R is an equivalence relation on Z.

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Question 575 Marks

Show that the exponential function $f : R \rightarrow R,$ given by $f(x) = e^x,$ is one-one but not onto. What happens if the co-domain is replaced by $R + 0R0 + ($set of all positive real numbers$)?$

Answer

Then the co-domain and range become the same and in that case,
f is onto and hence, it is a bijection.

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Question 585 Marks

Consider $f : {1, 2, 3} \rightarrow {a, b, c}$ and $\ce{g : \{a, b, c\} \rightarrow \{apple, ball, cat\}}$ defined as $f(1) = a, f(2) = b, f(3) = c, g(a) =$ apple, $g(b) =$ ball and $g(c) =$ cat. Show that $f, g$ and gof are invertible. Find $f^{-1}, g^{-1}$ and $gof^{-1}$ and show that $(gof)^{-1} = f^{-1}og^{-1}.$

Answer

$\ce{f = \{(1, a), (2, b), (3, c)\}}$ and $\ce{g = \{(a, apple), (b, ball), (c, cat)\}}$
Clearly, f and g are bijections.
So, $f$ and $g$ are invertible.
Now,
$\ce{f^{-1} = \{(a, 1), (b, 2), (c, 3)\}}$ and $\ce{g^{-1} = \{(apple, a), (ball, b), (cat, c)\}}$
So,$\ce{f^{-1}og^{-1} = \{(apple, 1), (ball, 2), (cat, 3)\} .....(1)}$
$\ce{f : \{1, 2, 3\} \rightarrow \{a, b, c\}}$ and $\ce{g : \{a, b, c\} \rightarrow \{apple, ball, cat\}}$
So, $\ce{gof : \{1, 2, 3\} \rightarrow \{apple, ball, cat\}}$
$\ce{\Rightarrow (gof)(1) = g(f(1)) = g(a) = apple}$
$\ce{(gof)(2) = g(f(2)) = g(b) = ball}$
and $\ce{(gof)(3) = g(f(3)) = g(c) = cat}$
$\ce{\therefore gof = \{(1, apple), (2, ball), (3, cat)\}}$
Clearly, gof is a bijection.
So, gof is invertible.
$\ce{(gof)^{-1} = \{(apple, 1), (ball, 2), (cat, 3)\} ......(2)}$
From $(1)$ and $(2)$, we get
$\ce{(gof)^{-1} = f^{-1}og^{-1}}$

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Question 595 Marks

Let $A = R - {3}, B = R - {1}.$ Let $f : A \rightarrow B$ be defined by $\text{f}(\text{x})=\frac{\text{x}-2}{\text{x}-3}\ \forall\ \text{x}\in\text{A}.$ Then show that $f$ is bijective.

Answer

We are given that, $A = R - {3}$ and $B = R - {1}.$ Consider, $\text{f}(\text{x})=\frac{\text{x}-2}{\text{x}-3}\ \forall\ \text{x}\in\text{A}$
Injectivity:
Let $f (x_1) = f (x_2) \Rightarrow\ \frac{\text{x}_1-2}{\text{x}_1-3}=\frac{\text{x}_2-2}{\text{x}_2-3} $
$\Rightarrow (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) $
$\Rightarrow x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6 $
$\Rightarrow -3x_1 - 2x_2 = -3x_2 - 2x_1 $
$\Rightarrow -x_1 = -x_2 $
$\Rightarrow x_1 = x_2$
Hence, $f(x)$ is an injective function.
Surjectivity:
Let $f(x) = y$
$\Rightarrow\ \frac{\text{x}-2}{\text{x}-3}=\text{y}$
$\Rightarrow x - 2 = xy - 3y $
$\Rightarrow x - xy = -3y + 2 $
$\Rightarrow x(1 - y) = 2 - 3y $
$\Rightarrow\ \text{x}=\frac{2-3\text{y}}{1-\text{y}} $
$\Rightarrow\ \text{x}=\frac{3\text{y}-2}{\text{x}-3}\in\text{A},\ \forall\ \text{y}\in\text{B}$
Hence, $f(x)$ is surjective function.
Since, $f(x)$ is both injective and surjective function.
Hence, $f(x)$ is a bijective function.

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Question 605 Marks

Give an example of a function:
Which is not one-one but onto.

Answer

Which is not one-one but onto.
$\text{f}:\text{Z}\rightarrow\text{N}\cup\{0\}$ given by f(x) = |x|
Infectivity: Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
Implies that |x| = |y|
Implies that $\text{x}=\pm\text{y}$
Therefore, different elements of domain f may give the same image.
Therefore, f is not one-one.
Subjectivity: Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
Implies that |x| = y
Implies that $\text{x}=\pm\text{y},$
Which is an element in Z (domain).
Therefore, for every element in the co-domain, there exists a pre-image in the domain.
Thus, f is onto.

