## NCERT Exemplar for Class 12 Maths - Relations and Functions - Free PDF Download

Free PDF download of NCERT Exemplar for Class 12 Maths Chapter 1 - Relations and Functions solved by expert Maths teachers on Vedantu.com as per NCERT (CBSE) Book guidelines. All Chapter 1 - Relations and Functions Exercise questions with solutions to help you to revise the complete syllabus and score more marks in your Examinations.

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Relations and Functions is the first Chapter of NCERT Exemplar Solutions for Class 12 Maths. This Chapter discusses several sorts of relationships and functions. Function composition, invertible functions, and binary operations are also explored. Students interested in learning simple methods and shortcuts for solving Exercise problems can get the NCERT Exemplar Solutions for Class 12 Maths Chapter 1 Relations and Functions solutions PDF from the link below.

## Access NCERT Exemplar Solutions for Class 12 Mathematics Chapter 1 - Relations and Functions

### Solved Examples

Example 1. Let A = {0, 1, 2, 3} and define a relation R on A as follows:

R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}. Is R reflexive? symmetric? Transitive?

Ans: R is reflexive and symmetric, but not transitive

Since, for (1, 0) ∈ R and (0, 3) ∈ R whereas (1, 3) ∉ R.

Example 2. For the set A = {1, 2, 3}, define a relation R in the set A as follows: R = {(1, 1), (2, 2), (3, 3), (1, 3)}. Write the ordered pairs to be added to R to make it the smallest equivalence relation.

Ans: (3, 1) is the single ordered pair that needs to be added to R to make it the smallest equivalence relation.

Example 3. Let R be the equivalence relation in the set Z of integers given by R = {(a, b):2 divides a – b}. Write the equivalence class [0].

Ans: $\left[0\right]$ = {0, ± 2, ± 4, ± 6,...}.

Example 4. Let the function f : R → R be defined by f (x) = 4x – 1, ∀ x ∈ R. Then, show that f is one-one.

Ans: For any two elements x1, x2 ∈ R such that f (x1) = f (x2),

we have 4${{{x}}_1}$ – 1 = 4${{{x}}_2}$ – 1

⇒ 4${{{x}}_1}$ = 4${{{x}}_2}$, i.e., ${{{x}}_1}$ = ${{{x}}_2}$

Hence, f is one-one.

Example 5 If f = {(5, 2), (6, 3)}, g = {(2, 5), (3, 6)}, write f o g.

Ans: Since, f o g = f(g(x))

Now, f(g(2)=f(5)=2 and f(g(3))=f(6)=3

Thus, fog = {(2,2), (3,3)}.

Example 6. Let f : R → R be the function defined by f (x) = 4x – 3 ∀ x ∈ R. Then write f-1.

Ans: Given that f(x) = 4 x- 3 = y (say), then

4 x = y + 3

x = $\dfrac{{{{y}} + {{3}}}}{{{4}}}$

Hence ${{{f}}^{ - 1}}\left( {{y}} \right){{}} = \dfrac{{{{y}} + {{3}}}}{{{4}}}$

$ \Rightarrow $ ${{{f}}^{ - 1}}\left( {{x}} \right) = \dfrac{{{{x}} + {{3}}}}{{{4}}}$

Example 7. Is the binary operation * defined on Z (set of integer) by m * n = m – n + mn ∀ m, n ∈ Z commutative?

Ans: No.

Since for 1, 2 ∈ Z, 1 * 2 = 1 – 2 + 1.2 = 1 while 2 * 1 = 2 – 1 + 2.1 = 3 so that 1 * 2 ≠ 2 * 1.

Example 8. If f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, write the range of f and g.

Ans: The range of f = {2, 3} and the range of g = {5, 6}.

Example 9. If A = {1, 2, 3} and f, g are relations corresponding to the subset of A × A indicated against them, which of f, g is a function? Why?

f = {(1, 3), (2, 3), (3, 2)}

g = {(1, 2), (1, 3), (3, 1)}

Ans: f is a function since each element of A in the first place in the ordered pairs is related to only one element of A in the second place while g is not a function because 1 is related to more than one element of A, namely, 2 and 3.

Example 10. If A = {a, b, c, d} and f = {(a, b), (b, d), (c, a), (d, c)}, show that f is one one from A onto A. Find f-1.

Ans: f is one-one since each element of A is assigned to a distinct element of the set A. Also, f is onto since f (A) = A.

Moreover, ${{{f}}^{ - 1}}$ = {(b, a), (d, b), (a, c), (c, d)}.

