NCERT Exemplar Solutions Class 12 Maths Relations and Functions — Objective Type Q28–47

NCERT Exemplar Problems, Class 12 Mathematics — Relations and Functions, Objective Type questions 28 to 47. All 20 questions solved.

Twenty multiple-choice questions. The first six test the properties of a given relation (28 to 33); then come counting questions on mappings (35, 36), one-one and onto tests (37, 39), composites and inverses (38, 40 to 43, 45, 47), a range (44) and a piecewise function (46). Question 34 is on binary operations, which the rationalised textbook and the current JEE syllabus no longer include; it needs only the definition of an identity element, given where it is used.

Three printing slips. In Questions 44 and 45 the fourth option is labelled (B) a second time, and in Question 45 the value $1$ is printed as both (A) and (B). The fourth option is shown here as (D), which is how the answer key refers to it.

Key insight. Before choosing an option about a function, check that the function is properly defined on the stated domain. Several of these questions turn on exactly that: $\frac1x$ on all of $\mathbb{R}$ (37), $\frac32$ fed to a function on $\mathbb{N}$ (45), and $\tan x$ on all of $\mathbb{R}$ (47).

Question 28

Let $T$ be the set of all triangles in the Euclidean plane, and let a relation $R$ on $T$ be defined as $aRb$ if $a$ is congruent to $b$, $\forall a, b \in T$. Then $R$ is

Solution. Every triangle is congruent to itself (reflexive); if $a \cong b$ then $b \cong a$ (symmetric); if $a \cong b$ and $b \cong c$ then $a \cong c$ (transitive).

Question 29

Consider the non-empty set consisting of children in a family and a relation $R$ defined as $aRb$ if $a$ is brother of $b$. Then $R$ is

Solution. Not symmetric: if $a$ is a boy and $b$ his sister, $a$ is the brother of $b$, but $b$ is not the brother of $a$.

Transitive: if $a$ is the brother of $b$ and $b$ is a sibling of $c$, then $a$ is a boy and a sibling of $c$, so $a$ is the brother of $c$.

There is one edge case. If $a$ and $b$ are both boys, then $aRb$ and $bRa$, and transitivity would ask for $aRa$, “$a$ is his own brother”, which is false. The question intends the usual reading in which a chain through two different people is what is being tested, and the key’s answer is (B). Be ready for a stricter examiner to argue (C) on exactly this point.

Question 30

The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ are

Solution. An equivalence relation on $A$ is the same thing as a way of splitting $A$ into classes. The splittings of $\{1, 2, 3\}$ are:

• one class: $\{1, 2, 3\}$ (the universal relation);
• two classes: $\{1, 2\}\{3\}$, $\{1, 3\}\{2\}$, $\{2, 3\}\{1\}$;
• three classes: $\{1\}\{2\}\{3\}$ (the identity relation).

That is $1 + 3 + 1 = 5$.

Question 31

If a relation $R$ on the set $\{1, 2, 3\}$ be defined by $R = \{(1, 2)\}$, then $R$ is

Solution. Not reflexive ($(1,1)$ is missing) and not symmetric ($(2,1)$ is missing). Transitive: there is no pair of the form $(2, z)$, so no chain $(1,2),(2,z)$ exists to break it. A condition with nothing to test holds automatically.

Question 32

Let us define a relation $R$ in $\mathbb{R}$ as $aRb$ if $a \ge b$. Then $R$ is

Solution. $a \ge a$ (reflexive); $a \ge b$ and $b \ge c$ give $a \ge c$ (transitive); but $2 \ge 1$ while $1 \not\ge 2$ (not symmetric).

Question 33

Let $A = \{1, 2, 3\}$ and consider the relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$. Then $R$ is

Solution. All three $(a, a)$ are present: reflexive. $(1, 2) \in R$ but $(2, 1) \notin R$: not symmetric. The only chain of two different pairs is $(1,2),(2,3)$, and $(1,3)$ is present, so it is transitive. (This is the relation $\le$ on $\{1, 2, 3\}$.) Option (A) is the one that is true.