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Question 615 Marks

Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Answer

It is given that
R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin},
Now, it is clear that
$(\text{P},\text{P})\in\text{R}$ since the distance of point P from origin is always the same as the distance of the same point P from the origin.
Therefore, R is reflexive.
Now, Let us take $(\text{P},\text{Q})\in\text{R},$
⇒ The distance of point P from origin is always the same as the distance of the same point Q from the origin.
⇒ The distance of point Q from origin is always the same as the distance of the same point P from the origin.
$\Rightarrow(\text{Q},\text{P})\in\text{R}$
Therefore, R is symmetric.
Now, Let $(\text{P},\text{Q}),(\text{Q},\text{S})\in\text{R}$
⇒ The distance of point P and Q from the origin is always the same as the distance of the same point Q and S
from the origin.
⇒ The distance of points P and S from the origin is the same.
$\Rightarrow(\text{P},\text{S})\in\text{R}$
Therefore, R is transitive.
Therefore, R is equivalence relation.
The set of all points related to $\text{P}\neq(0,0)$ will be those the set of all points related to $\text{P}\neq(0,0)$ will be those points whose distance from the origin is the same as the distance of point P from the origin.
So, if O(0,0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.
Therefore, this set of points forms a circle with the centre as the origin and this circle passes through point P.

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Question 625 Marks

Show that the relation $R,$ defined on the set $A$ of all polygons as $R = (P_1, P_2): P_1$ and $P_2$ have same number of sides is an equivalence relation. What is the set of all elements in $A$ related to the right angle triangle $T$ with sides $3, 4$ and $5$?

Answer

We observe the following properties on $R.$
Reflexivity: Consider $P_1$ be an arbitrary element of $A.$
Then, polygon $P_1$ and $P_1$ have the same number of sides.
Since they are one and the same.
Implies that $\text{P}_1, \text{P}_1\in\text{R}$ for all $\text{P}_1\in\text{A.}$
So,$R$ is reflexive on $A.$
Symmetry: Consider $\text{P}_1,\text{ P}_2\in\text{R}$
Implies that $P_1$ and $P_2$ have the same number of sides.
Implies that $P_2$ and $P_1$ have the same number of sides.
Implies that $\text{P}_2,\text{ P}_1\in\text{R}$ for all $\text{P}_1,\text{ P}_2\in\text{A}$
So, $R$ is symmetric on $A.$
Transitivity: Consider $\text{P}_1, \text{P}_2, \text{P}_3\in\text{R}$
Implies that $P_1$ and $P_2$ have the same number of sides and $P_2$ and $P_3$ have the same number of sides
Implies that $P_1, P_2$ and $P_3$ have the same number of sides.
Implies that $P_1$ and $P_3$ have the same number of sides.
Implies that $\text{P}_1,\text{ P}_3\in\text{R}$ for all $\text{P}_1,\text{ P}_3\in\text{A.}$
So, $R$ is transitive on $A.$
Hence, $R$ is an equivalence relation on the set $A.$
Also, the set of all the triangles $\in\text{A}$ is related to the right angle triangle $T$ with the sides $3, 4, 5.$

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Question 635 Marks

Let $Z$ be the set of all integers and $Z_0$ be the set of all non$-$zero integers. Let a relation $R$ on $Z \times Z_0$ be defined as $(a, b)R(c, d) \leftrightarrow ad = bc$ for all $(a, b), (c, d) \in Z \times Z_0,$ Prove that $R$ is an equivalence relation on $Z \times Z_0.$

Answer

We have, $Z$ be the set of integers and $Z_0$ be the set of non$-$zero integers.
$R = \{(a, b), (c, d): ad = bc\}$ be a relation on $Z \times Z_0$
Now,
Reflexivity: $(a, b) \in Z \times Z_0$
$\Rightarrow ab = ba$
$\Rightarrow \{(a, b), (a, b)\} \in R$
$\Rightarrow R$ is reflexive.
Symmetric: Let $\{(a, b), (c, d)\} \in R$
$\Rightarrow ad = bc$
$\Rightarrow cd = da$
$\Rightarrow \{(c, d), (a, b)\} \in R$
$\Rightarrow R$ is symmetric.
Transitive: Let $(a, b), (c, d) \in R$ and $(c, d), (e, f) \in R$
$\Rightarrow ad = bc$ and $cf = de$
$\Rightarrow\ \frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}$ and $\frac{\text{c}}{\text{d}}=\frac{\text{e}}{\text{f}}$
$\Rightarrow\ \frac{\text{a}}{\text{b}}=\frac{\text{e}}{\text{f}}$
$\Rightarrow\ \text{af}=\text{be}$
$\Rightarrow\ (\text{a, b})(\text{e, f})\in\text{R}$
$\Rightarrow R$ is transitive.
Hence$, R$ is an equivalence relation on $Z \times Z_0$

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Question 645 Marks

Let f : N → N be defined by $\text{f(n)}=\begin{cases}\text{n}+1,&\text{if n is odd}\\\text{n}-1,&\text{if n is even}\end{cases}$ Show that f is a bijection.

