NCERT Solutions: Class 12 Maths Chapter 1 - Relations and Functions

(i) $R = \{(1,3),(2,6),(3,9),(4,12)\}$. $(1,1)\notin R$ so not reflexive; $(1,3)\in R$ but $(3,1)\notin R$ so not symmetric; $(1,3),(3,9)\in R$ but $(1,9)\notin R$ so not transitive.
(ii) $R=\{(1,6),(2,7),(3,8)\}$. $(1,1)\notin R$ (not reflexive); $(1,6)\in R$ but $(6,1)\notin R$ (not symmetric); no pair has a first element equal to another pair's second element, so transitivity is satisfied vacuously.
(iii) $x$ divides $x$, so reflexive; $(1,2)\in R$ but $(2,1)\notin R$, not symmetric; if $x\mid y$ and $y\mid z$ then $x\mid z$, so transitive.
(iv) $x-x=0$ is an integer (reflexive); $x-y$ integer $\Rightarrow y-x$ integer (symmetric); $x-y$ and $y-z$ integers $\Rightarrow x-z$ integer (transitive). Hence equivalence.
(v) (a),(b): 'same place'/'same locality' are reflexive, symmetric and transitive, hence equivalence relations. (c) $x$ is not 7 cm taller than himself (not reflexive); if $x$ is 7 cm taller than $y$ then $y$ is not 7 cm taller than $x$ (not symmetric); $x$ 7 cm taller than $y$ and $y$ 7 cm taller than $z$ gives $x$ 14 cm taller than $z$ (not transitive). (d) $x$ is not her own wife (not reflexive); if $x$ is wife of $y$, $y$ is not wife of $x$ (not symmetric); no $y$ is simultaneously a wife and has a wife, so transitivity holds vacuously. (e) $x$ is not his own father (not reflexive); not symmetric; if $x$ is father of $y$ and $y$ is father of $z$, then $x$ is the grandfather, not father, of $z$ (not transitive).

Reflexive: take $a = \dfrac12$. Then $a \le a^2$ means $\dfrac12 \le \dfrac14$, which is false, so $\left(\dfrac12, \dfrac12\right)\notin R$; $R$ is not reflexive.
Symmetric: $(1, 2)\in R$ since $1 \le 2^2 = 4$, but $(2, 1)\notin R$ since $2 \le 1^2 = 1$ is false; not symmetric.
Transitive: $(3, 2)\in R$ since $3 \le 4$, and $(2, 1.5)\in R$ since $2 \le (1.5)^2 = 2.25$, but $(3, 1.5)\notin R$ since $3 \le 2.25$ is false; not transitive.

$R = \{(1,2),(2,3),(3,4),(4,5),(5,6)\}$. $(1,1)\notin R$, so not reflexive. $(1,2)\in R$ but $(2,1)\notin R$, so not symmetric. $(1,2),(2,3)\in R$ but $(1,3)\notin R$, so not transitive.

Reflexive: $a \le a$ holds for every real $a$, so $(a,a)\in R$. Transitive: if $a \le b$ and $b \le c$ then $a \le c$, so $(a,b),(b,c)\in R \Rightarrow (a,c)\in R$. Not symmetric: $(1,2)\in R$ since $1 \le 2$, but $(2,1)\notin R$ since $2 \le 1$ is false.

Reflexive: for $a = \dfrac12$, $a \le a^3$ means $\dfrac12 \le \dfrac18$, which is false; not reflexive.
Symmetric: $(1, 2)\in R$ since $1 \le 2^3 = 8$, but $(2, 1)\notin R$ since $2 \le 1^3 = 1$ is false; not symmetric.
Transitive: $(10, 3)\in R$ since $10 \le 3^3 = 27$, and $(3, 1.5)\in R$ since $3 \le (1.5)^3 = 3.375$, but $(10, 1.5)\notin R$ since $10 \le 3.375$ is false; not transitive.

