NCERT Exemplar Solutions Class 12 Maths Relations and Functions — Fill in the Blanks and True or False Q48–62

NCERT Exemplar Problems, Class 12 Mathematics — Relations and Functions, Fill in the Blanks questions 48 to 52 and True or False questions 53 to 62. All 15 questions solved.

Fifteen short questions. The five blanks ask you to list a relation (48, 49), compose two functions given as pairs (50), compose a formula with itself (51) and invert one (52). The ten statements test relation properties (53, 55, 56, 58), whether a function is onto or invertible (54, 57, 61), and two general facts about composition (59, 60). The book’s instruction says “Exercises 53 to 63”, but the unit ends at Question 62.

Question 62 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.

Key insight. A True or False statement about “every” relation or function is false as soon as one counterexample exists — so look for the smallest one, usually on a set of two or three elements. A statement about one specific relation is settled by checking its listed pairs, not by intuition about the rule.

Question 48

Let the relation $R$ be defined in $\mathbb{N}$ by $aRb$ if $2a + 3b = 30$. Then $R = $ ______.

Solution. $3b = 30 – 2a$ must be a positive multiple of $3$, so $a$ must be a multiple of $3$ and less than $15$: $a = 3, 6, 9, 12$, giving $b = 8, 6, 4, 2$. ($a = 15$ gives $b = 0 \notin \mathbb{N}$.)

Question 49

Let the relation $R$ be defined on the set $A = \{1, 2, 3, 4, 5\}$ by $R = \{(a, b) : |a^2 – b^2| < 8\}$. Then $R$ is given by ______.

Solution. The squares are $1, 4, 9, 16, 25$. Two of them differ by less than $8$ only when they are equal or neighbours among $1, 4, 9, 16$: $|1 – 4| = 3$, $|4 – 9| = 5$, $|9 – 16| = 7$. Every other gap is $8$ or more ($|1 – 9| = 8$, $|16 – 25| = 9$). So $R$ has the five pairs $(a, a)$ and both orders of $(1,2)$, $(2,3)$, $(3,4)$.

Question 50

Let $f = \{(1, 2), (3, 5), (4, 1)\}$ and $g = \{(2, 3), (5, 1), (1, 3)\}$. Then $g \circ f = $ __ and $f \circ g = $ ____.

Solution. $g \circ f$ applies $f$ first, starting from $f$’s domain $\{1, 3, 4\}$: $g(f(1)) = g(2) = 3$, $g(f(3)) = g(5) = 1$, $g(f(4)) = g(1) = 3$.

$f \circ g$ applies $g$ first, starting from $g$’s domain $\{2, 5, 1\}$: $f(g(2)) = f(3) = 5$, $f(g(5)) = f(1) = 2$, $f(g(1)) = f(3) = 5$.

Question 51

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \dfrac{x}{\sqrt{1 + x^2}}$. Then $(f \circ f \circ f)(x) = $ ______.

Solution. First $f \circ f$:

$$f(f(x)) = \frac{\dfrac{x}{\sqrt{1+x^2}}}{\sqrt{1 + \dfrac{x^2}{1+x^2}}} = \frac{\dfrac{x}{\sqrt{1+x^2}}}{\dfrac{\sqrt{1+2x^2}}{\sqrt{1+x^2}}} = \frac{x}{\sqrt{1 + 2x^2}}.$$

Each application adds one more $x^2$ under the root. Applying $f$ once more, the same way:

$$f\!\left(\frac{x}{\sqrt{1+2x^2}}\right) = \frac{\dfrac{x}{\sqrt{1+2x^2}}}{\sqrt{\dfrac{1+3x^2}{1+2x^2}}} = \frac{x}{\sqrt{1 + 3x^2}}.$$

Check at $x = 1$: $f(1) = \frac{1}{\sqrt2}$, $f\left(\frac1{\sqrt2}\right) = \frac{1}{\sqrt3}$, $f\left(\frac1{\sqrt3}\right) = \frac12$, and $\frac{1}{\sqrt{1+3}} = \frac12$.

Question 52

If $f(x) = 4 – (x – 7)^3$, then $f^{-1}(x) = $ ______.

Solution. Put $y = 4 – (x – 7)^3$. Then $(x – 7)^3 = 4 – y$, so $x – 7 = (4 – y)^{1/3}$ (the real cube root, defined for every real number) and $x = 7 + (4 – y)^{1/3}$. The function is a cubic shifted and reflected, so it is one-one and onto $\mathbb{R}$, and this inverse is valid everywhere.

Question 53

Let $R = \{(3, 1), (1, 3), (3, 3)\}$ be a relation defined on the set $A = \{1, 2, 3\}$. Then $R$ is symmetric, transitive but not reflexive.