Question 34

The identity element for the binary operation $*$ defined on $\mathbb{Q} – \{0\}$ as $a * b = \dfrac{ab}{2}$, $\forall a, b \in \mathbb{Q} – \{0\}$ is

Solution. An identity $e$ satisfies $a * e = e * a = a$ for every $a$. Here $\dfrac{ae}{2} = a$, and since $a \ne 0$, $e = 2$. Check: $a * 2 = \frac{2a}{2} = a$.

Question 35

If the set $A$ contains $5$ elements and the set $B$ contains $6$ elements, then the number of one-one and onto mappings from $A$ to $B$ is

Solution. A one-one map from $A$ uses $5$ different images, so it reaches only $5$ of the $6$ elements of $B$ and cannot be onto. A bijection needs the two sets to have the same size. ($720 = 6!$ and $120 = {}^6P_5$ count other things — the latter is the number of one-one maps.)

Question 36

Let $A = \{1, 2, 3, \ldots, n\}$ and $B = \{a, b\}$. Then the number of surjections from $A$ into $B$ is

Solution. Each of the $n$ elements can go to $a$ or $b$: $2^n$ maps in all. Exactly two are not onto — everything to $a$, and everything to $b$. So $2^n – 2$ are onto (for $n = 3$: $6$, which can be listed by hand).

Question 37

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \dfrac1x$, $\forall x \in \mathbb{R}$. Then $f$ is

Solution. $\frac10$ has no value, so the rule does not assign an image to $0 \in \mathbb{R}$. As stated, $f$ is not a function on $\mathbb{R}$ at all. (On $\mathbb{R} – \{0\}$ it would be one-one, but the question did not say that.)

Question 38

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = 3x^2 – 5$ and $g : \mathbb{R} \to \mathbb{R}$ by $g(x) = \dfrac{x}{x^2 + 1}$. Then $g \circ f$ is

Solution. $g(f(x)) = \dfrac{3x^2 – 5}{(3x^2 – 5)^2 + 1}$, and $(3x^2 – 5)^2 + 1 = 9x^4 – 30x^2 + 25 + 1 = 9x^4 – 30x^2 + 26$.

Question 39

Which of the following functions from $\mathbb{Z}$ into $\mathbb{Z}$ are bijections?

Solution. (B) has the inverse $y \mapsto y – 2$, which takes integers to integers, so it is a bijection of $\mathbb{Z}$. The others fail onto in $\mathbb{Z}$: $x^3 = 2$ has no integer solution; $2x + 1$ is always odd, so $2$ is missed; $x^2 + 1 \ge 1$ (and $x^2 + 1$ is not one-one either).

Question 40

Let $f : \mathbb{R} \to \mathbb{R}$ be the function defined by $f(x) = x^3 + 5$. Then $f^{-1}(x)$ is

Solution. $y = x^3 + 5 \Rightarrow x^3 = y – 5 \Rightarrow x = (y – 5)^{1/3}$, the real cube root, which exists for every real $y$.

Question 41

Let $f : A \to B$ and $g : B \to C$ be the bijective functions. Then $(g \circ f)^{-1}$ is

Solution. $g \circ f$ goes $A \to B \to C$. To undo it, undo the last step first: $g^{-1}$ takes $C \to B$, then $f^{-1}$ takes $B \to A$. That is $f^{-1} \circ g^{-1}$. Check: $(f^{-1} \circ g^{-1}) \circ (g \circ f) = f^{-1} \circ (g^{-1} \circ g) \circ f = f^{-1} \circ f = I_A$. (Option (C) does not even make sense: $f^{-1}$ lands in $A$, where $g^{-1}$ is not defined.)

Question 42

Let $f : \mathbb{R} – \left\{\dfrac35\right\} \to \mathbb{R}$ be defined by $f(x) = \dfrac{3x + 2}{5x – 3}$. Then

Solution. Compute $f(f(x))$:

$$f(f(x)) = \frac{3\cdot\frac{3x+2}{5x-3} + 2}{5\cdot\frac{3x+2}{5x-3} – 3} = \frac{9x + 6 + 10x – 6}{15x + 10 – 15x + 9} = \frac{19x}{19} = x.$$

So $f \circ f$ is the identity, which means $f$ is its own inverse. (Any $\frac{ax+b}{cx-a}$ has this property.)