Symmetric: the only pairs are $(1,2)$ and $(2,1)$, and each one's reverse is also present, so $R$ is symmetric. Not reflexive: $(1,1)\notin R$. Not transitive: $(1,2),(2,1)\in R$ but $(1,1)\notin R$.

Reflexive: a book $x$ has the same number of pages as itself, so $(x,x)\in R$. Symmetric: if $x$ has the same number of pages as $y$, then $y$ has the same number of pages as $x$, so $(x,y)\in R \Rightarrow (y,x)\in R$. Transitive: if $x$ and $y$ have the same number of pages and $y$ and $z$ have the same number of pages, then $x$ and $z$ have the same number of pages, so $(x,y),(y,z)\in R \Rightarrow (x,z)\in R$. Hence $R$ is an equivalence relation.

Reflexive: $|a-a| = 0$ is even, so $(a,a)\in R$. Symmetric: $|a-b| = |b-a|$, so $(a,b)\in R \Rightarrow (b,a)\in R$. Transitive: $|a-b|$ even means $a,b$ have the same parity; if also $|b-c|$ is even then $b,c$ have the same parity, so $a,c$ have the same parity and $|a-c|$ is even. Hence $R$ is an equivalence relation.
The elements $1, 3, 5$ are all odd, so the difference of any two is even — they are mutually related. The elements $2, 4$ are even, so they are related. An odd number and an even number differ by an odd amount, so no element of $\{1,3,5\}$ is related to any element of $\{2,4\}$.

(i) Reflexive: $|a-a| = 0 = 4\cdot 0$ is a multiple of 4. Symmetric: $|a-b| = |b-a|$. Transitive: if $|a-b| = 4m$ and $|b-c| = 4n$, then $a-c$ is a sum/difference of these, so $|a-c|$ is a multiple of 4. Hence equivalence. Elements related to 1 satisfy $|a-1|$ is a multiple of 4 with $0\le a\le 12$: $a \in \{1, 5, 9\}$.
(ii) $a = b$ is reflexive, symmetric and transitive, so it is an equivalence relation (the identity relation). The only element related to 1 is 1 itself, i.e. $\{1\}$.

(i) On $\{1,2,3\}$, $R=\{(1,2),(2,1)\}$ is symmetric, but $(1,1)\notin R$ (not reflexive) and $(1,2),(2,1)\in R$ while $(1,1)\notin R$ (not transitive).
(ii) On $\mathbb{R}$, $R=\{(a,b):a<b\}$ is transitive, but $a<a$ is false (not reflexive) and $1<2$ while $2<1$ is false (not symmetric).
(iii) On $\{1,2,3\}$, $R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}$ is reflexive and symmetric, but $(1,2),(2,3)\in R$ and $(1,3)\notin R$ (not transitive).
(iv) On $\mathbb{R}$, $R=\{(a,b):a\le b\}$ is reflexive and transitive but not symmetric (Q4).
(v) On $\{1,2,3\}$, $R=\{(1,1),(2,2),(1,2),(2,1)\}$ is symmetric and transitive, but $(3,3)\notin R$ (not reflexive).

Let $d(P)$ denote the distance of $P$ from the origin. Reflexive: $d(P) = d(P)$. Symmetric: $d(P) = d(Q) \Rightarrow d(Q) = d(P)$. Transitive: $d(P) = d(Q)$ and $d(Q) = d(S) \Rightarrow d(P) = d(S)$. Hence equivalence.
For a fixed $P \ne (0,0)$ with $d(P) = r > 0$, the points $Q$ related to $P$ are exactly those with $d(Q) = r$, i.e. all points at distance $r$ from the origin. That set is the circle of radius $r$ centred at the origin, which passes through $P$.

