NCERT Exemplar Solutions Class 12 Maths Relations and Functions — Short Answer Q1–15

NCERT Exemplar Problems, Class 12 Mathematics — Relations and Functions, Short Answer questions 1 to 15. All 15 questions solved.

Fifteen short questions on three skills. The first is reading a relation or function written as a set of ordered pairs — deciding whether it is a function at all, and finding its inverse or a composite pair by pair (Questions 1, 5, 7, 9, 12). The second is composing and inverting formulas (3, 4, 6). The third is the one-one and onto tests, with domain and range (2, 8, 10, 11, 13, 14), and one equivalence-relation proof (15).

Key insight. To show a function is not one-one, one pair of different inputs with the same output is enough; to show it is not onto, one element of the codomain that is never reached is enough. A counterexample is a complete proof — you never need a general argument to show a property fails.

Question 1

Let $A = \{a, b, c\}$ and the relation $R$ be defined on $A$ as follows: $R = \{(a, a), (b, c), (a, b)\}$. Then, write the minimum number of ordered pairs to be added in $R$ to make $R$ reflexive and transitive.

Solution. Reflexive needs $(x, x)$ for every $x \in A$. $(a, a)$ is already there, so $(b, b)$ and $(c, c)$ must be added.

Transitive needs: whenever $(x, y)$ and $(y, z)$ are in $R$, so is $(x, z)$. Here $(a, b)$ and $(b, c)$ are in $R$, so $(a, c)$ must be added.

Check the new relation $\{(a,a), (b,b), (c,c), (a,b), (b,c), (a,c)\}$: every chain $x \to y \to z$ now has its shortcut $x \to z$, so it is transitive. Nothing smaller works, because each of the three added pairs was forced.

Question 2

Let $D$ be the domain of the real valued function $f$ defined by $f(x) = \sqrt{25 – x^2}$. Then, write $D$.

Solution. A square root is real only when what is under it is non-negative: $25 – x^2 \ge 0$, so $x^2 \le 25$, which means $-5 \le x \le 5$.

Question 3

Let $f, g : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = 2x + 1$ and $g(x) = x^2 – 2$, $\forall x \in \mathbb{R}$, respectively. Then, find $g \circ f$.

Solution. $g \circ f$ means “apply $f$ first, then $g$”:

$$(g \circ f)(x) = g(2x + 1) = (2x+1)^2 – 2 = 4x^2 + 4x – 1.$$

Question 4

Let $f : \mathbb{R} \to \mathbb{R}$ be the function defined by $f(x) = 2x – 3$, $\forall x \in \mathbb{R}$. Write $f^{-1}$.

Solution. $f$ is one-one and onto (a non-constant linear function), so it has an inverse. Put $y = 2x – 3$ and solve for $x$: $x = \dfrac{y+3}{2}$.

Check: $f\!\left(\frac{x+3}{2}\right) = (x + 3) – 3 = x$.

Question 5

If $A = \{a, b, c, d\}$ and the function $f = \{(a, b), (b, d), (c, a), (d, c)\}$, write $f^{-1}$.

Solution. Each element of $A$ appears exactly once as a second entry, so $f$ is one-one and onto $A$ and has an inverse. The inverse undoes each pair, so reverse every ordered pair.

Question 6

If $f : \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = x^2 – 3x + 2$, write $f(f(x))$.

Solution. Substitute $f(x)$ for $x$:

$$f(f(x)) = (x^2 – 3x + 2)^2 – 3(x^2 – 3x + 2) + 2.$$

The square is $x^4 – 6x^3 + 13x^2 – 12x + 4$. Subtracting $3x^2 – 9x + 6$ and adding $2$:

$$f(f(x)) = x^4 – 6x^3 + 10x^2 – 3x.$$

Question 7

Is $g = \{(1, 1), (2, 3), (3, 5), (4, 7)\}$ a function? If $g$ is described by $g(x) = \alpha x + \beta$, then what value should be assigned to $\alpha$ and $\beta$?

Solution. Yes: each of $1, 2, 3, 4$ appears exactly once as a first entry, so each input has exactly one image.

From $g(1) = 1$ and $g(2) = 3$: $\alpha + \beta = 1$ and $2\alpha + \beta = 3$, so $\alpha = 2$, $\beta = -1$. The other two pairs confirm it: $2(3) – 1 = 5$ and $2(4) – 1 = 7$.

Question 8

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.

(i) $\{(x, y) : x \text{ is a person}, y \text{ is the mother of } x\}$.

(ii) $\{(a, b) : a \text{ is a person}, b \text{ is an ancestor of } a\}$.

Solution. (i) Every person has exactly one mother, so each $x$ has exactly one $y$: this is a function. It is not injective, because two siblings have the same mother — two different inputs, one output. Whether it is surjective depends on the codomain, which the question does not state. Taking the codomain as the set of mothers (the set of second entries), every mother is the mother of someone, so the map is onto; that is the intended reading. (If the codomain were all people it would not be onto, since not everyone is a mother.)

(ii) A person has many ancestors (two parents, four grandparents, and so on), so one $a$ is paired with many $b$. This is not a function.

Question 9

If the mappings $f$ and $g$ are given by $f = \{(1, 2), (3, 5), (4, 1)\}$ and $g = \{(2, 3), (5, 1), (1, 3)\}$, write $f \circ g$.

