NCERT Exemplar Solutions Class 12 Maths Relations and Functions — Long Answer Q16–27

NCERT Exemplar Problems, Class 12 Mathematics — Relations and Functions, Long Answer questions 16 to 27. All 12 questions solved.

Twelve longer questions. Four ask you to test or construct relations with given properties (16, 17, 22, 23); four ask for examples or proofs about one-one and onto maps (18 to 21, 24); one is a set of composites (25); and the last two are on binary operations (26, 27).

A note on the syllabus. The rationalised NCERT textbook has dropped binary operations, and neither the current JEE Main nor JEE Advanced syllabus lists them. Questions 26 and 27 are solved for completeness; each needs only the two definitions stated where they are used. Everything else on this page is inside the current textbook.

Key insight. To show a relation has a property, argue for every element; to show it lacks one, a single counterexample is enough. For each relation, test the three properties separately and look for the smallest counterexample first — it is usually two or three elements.

Question 16

If $A = \{1, 2, 3, 4\}$, define relations on $A$ which have properties of being:

(a) reflexive, transitive but not symmetric

(b) symmetric but neither reflexive nor transitive

(c) reflexive, symmetric and transitive.

Solution. Many answers are possible; here is the smallest of each kind.

(a) Start with the four pairs $(1,1), (2,2), (3,3), (4,4)$ that reflexivity needs, and add one non-symmetric pair:

$$R_1 = \{(1,1), (2,2), (3,3), (4,4), (1,2)\}.$$

It is reflexive. It is transitive: the only chains are $(1,1),(1,2)$ and $(1,2),(2,2)$, both giving $(1,2) \in R_1$. It is not symmetric: $(1,2) \in R_1$ but $(2,1) \notin R_1$.

(b) $R_2 = \{(1, 2), (2, 1)\}$. It is symmetric. It is not reflexive ($(1,1) \notin R_2$). It is not transitive: $(1,2)$ and $(2,1)$ are in $R_2$, but $(1,1)$ is not.

(c) The identity relation $R_3 = \{(1,1), (2,2), (3,3), (4,4)\}$ is reflexive, symmetric and transitive. (The universal relation $A \times A$ also works.)

Question 17

Let $R$ be relation defined on the set of natural number $\mathbb{N}$ as follows: $R = \{(x, y) : x \in \mathbb{N}, y \in \mathbb{N}, 2x + y = 41\}$. Find the domain and range of the relation $R$. Also verify whether $R$ is reflexive, symmetric and transitive.

Solution. $y = 41 – 2x$ must be a natural number, so $41 – 2x \ge 1$, giving $x \le 20$. As $x$ runs through $1, 2, \ldots, 20$, $y$ runs through $39, 37, \ldots, 1$:

$$R = \{(1, 39), (2, 37), (3, 35), \ldots, (20, 1)\}.$$

Domain $= \{1, 2, 3, \ldots, 20\}$; range $= \{1, 3, 5, \ldots, 39\}$.

• Not reflexive: $(1, 1) \notin R$, since $2 + 1 \ne 41$.
• Not symmetric: $(1, 39) \in R$, but $(39, 1) \notin R$, since $78 + 1 \ne 41$.
• Not transitive: $(20, 1) \in R$ and $(1, 39) \in R$, but $(20, 39) \notin R$, since $40 + 39 \ne 41$.

Question 18

Given $A = \{2, 3, 4\}$, $B = \{2, 5, 6, 7\}$. Construct an example of each of the following:

(a) an injective mapping from $A$ to $B$

(b) a mapping from $A$ to $B$ which is not injective

(c) a mapping from $B$ to $A$.

Solution. A mapping from $A$ must send each of $2, 3, 4$ to exactly one element of $B$.

(a) Send them to different elements: $f = \{(2, 2), (3, 5), (4, 6)\}$. No two inputs share an image, so $f$ is injective.

(b) Let two inputs share an image: $g = \{(2, 2), (3, 2), (4, 5)\}$. Here $g(2) = g(3)$, so $g$ is not injective.

(c) Each of $2, 5, 6, 7$ needs one image in $A$: $h = \{(2, 2), (5, 3), (6, 4), (7, 4)\}$. (Since $B$ has four elements and $A$ three, no mapping from $B$ to $A$ can be injective.)

Question 19

Give an example of a map

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which is neither one-one nor onto.

Solution. (i) $f : \mathbb{N} \to \mathbb{N}$, $f(x) = 2x$. If $2x_1 = 2x_2$ then $x_1 = x_2$, so it is one-one; odd numbers are never reached, so it is not onto.

(ii) $g : \mathbb{N} \to \mathbb{N}$, $g(n) = \dfrac{n+1}{2}$ for odd $n$ and $g(n) = \dfrac{n}{2}$ for even $n$. Then $g(1) = g(2) = 1$, so it is not one-one; any $m \in \mathbb{N}$ is $g(2m)$, so it is onto.

(iii) $h : \mathbb{R} \to \mathbb{R}$, $h(x) = x^2$. $h(-1) = h(1)$, so it is not one-one; $-1$ is never reached, so it is not onto.