Similarity of triangles is reflexive (every triangle is similar to itself), symmetric (if $T_1\sim T_2$ then $T_2\sim T_1$) and transitive (if $T_1\sim T_2$ and $T_2\sim T_3$ then $T_1\sim T_3$); hence $R$ is an equivalence relation.
The sides of $T_3$ are $6, 8, 10 = 2\times(3, 4, 5)$, the sides of $T_1$, so corresponding sides are proportional and $T_1 \sim T_3$. The ratios $\dfrac{3}{5}, \dfrac{4}{12}, \dfrac{5}{13}$ are unequal, so $T_2$ is not similar to $T_1$ (nor to $T_3$). Thus only $T_1$ and $T_3$ are related.

Reflexive: a polygon has the same number of sides as itself. Symmetric: if $P_1$ and $P_2$ have the same number of sides, so do $P_2$ and $P_1$. Transitive: equal side-counts pass through a middle polygon. Hence equivalence.
The triangle $T$ has 3 sides, so the polygons related to it are exactly those with 3 sides — that is, the set of all triangles.

Taking a line to be parallel to itself, $R$ is reflexive; if $L_1 \parallel L_2$ then $L_2 \parallel L_1$ (symmetric); if $L_1 \parallel L_2$ and $L_2 \parallel L_3$ then $L_1 \parallel L_3$ (transitive). Hence equivalence.
Lines parallel to $y = 2x + 4$ are exactly the lines with the same slope, $2$. These are $y = 2x + c$ for any real constant $c$.

1. i. $R$ is reflexive and symmetric but not transitive.
2. ii. $R$ is reflexive and transitive but not symmetric.
3. iii. $R$ is symmetric and transitive but not reflexive.
4. iv. $R$ is an equivalence relation.

All of $(1,1),(2,2),(3,3),(4,4)$ are in $R$, so $R$ is reflexive. $(1,2)\in R$ but $(2,1)\notin R$, so $R$ is not symmetric. Checking the chains: $(1,3),(3,2)\in R$ gives $(1,2)\in R$; $(1,2),(2,2)$ gives $(1,2)$; $(3,2),(2,2)$ gives $(3,2)$ — every required pair is present, so $R$ is transitive. Hence option (ii).

1. i. $(2, 4) \in R$
2. ii. $(3, 8) \in R$
3. iii. $(6, 8) \in R$
4. iv. $(8, 7) \in R$

A pair $(a,b)$ is in $R$ when $a = b - 2$ and $b > 6$. (2,4): $b=4 \not> 6$. (3,8): $a = 8-2 = 6 \ne 3$. (6,8): $a = 8-2 = 6$ and $b = 8 > 6$ — both conditions hold. (8,7): $a = 7-2 = 5 \ne 8$. Hence $(6,8)\in R$, option (iii).

One-one: $f(x) = f(y) \Rightarrow \dfrac{1}{x} = \dfrac{1}{y} \Rightarrow x = y$. Onto: for any $y \in \mathbb{R}_*$, take $x = \dfrac{1}{y} \in \mathbb{R}_*$; then $f(x) = y$. So $f$ is a bijection on $\mathbb{R}_*$.
With domain $\mathbb{N}$, $g(x) = \dfrac{1}{x}$ is still one-one. But it is not onto: e.g. $\dfrac{2}{3} \in \mathbb{R}_*$ has no preimage, since $\dfrac{1}{n} = \dfrac{2}{3}$ gives $n = \dfrac{3}{2}\notin \mathbb{N}$. Hence the result fails for domain $\mathbb{N}$.

(i) On $\mathbb{N}$, $x^2 = y^2 \Rightarrow x = y$ (positive), so injective; but $2 \in \mathbb{N}$ is not a perfect square, so not surjective.
(ii) On $\mathbb{Z}$, $f(-1) = f(1) = 1$, not injective; negative integers (and $2$) are not attained, not surjective.
(iii) On $\mathbb{R}$, $f(-1) = f(1)$, not injective; negative reals are never values of $x^2$, not surjective.
(iv) On $\mathbb{N}$, $x^3 = y^3 \Rightarrow x = y$, injective; $2$ is not a perfect cube, not surjective.
(v) On $\mathbb{Z}$, $x \mapsto x^3$ is strictly increasing, so injective; $2$ is not a perfect cube, not surjective.