Answer

We have, $\text{f(n)}=\begin{cases}\text{n}+1,&\text{if n is odd}\\\text{n}-1,&\text{if n is even}\end{cases}$Injection test:
Case I: If n is odd,
Let $\text{x, y}\in\text{N}$ such that f(x) = f(y) As, f(x) = f(y) ⇒ x + 1 = y + 1 ⇒ x = yCase II: If n is even,
Let $\text{x, y}\in\text{N}$ such that f(x) = f(y) As, f(x) = f(y) ⇒ x - 1 = y - 1 ⇒ x = y So, f is injective. Surjection test: Case I: If n is odd, As, for every $\text{n}\in\text{N},$ there exists y = n - 1 in N such that f(y) = f(n - 1) = n - 1 + 1 = nCase II: If n is even,
As, for every $\text{n}\in\text{N},$ there exists y = n + 1 in N such that f(y) = f(n + 1) = n + 1 - 1 = n So, f is surjective. So, f is bijection.

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Question 655 Marks

Let R be the relation defined on the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b): both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each other and all the elements of the subset {2, 4, 6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.

Answer

We observe the following properties of R.
Reflexivity: Let a be an arbitrary element of R. Then,
$\text{a}\in\text{R}$
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{A}$
So, R is reflexive on A.
Symmetry: Let $(\text{a, b})\in\text{R}$
⇒ Both a and b are either even or odd.
⇒ Both b and a are either even or odd.
$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{A}$
So, R is symmetric on A.
Transitivity: Let (a, b) and (b, c) $\in\text{R}$
⇒ Both a and b are either even or odd and both b and c are either even or odd.
⇒ a, b and c are either even or odd.
⇒ a and c both are either even or odd.
$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a, c}\in\text{A}$
So, R is transitive on A.
Thus, R is an equivalence relation on A.
We observe that all the elements of the subset {1, 3, 5, 7} are odd. Thus, they are related to each other.
This is because the relation R on A is an equivalence relation.
Similarly, the elements of the subset {2, 4, 6} are even. Thus, they are related to each other because every element is even.
Hence proved.

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Question 665 Marks

Let $C$ be the set of all complex numbers and $C_{0 }$ be the set of all no$-$zero complex numbers. Let a relation $R$ on $C_{0 }$ be defined as $\text{z}_1\text{R z}_2\Leftrightarrow\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}$ is real for all $\text{z}_1,\ \text{z}_2\in\text{C}_0.$ Show that $R$ is an equivalence relation.

Answer

Test for reflexivity: Since, $\frac{\text{z}_1-\text{z}_1}{\text{z}_1+\text{z}_1}=0,$ which is a real number.

So, $(\text{z}_1,\text{ z}_1)\in\text{R}$
Hence, $R$ is relexive relation.

Test for symmetric: Let $(\text{z}_1,\text{ z}_2)\in\text{R.}$

Then, $\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}=\text{x,}$ where $x$ is real
$\Rightarrow\ -\Big(\frac{\text{z}_1-\text{ z}_2}{\text{z}_1+\text{ z}_2}\Big)=-\text{x}$
$\Rightarrow\ \Big(\frac{\text{z}_2-\text{ z}_1}{\text{z}_2+\text{ z}_1}\Big)=-\text{x},$ is also a real number
So, $(\text{z}_2,\text{ z}_1)\in\text{R}$
Hence, $R$ is symmetric relation.

Test for transivity: Let $(\text{z}_1,\text{ z}_2)\in\text{R}$ and $(\text{z}_2,\text{ z}_3)\in\text{R}.$

Then,
$\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}=\text{x},$ where $x$ is a real number.
$\Rightarrow\ \text{z}_1-\text{z}_2=\text{xz}_1+\text{xz}_2$
$\Rightarrow\ \text{z}_1-\text{xz}_1=\text{z}_2+\text{xz}_2$
$\Rightarrow\ \text{z}_1(1-\text{x})=\text{z}_2(1+\text{x})$
$\Rightarrow\ \frac{\text{z}_1}{\text{z}_2}=\frac{(1+\text{x})}{(1-\text{x})}\ .....(1)$
Also,
$\frac{\text{z}_2-\text{z}_3}{\text{z}_2+\text{z}_3}=\text{y},$ where $y$ is a real number.
$\Rightarrow\ \text{z}_2-\text{z}_3=\text{yz}_2+\text{yz}_3$
$\Rightarrow\ \text{z}_2-\text{yz}_2=\text{z}_3+\text{yz}_3$
$\Rightarrow\ \text{z}_2(1-\text{y})=\text{z}_3(1+\text{y})$
$\Rightarrow\ \frac{\text{z}_2}{\text{z}_3}=\frac{(1+\text{y})}{(1-\text{y})}\ .....(2)$
Dividing $(1)$ and $(2),$ we get
$\frac{\text{z}_1}{\text{z}_3}=\Big(\frac{1+\text{x}}{1-\text{x}}\Big)\times\Big(\frac{1-\text{y}}{1+\text{y}}\Big)=\text{z,}$ where $z$ is a real number.
$\Rightarrow\ \frac{\text{z}_1-\text{z}_3}{\text{z}_1+\text{z}_3}=\frac{\text{z}-1}{\text{z}+1},$ which is real
$\Rightarrow\ (\text{z}_1,\text{ z}_3)\in\text{R}$
Hence, $R$ is transitive relation.
From $(i), (ii)$ and $(iii), R$ is an equivalenve relation.

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