Solution. $f \circ g$ applies $g$ first, so start from the domain of $g$, which is $\{2, 5, 1\}$:

• $f(g(2)) = f(3) = 5$
• $f(g(5)) = f(1) = 2$
• $f(g(1)) = f(3) = 5$

Question 10

Let $\mathbb{C}$ be the set of complex numbers. Prove that the mapping $f : \mathbb{C} \to \mathbb{R}$ given by $f(z) = |z|$, $\forall z \in \mathbb{C}$, is neither one-one nor onto.

Solution. Not one-one: $f(1) = |1| = 1$ and $f(-1) = |-1| = 1$, but $1 \ne -1$. Two different inputs share an output.

Not onto: a modulus is never negative, so $-1 \in \mathbb{R}$ is not $|z|$ for any $z \in \mathbb{C}$. The range is $[0, \infty)$, not all of $\mathbb{R}$.

Question 11

Let the function $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \cos x$, $\forall x \in \mathbb{R}$. Show that $f$ is neither one-one nor onto.

Solution. Not one-one: $\cos 0 = \cos 2\pi = 1$, with $0 \ne 2\pi$.

Not onto: $-1 \le \cos x \le 1$ for every real $x$, so $2 \in \mathbb{R}$ is never an output. The range is $[-1, 1]$.

Question 12

Let $X = \{1, 2, 3\}$ and $Y = \{4, 5\}$. Find whether the following subsets of $X \times Y$ are functions from $X$ to $Y$ or not.

(i) $f = \{(1, 4), (1, 5), (2, 4), (3, 5)\}$ (ii) $g = \{(1, 4), (2, 4), (3, 4)\}$

(iii) $h = \{(1, 4), (2, 5), (3, 5)\}$ (iv) $k = \{(1, 4), (2, 5)\}$

Solution. A function from $X$ to $Y$ must give every element of $X$ exactly one image.

• (i) $1$ has two images, $4$ and $5$: not a function.
• (ii) Each of $1, 2, 3$ has one image (all $4$): a function (a constant one).
• (iii) Each of $1, 2, 3$ has one image: a function.
• (iv) $3$ has no image: not a function from $X$.

Question 13

If functions $f : A \to B$ and $g : B \to A$ satisfy $g \circ f = I_A$, then show that $f$ is one-one and $g$ is onto.

Solution. $f$ is one-one. Suppose $f(x_1) = f(x_2)$. Apply $g$ to both sides: $g(f(x_1)) = g(f(x_2))$, that is, $I_A(x_1) = I_A(x_2)$, so $x_1 = x_2$.

$g$ is onto. Take any $a \in A$. Then $f(a) \in B$, and $g(f(a)) = I_A(a) = a$. So $b = f(a)$ is an element of $B$ with $g(b) = a$: every element of $A$ has a pre-image under $g$.

The idea: $g$ undoes $f$, so $f$ cannot have merged two inputs (or $g$ could not tell them apart), and $g$ must reach every element of $A$ (because $g$ returns each $a$ after $f$).

Question 14

Let $f : \mathbb{R} \to \mathbb{R}$ be the function defined by $f(x) = \dfrac{1}{2 – \cos x}$, $\forall x \in \mathbb{R}$. Then, find the range of $f$.

Solution. Since $-1 \le \cos x \le 1$, the denominator satisfies $1 \le 2 – \cos x \le 3$, and it takes every value in between. It is always positive, so taking reciprocals reverses the inequality:

$$\frac13 \le \frac{1}{2 – \cos x} \le 1.$$

The value $1$ comes at $x = 0$ and $\frac13$ at $x = \pi$.

Question 15

Let $n$ be a fixed positive integer. Define a relation $R$ in $\mathbb{Z}$ as follows: $\forall a, b \in \mathbb{Z}$, $aRb$ if and only if $a – b$ is divisible by $n$. Show that $R$ is an equivalence relation.

Solution. Reflexive: $a – a = 0 = 0 \cdot n$, divisible by $n$, so $aRa$ for every $a$.

Symmetric: if $aRb$, then $a – b = kn$ for some integer $k$. So $b – a = (-k)n$, divisible by $n$, and $bRa$.

Transitive: if $aRb$ and $bRc$, then $a – b = kn$ and $b – c = mn$. Adding, $a – c = (k + m)n$, so $aRc$.

All three hold, so $R$ is an equivalence relation. (This is “congruence modulo $n$”; its equivalence classes are the $n$ remainder classes $0, 1, \ldots, n-1$.)

Common mistakes

• Reading $g \circ f$ the wrong way round. $g \circ f$ applies $f$ first. In Question 9, $f \circ g$ starts from the domain of $g$, not of $f$.
• Forgetting the forced transitive pair. In Question 1 the reflexive pairs are obvious; $(a, c)$, forced by $(a, b)$ and $(b, c)$, is the one most often missed.
• Calling a relation a function when an input has no image. Question 12(iv) gives each listed input one image, but $3 \in X$ has none, so it is not a function from $X$.
• Ignoring the codomain when testing onto. Question 8(i) is onto or not depending on what the codomain is; Questions 10 and 11 fail onto because the codomain is all of $\mathbb{R}$.
• Inverting the inequality wrongly in a reciprocal. In Question 14, larger denominators give smaller values, so the range is $[\frac13, 1]$, not $[1, 3]$.

Practise next

• Exemplar Long Answer Q16–27, which builds relations with given properties and proves bijections.
• Exemplar Objective Type Q28–47, the multiple-choice questions from this unit.
• NCERT textbook Exercise 1.1 for more reflexive, symmetric and transitive checks, as in Questions 1 and 15.
• NCERT textbook Exercise 1.2 for one-one and onto proofs like Questions 10 and 11.