Question 43

Let $f : [0, 1] \to [0, 1]$ be defined by $f(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1 – x, & \text{if } x \text{ is irrational} \end{cases}$. Then $(f \circ f)\,x$ is

Solution. If $x$ is rational, $f(x) = x$ is rational, so $f(f(x)) = x$. If $x$ is irrational, $f(x) = 1 – x$ is also irrational (a rational minus an irrational), so $f(f(x)) = 1 – (1 – x) = x$. Either way, $(f \circ f)(x) = x$.

Question 44

Let $f : [2, \infty) \to \mathbb{R}$ be the function defined by $f(x) = x^2 – 4x + 5$, then the range of $f$ is

Solution. Complete the square: $f(x) = (x – 2)^2 + 1$. On $[2, \infty)$, $(x – 2)^2$ takes every value in $[0, \infty)$, so $f$ takes every value in $[1, \infty)$. The trap is $[5, \infty)$, from putting $x = 0$, which is outside the domain.

Question 45

Let $f : \mathbb{N} \to \mathbb{R}$ be the function defined by $f(x) = \dfrac{2x – 1}{2}$ and $g : \mathbb{Q} \to \mathbb{R}$ be another function defined by $g(x) = x + 2$. Then $(g \circ f)\left(\dfrac32\right)$ is

Solution. $g \circ f$ applies $f$ first, and $f$ is defined only on $\mathbb{N}$. Since $\frac32 \notin \mathbb{N}$, $(g \circ f)\left(\frac32\right)$ is not defined.

Even if you ignored the domain and used the formula, $f\left(\frac32\right) = 1$ and $g(1) = 3$, which is not among the options either. So the answer is “none of these” on both readings.

Question 46

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \begin{cases} 2x, & x > 3 \\ x^2, & 1 < x \le 3 \\ 3x, & x \le 1 \end{cases}$. Then $f(-1) + f(2) + f(4)$ is

Solution. Pick the right piece for each input: $-1 \le 1$, so $f(-1) = -3$; $1 < 2 \le 3$, so $f(2) = 4$; $4 > 3$, so $f(4) = 8$. The sum is $-3 + 4 + 8 = 9$.

Question 47

Let $f : \mathbb{R} \to \mathbb{R}$ be given by $f(x) = \tan x$. Then $f^{-1}(1)$ is

Solution. $\tan x$ is not one-one (it repeats every $\pi$), so $f$ has no inverse function, and $f^{-1}(1)$ here means the set of pre-images of $1$: every $x$ with $\tan x = 1$. Those are $x = n\pi + \frac\pi4$, $n \in \mathbb{Z}$.

(Strictly, $\tan x$ is not defined at odd multiples of $\frac\pi2$, so “$f : \mathbb{R} \to \mathbb{R}$” is itself loose; that does not change which $x$ give $\tan x = 1$. Option (A), $\frac\pi4$, is the principal value $\tan^{-1} 1$ — the answer to a different question, about the inverse trigonometric function.)

Common mistakes

• Reversing the order in $(g \circ f)^{-1}$. It is $f^{-1} \circ g^{-1}$ — undo the last step first (Question 41).
• Using a value outside the domain. $[5, \infty)$ in Question 44 comes from $x = 0$, and $3$ in Question 45 comes from feeding $\frac32$ to a function on $\mathbb{N}$.
• Forgetting that a vacuous condition holds. $\{(1, 2)\}$ in Question 31 is transitive because there is no chain to test.
• Counting all maps instead of onto maps. $2^n$ counts every map to $\{a, b\}$; subtract the two constant ones (Question 36).
• Confusing a pre-image set with a principal value. $\tan x = 1$ has infinitely many solutions; $\tan^{-1} 1 = \frac\pi4$ is one chosen value (Question 47).

Practise next

• Exemplar Long Answer Q16–27, where the same properties are proved rather than chosen.
• Exemplar Fill in the Blanks and True or False Q48–62, the rest of the unit.
• NCERT textbook Exercise 1.1 for relation-property questions like 28 to 33.
• NCERT textbook Exercise 1.2 for one-one and onto checks like Questions 37 and 39.