Not one-one: $[1.2] = 1 = [1.5]$, so $f(1.2) = f(1.5)$ although $1.2 \ne 1.5$. Not onto: the values of $[x]$ are integers only, so a non-integer such as $0.5 \in \mathbb{R}$ has no preimage. Hence $f$ is neither one-one nor onto.

Not one-one: $|-1| = 1 = |1|$, so $f(-1) = f(1)$ while $-1 \ne 1$. Not onto: $|x| \ge 0$ for all $x$, so a negative number such as $-1 \in \mathbb{R}$ is not a value of $f$. Hence $f$ is neither one-one nor onto.

Not one-one: $f(1) = 1 = f(2)$ although $1 \ne 2$ (every positive number maps to $1$). Not onto: the range of $f$ is $\{-1, 0, 1\}$, so any other real number, for example $2$, has no preimage. Hence $f$ is neither one-one nor onto.

The images $f(1) = 4$, $f(2) = 5$, $f(3) = 6$ are all distinct. Since distinct elements of $A$ have distinct images in $B$, $f$ is one-one (injective).

(i) One-one: $3 - 4x_1 = 3 - 4x_2 \Rightarrow x_1 = x_2$. Onto: for any $y \in \mathbb{R}$, $x = \dfrac{3 - y}{4} \in \mathbb{R}$ gives $f(x) = y$. Hence bijective.
(ii) Not one-one: $f(1) = 2 = f(-1)$. Not onto: $1 + x^2 \ge 1$, so any value less than $1$ (e.g. $0$) has no preimage. Hence neither one-one nor onto.

One-one: $f(a_1, b_1) = f(a_2, b_2) \Rightarrow (b_1, a_1) = (b_2, a_2) \Rightarrow a_1 = a_2$ and $b_1 = b_2$, so $(a_1,b_1) = (a_2,b_2)$. Onto: any element of $B \times A$ has the form $(b, a)$, and $f(a, b) = (b, a)$, so it has a preimage. Hence $f$ is bijective.

$f(1) = \dfrac{1+1}{2} = 1$ (since $1$ is odd) and $f(2) = \dfrac{2}{2} = 1$ (since $2$ is even). Thus $f(1) = f(2) = 1$ with $1 \ne 2$, so $f$ is not one-one. Therefore $f$ is not bijective. (It is, however, onto, since every $m \in \mathbb{N}$ is attained, e.g. $f(2m) = m$.)

One-one: suppose $\dfrac{x_1 - 2}{x_1 - 3} = \dfrac{x_2 - 2}{x_2 - 3}$. Cross-multiplying, $(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$. Expanding, $x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6$, which gives $-3x_1 - 2x_2 = -3x_2 - 2x_1$, i.e. $-x_1 = -x_2$, so $x_1 = x_2$.
Onto: let $y \in B$, so $y \ne 1$. Solving $y = \dfrac{x-2}{x-3}$ gives $y(x-3) = x-2$, so $x(y-1) = 3y - 2$ and $x = \dfrac{3y - 2}{y - 1}$, which is defined for $y \ne 1$. Also $x \ne 3$, because $x = 3$ would force $3y - 2 = 3y - 3$, i.e. $-2 = -3$, impossible. So every $y \in B$ has a preimage $x \in A$. Hence $f$ is one-one and onto.

1. i. $f$ is one-one onto
2. ii. $f$ is many-one onto
3. iii. $f$ is one-one but not onto
4. iv. $f$ is neither one-one nor onto

$f(1) = 1 = f(-1)$ with $1 \ne -1$, so $f$ is many-one (not one-one). Also $x^4 \ge 0$ for all $x$, so negative reals are not attained and $f$ is not onto. Hence $f$ is neither one-one nor onto, option (iv).

1. i. $f$ is one-one onto
2. ii. $f$ is many-one onto
3. iii. $f$ is one-one but not onto
4. iv. $f$ is neither one-one nor onto

One-one: $3x_1 = 3x_2 \Rightarrow x_1 = x_2$. Onto: for any $y \in \mathbb{R}$, $x = \dfrac{y}{3} \in \mathbb{R}$ gives $f(x) = y$. Hence $f$ is one-one and onto, option (i).