Question 20

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

Solution. One-one. Suppose $f(x_1) = f(x_2)$:

$$\frac{x_1 – 2}{x_1 – 3} = \frac{x_2 – 2}{x_2 – 3} \;\Rightarrow\; (x_1 – 2)(x_2 – 3) = (x_2 – 2)(x_1 – 3).$$

Expanding, $x_1x_2 – 3x_1 – 2x_2 + 6 = x_1x_2 – 3x_2 – 2x_1 + 6$, so $-x_1 = -x_2$ and $x_1 = x_2$.

Onto. Take any $y \in B$, so $y \ne 1$. Solve $\dfrac{x-2}{x-3} = y$: $x – 2 = xy – 3y$, so $x(1 – y) = 2 – 3y$ and

$$x = \frac{2 – 3y}{1 – y} = \frac{3y – 2}{y – 1}.$$

This is defined because $y \ne 1$, and $x \ne 3$ (if $\frac{3y-2}{y-1} = 3$ then $3y – 2 = 3y – 3$, impossible). So $x \in A$ and $f(x) = y$. Every $y \in B$ has a pre-image.

That is why the question removes $1$ from $B$: $1$ is the one value $f$ never takes.

Question 21

Let $A = [-1, 1]$. Then, discuss whether the following functions defined on $A$ are one-one, onto or bijective:

(i) $f(x) = \dfrac{x}{2}$ (ii) $g(x) = |x|$ (iii) $h(x) = x|x|$ (iv) $k(x) = x^2$.

Solution. Each function is from $A$ to $A$.

(i) $\frac{x_1}{2} = \frac{x_2}{2} \Rightarrow x_1 = x_2$: one-one. But the outputs lie in $[-\frac12, \frac12]$, so $1 \in A$ is never reached: not onto.

(ii) $g(-1) = g(1)$: not one-one. Outputs are never negative, so $-1$ is not reached: not onto.

(iii) $h(x) = x^2$ for $x \ge 0$ and $-x^2$ for $x < 0$. It is strictly increasing on $[-1, 1]$, so one-one. Every $y \in [0, 1]$ is $h(\sqrt y)$ and every $y \in [-1, 0)$ is $h(-\sqrt{-y})$, so it is onto. Hence bijective.

(iv) $k(-1) = k(1)$: not one-one. Outputs are never negative: not onto.

Question 22

Each of the following defines a relation on $\mathbb{N}$:

(i) $x$ is greater than $y$, $x, y \in \mathbb{N}$

(ii) $x + y = 10$, $x, y \in \mathbb{N}$

(iii) $xy$ is square of an integer, $x, y \in \mathbb{N}$

(iv) $x + 4y = 10$, $x, y \in \mathbb{N}$.

Determine which of the above relations are reflexive, symmetric and transitive.

Solution. (i) Not reflexive ($1 > 1$ is false). Not symmetric ($2 > 1$ but not $1 > 2$). Transitive: $x > y$ and $y > z$ give $x > z$.

(ii) Not reflexive ($1 + 1 \ne 10$). Symmetric: $x + y = 10 \Rightarrow y + x = 10$. Not transitive: $(1, 9)$ and $(9, 1)$ are in the relation, but $(1, 1)$ is not.

(iii) Reflexive: $x \cdot x = x^2$. Symmetric: $xy = yx$. Transitive: if $xy = a^2$ and $yz = b^2$, then $xz \cdot y^2 = a^2b^2$, so $xz = \left(\frac{ab}{y}\right)^2$. A natural number that is the square of a rational is the square of an integer, so $xz$ is a perfect square. This one is an equivalence relation.

(iv) The only solutions in $\mathbb{N}$ are $y = 1, x = 6$ and $y = 2, x = 2$, so the relation is $\{(6, 1), (2, 2)\}$. Not reflexive ($(1,1)$ is missing). Not symmetric ($(1, 6)$ is missing). Transitive: the only chain is $(2,2),(2,2)$, giving $(2,2)$, which is there; $(6,1)$ cannot be continued because nothing starts with $1$.

Question 23

Let $A = \{1, 2, 3, \ldots, 9\}$ and $R$ be the relation in $A \times A$ defined by $(a, b)\,R\,(c, d)$ if $a + d = b + c$ for $(a, b), (c, d)$ in $A \times A$. Prove that $R$ is an equivalence relation and also obtain the equivalence class $[(2, 5)]$.

Solution. The condition $a + d = b + c$ is the same as $a – b = c – d$: two pairs are related when they have the same difference. Written this way, the three properties are clear.

Reflexive: $a – b = a – b$, so $(a, b)\,R\,(a, b)$.

Symmetric: $a – b = c – d \Rightarrow c – d = a – b$.

Transitive: $a – b = c – d$ and $c – d = e – f$ give $a – b = e – f$.

So $R$ is an equivalence relation. The class of $(2, 5)$ is every pair in $A \times A$ with $a – b = 2 – 5 = -3$, that is $b = a + 3$, with both in $\{1, \ldots, 9\}$: $a = 1, 2, \ldots, 6$.