Solution. Symmetric: yes, $(3,1)$ and $(1,3)$ come as a pair. Not reflexive: yes, $(1,1)$ is missing. But transitive fails: $(1, 3)$ and $(3, 1)$ are in $R$, so transitivity needs $(1, 1)$, which is not there. The statement claims transitivity, so it is false.

Question 54

Let $f : \mathbb{R} \to \mathbb{R}$ be the function defined by $f(x) = \sin(3x + 2)$, $\forall x \in \mathbb{R}$. Then $f$ is invertible.

Solution. Invertible means one-one and onto. $\sin$ repeats every $2\pi$, so $f(x) = f\left(x + \frac{2\pi}{3}\right)$: not one-one. Also its values lie in $[-1, 1]$, so it is not onto $\mathbb{R}$.

Question 55

Every relation which is symmetric and transitive is also reflexive.

Solution. Counterexample: on $A = \{1, 2\}$, take $R = \{(1, 1)\}$. It is symmetric and transitive, but $(2, 2) \notin R$, so it is not reflexive. (The empty relation on any non-empty set is another.) The tempting “proof” — $aRb$ and $bRa$ give $aRa$ — only works for an $a$ that is related to something.

Question 56

An integer $m$ is said to be related to another integer $n$ if $m$ is an integral multiple of $n$. This relation in $\mathbb{Z}$ is reflexive, symmetric and transitive.

Solution. Reflexive ($m = 1 \cdot m$) and transitive (if $m = kn$ and $n = lp$ then $m = (kl)p$) both hold. But it is not symmetric: $2$ is a multiple of $1$, while $1$ is not an integral multiple of $2$. So the statement is false.

Question 57

Let $A = \{0, 1\}$ and $\mathbb{N}$ be the set of natural numbers. Then the mapping $f : \mathbb{N} \to A$ defined by $f(2n – 1) = 0$, $f(2n) = 1$, $\forall n \in \mathbb{N}$, is onto.

Solution. $f$ sends every odd number to $0$ and every even number to $1$. Both elements of $A$ are reached ($f(1) = 0$, $f(2) = 1$), so $f$ is onto. (It is far from one-one, but that is not what is asked.)

Question 58

The relation $R$ on the set $A = \{1, 2, 3\}$ defined as $R = \{(1, 1), (1, 2), (2, 1), (3, 3)\}$ is reflexive, symmetric and transitive.

Solution. $(2, 2)$ is missing, so $R$ is not reflexive. (It also fails transitivity: $(2,1)$ and $(1,2)$ need $(2,2)$.) It is symmetric, but the statement claims all three.

Question 59

The composition of functions is commutative.

Solution. Take $f(x) = x + 1$ and $g(x) = 2x$. Then $(g \circ f)(x) = 2x + 2$ but $(f \circ g)(x) = 2x + 1$. One counterexample settles it; Question 25 on the Long Answer page is another.

Question 60

The composition of functions is associative.

Solution. For any $f$, $g$, $h$ for which the composites are defined,

$$\big(h \circ (g \circ f)\big)(x) = h\big(g(f(x))\big) = \big((h \circ g) \circ f\big)(x).$$

Both sides apply $f$, then $g$, then $h$, so they always agree.

Question 61

Every function is invertible.

Solution. Only functions that are one-one and onto are invertible. $f(x) = x^2$ on $\mathbb{R}$ is neither, so it has no inverse.

Question 62

A binary operation on a set has always the identity element.

Solution. Subtraction on $\mathbb{R}$ is a binary operation with no identity: $a – 0 = a$, but $0 – a = -a \ne a$, and no other $e$ satisfies $a – e = a$. Example 34 in the unit’s worked examples (addition on $\mathbb{N}$, which has no identity because $0 \notin \mathbb{N}$) is another.

Common mistakes

• Accepting “symmetric and transitive implies reflexive”. The argument only reaches elements that are related to something (Question 55); an element on its own breaks it.
• Checking only the properties the statement gets right. Question 53 is symmetric and not reflexive, as claimed, but its transitivity claim is false, so the whole statement is false.
• Composing in the wrong order with pairs. $g \circ f$ starts from the domain of $f$ (Question 50).
• Taking the cube root only for positive numbers. $(4 – x)^{1/3}$ is the real cube root and exists for every $x$, so the inverse in Question 52 needs no restriction.
• Forgetting $0 \notin \mathbb{N}$. It is why $a = 15$ is excluded in Question 48.

Practise next

• Exemplar Short Answer Q1–15, which has the same $f \circ g$ as Question 50 and more inverses.
• Exemplar Objective Type Q28–47, the multiple-choice questions from this unit.
• NCERT textbook Exercise 1.1 for listing relations and testing their properties.
• NCERT textbook Miscellaneous Exercise on Relations and Functions for more composites and inverses.