For $x \ge 0$, $f(x) = \dfrac{x}{1+x}$, so $f$ is increasing and has values in $[0,1)$. For $x \lt 0$, $f(x) = \dfrac{x}{1-x}$, so $f$ is increasing and has values in $(-1,0)$. Hence two unequal inputs on the same side of $0$ cannot have the same image, and inputs on opposite sides of $0$ have images of opposite signs. Also $f(0)=0$, so $f$ is one-one.
To prove onto, take any $y \in \{x \in \mathbb{R}: -1 \lt x \lt 1\}$. If $0 \le y \lt 1$, put $x=\dfrac{y}{1-y}$. Then $x \ge 0$ and $f(x)=\dfrac{x}{1+x}=y$. If $-1 \lt y \lt 0$, put $x=\dfrac{y}{1+y}$. Then $x \lt 0$ and $f(x)=\dfrac{x}{1-x}=y$. Thus every element of the codomain has a preimage, so $f$ is onto. Therefore $f$ is one-one and onto.

Let $x_1, x_2 \in \mathbb{R}$ and suppose $f(x_1)=f(x_2)$. Then $x_1^3=x_2^3$, so $x_1^3-x_2^3=0$. Hence $(x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0$. If $x_1 \ne x_2$, the second factor is positive, which is impossible. Therefore $x_1=x_2$, and $f$ is injective.

For an equivalence relation, the relation must be reflexive, symmetric and transitive. Since $X$ is non-empty, choose an element $a \in X$. Then $\{a\} \subset X$, so $\{a\}RX$. But $X \subset \{a\}$ is false unless $X=\{a\}$, and in general the subset relation is not symmetric; for example $\varnothing \subset X$ but $X \subset \varnothing$ is false. Hence $R$ is not symmetric, so it is not an equivalence relation.

The domain and codomain are the same finite set with $n$ elements. Any onto function from this set to itself must also be one-one, because all $n$ elements of the codomain are already hit by the $n$ elements of the domain. Thus the onto functions are exactly the permutations of the set. The number of such functions is $n!$.

Compare the values at every element of $A$. For $f(x)=x^2-x$: $f(-1)=2$, $f(0)=0$, $f(1)=0$ and $f(2)=2$. For $g(x)=2\left|x-\dfrac{1}{2}\right|-1$: $g(-1)=2\left| -\dfrac{3}{2}\right|-1=2$, $g(0)=2\left| -\dfrac{1}{2}\right|-1=0$, $g(1)=2\left|\dfrac{1}{2}\right|-1=0$ and $g(2)=2\left|\dfrac{3}{2}\right|-1=2$. Thus $f(a)=g(a)$ for every $a \in A$. Therefore $f$ and $g$ are equal.

1. i. $1$
2. ii. $2$
3. iii. $3$
4. iv. $4$

A reflexive relation must contain $(1,1),(2,2),(3,3)$. Since it contains $(1,2)$ and $(1,3)$ and is symmetric, it must also contain $(2,1)$ and $(3,1)$. The only remaining possible symmetric pair is $(2,3)$ together with $(3,2)$. If this pair is included, the relation becomes the universal relation on $A$, which is transitive. If this pair is not included, the relation is not transitive because $(2,1)$ and $(1,3)$ are present but $(2,3)$ is absent. Hence exactly one relation satisfies the conditions, option (i).

1. i. $1$
2. ii. $2$
3. iii. $3$
4. iv. $4$

An equivalence relation on $A$ is determined by a partition of $A$. Since $(1,2)$ must be in the relation, $1$ and $2$ must lie in the same equivalence class. There are two possible partitions: $\{\{1,2\},\{3\}\}$ and $\{\{1,2,3\}\}$. Therefore there are exactly two such equivalence relations, option (ii).