Question 24

Using the definition, prove that the function $f : A \to B$ is invertible if and only if $f$ is both one-one and onto.

Solution. By definition, $f$ is invertible if there is a $g : B \to A$ with $g \circ f = I_A$ and $f \circ g = I_B$.

If $f$ is invertible, it is one-one and onto. Suppose $f(x_1) = f(x_2)$. Applying $g$, $x_1 = g(f(x_1)) = g(f(x_2)) = x_2$: one-one. For any $y \in B$, the element $x = g(y) \in A$ has $f(x) = f(g(y)) = y$: onto.

If $f$ is one-one and onto, it is invertible. Because $f$ is onto, each $y \in B$ has at least one pre-image; because $f$ is one-one, it has exactly one. Define $g(y)$ to be that unique $x$ with $f(x) = y$. Then:

• for $x \in A$, $g(f(x))$ is the unique pre-image of $f(x)$, which is $x$; so $g \circ f = I_A$;
• for $y \in B$, $f(g(y)) = y$ by the definition of $g$; so $f \circ g = I_B$.

So $g$ is an inverse of $f$.

Question 25

Functions $f, g : \mathbb{R} \to \mathbb{R}$ are defined, respectively, by $f(x) = x^2 + 3x + 1$, $g(x) = 2x – 3$, find

(i) $f \circ g$ (ii) $g \circ f$ (iii) $f \circ f$ (iv) $g \circ g$

Solution. (i) $f(g(x)) = (2x-3)^2 + 3(2x-3) + 1 = 4x^2 – 12x + 9 + 6x – 9 + 1 = 4x^2 – 6x + 1$.

(ii) $g(f(x)) = 2(x^2 + 3x + 1) – 3 = 2x^2 + 6x – 1$.

(iii) $f(f(x)) = (x^2 + 3x + 1)^2 + 3(x^2 + 3x + 1) + 1$. The square is $x^4 + 6x^3 + 11x^2 + 6x + 1$; adding $3x^2 + 9x + 3$ and $1$ gives $x^4 + 6x^3 + 14x^2 + 15x + 5$.

(iv) $g(g(x)) = 2(2x – 3) – 3 = 4x – 9$.

Parts (i) and (ii) differ, which shows directly that composition is not commutative.

Question 26

Let $*$ be the binary operation defined on $\mathbb{Q}$. Find which of the following binary operations are commutative:

(i) $a * b = a – b$ (ii) $a * b = a^2 + b^2$ (iii) $a * b = a + ab$ (iv) $a * b = (a – b)^2$, $\forall a, b \in \mathbb{Q}$.

Solution. An operation is commutative if $a * b = b * a$ for all $a, b$.

• (i) $1 * 2 = -1$ but $2 * 1 = 1$: not commutative.
• (ii) $a^2 + b^2 = b^2 + a^2$: commutative.
• (iii) $1 * 2 = 1 + 2 = 3$ but $2 * 1 = 2 + 2 = 4$: not commutative.
• (iv) $(a – b)^2 = (b – a)^2$: commutative.

Question 27

Let $*$ be binary operation defined on $\mathbb{R}$ by $a * b = 1 + ab$, $\forall a, b \in \mathbb{R}$. Then the operation $*$ is

(i) commutative but not associative

(ii) associative but not commutative

(iii) neither commutative nor associative

(iv) both commutative and associative

Solution. Commutative: $1 + ab = 1 + ba$.

Associative would need $(a * b) * c = a * (b * c)$ for all $a, b, c$:

$$(a * b) * c = 1 + (1 + ab)c = 1 + c + abc, \qquad a * (b * c) = 1 + a(1 + bc) = 1 + a + abc.$$

These differ whenever $a \ne c$; for example $(1 * 2) * 3 = 10$ but $1 * (2 * 3) = 8$. Not associative.

Common mistakes

• Testing symmetry or transitivity on pairs that are not in the relation. In Question 22(iv) the relation has only two pairs; work from the list $\{(6,1),(2,2)\}$, not from the equation.
• Missing the transitivity counterexample $(a,b),(b,a) \Rightarrow (a,a)$. It breaks Questions 16(b) and 22(ii), and it is the first thing to try on any symmetric relation that is not reflexive.
• Forgetting the excluded point in an onto proof. In Question 20 you must check that the pre-image $\frac{3y-2}{y-1}$ is not $3$, or it would lie outside $A$.
• Calling $x|x|$ “not one-one” because it contains $|x|$. Question 21(iii) is strictly increasing, so it is one-one; $|x|$ alone is not.
• Proving associativity with a single example. One example can only disprove it, as in Question 27.

Practise next

• Exemplar Short Answer Q1–15, the quicker questions on composites, inverses and range.
• Exemplar Fill in the Blanks and True or False Q48–62, short checks on the same properties.
• NCERT textbook Exercise 1.1 for equivalence relations like Questions 22(iii) and 23.
• NCERT textbook Miscellaneous Exercise on Relations and Functions for composites and inverses as in Questions 20 and 25.