Example 11. In the set N of natural numbers, define the binary operation * by m * n = g.c.d (m, n), m, n ∈ N. Is the operation * commutative and associative?

Ans: The operation is clearly commutative since

m * n = g.c.d (m, n) = g.c.d (n, m) = n * m ∀ m, n ∈ N.

It is also associative because for l, m, n ∈ N, we have

l * (m * n) = g. c. d (l, g.c.d (m, n))

= g.c.d. (g. c. d (l, m), n)

= (l * m) * n.

Example 12. In the set of natural numbers N, define a relation R as follows: ∀ n, m ∈ N, nRm if on division by 5 each of the integers n and m leaves the remainder less than 5, i.e. one of the numbers 0, 1, 2, 3 and 4. Show that R is an equivalence relation. Also, obtain the pairwise disjoint subsets determined by R.

Ans: R is reflexive since for each a ∈ N, aRa. R is symmetric since if aRb, then bRa for a, b ∈ N. Also, R is transitive since for a, b, c ∈ N, if aRb and bRc, then aRc. Hence R is an equivalence relation in N which will partition the set N into the pairwise disjoint subsets.

The equivalent classes are as mentioned below:

A0 = {5, 10, 15, 20 ...}

A1 = {1, 6, 11, 16, 21 ...}

A2 = {2, 7, 12, 17, 22, ...}

A3 = {3, 8, 13, 18, 23, ...}

A4 = {4, 9, 14, 19, 24, ...}

It is evident that the above five sets are pairwise disjoint and

A0 ∪ A1 ∪ A2 ∪ A3 ∪ A4 = \[ \cup_{{i} = 0}^4 {{A}_{i}} = N \].

Example 13. Show that the function f : R → R defined by f (x) = $\dfrac{{\mathbf{x}}}{{{{\mathbf{x}}^2} + 1}}$ , ∀ x ∈ R , is neither one-one nor onto.

Ans: For ${{{x}}_1}$ , ${{{x}}_2}$ ∈ R, consider

f (${{{x}}_1}$) = f (${{{x}}_2}$)

⇒ $\dfrac{{{{{x}}_1}}}{{{{{x}}_1}^2 + {{1}}}}$ = $\dfrac{{{{{x}}_2}}}{{{{{x}}_2}^2 + {{1}}}}$

⇒${{{x}}_1}{{{x}}_2}^2$ + ${{{x}}_1}$ = ${{{x}}_2}{{{x}}_1}^2$ +${{{x}}_2}$

⇒ ${{{x}}_1}{{{x}}_2}$ (${{{x}}_2}$ - ${{{x}}_1}$) = ${{{x}}_2}$ - ${{{x}}_1}$

⇒ ${{{x}}_1}$ = ${{{x}}_2}$ or ${{{x}}_1}{{{x}}_2}$ =1

We note that there are point, ${{{x}}_1}$ and ${{{x}}_2}$ with ${{{x}}_1}$ ≠ ${{{x}}_2}$ and f (${{{x}}_1}$) = f (${{{x}}_2}$), for instance, if we take ${{{x}}_1}$ = 2 and ${{{x}}_2}$= $\dfrac{{{1}}}{{{2}}}$, then we have f(${{{x}}_1}$) = $\dfrac{{{2}}}{{{5}}}$ and

f (${{{x}}_2}$) = $\dfrac{{{2}}}{{{5}}}$ but 2 ≠ $\dfrac{{{1}}}{{{2}}}$ .

Hence f is not one-one. Also, f is not onto for if so then for 1∈R ∃ x ∈ R such that f (x) = 1

Which gives $\dfrac{{{x}}}{{{{{x}}^2} + {{1}}}}$ = 1. But there is no such x in the domain R, since the equation ${{{x}}^2} - {{x}} + {{1}}$ =0 does not give any real value of x.

Example 14. Let f, g : R → R be two functions defined as f (x) = $\left| {\mathbf{x}} \right|$ + x and g (x) = $\left| {\mathbf{x}} \right|$ - x ∀ x ∈ R. Then, find f o g and g o f.

Ans: Here f (x) = $\left| {{x}} \right|$ + x which can be redefined as

$f\left( x \right)=\left\{ \begin{align} & 2x,\text{ if }x\ge 0 \\ & 0,\text{ if }\text{ }x<0 \\ \end{align} \right.$

Similarly, the function g defined by \[ g (x) = \left| {{x}} \right|\] - x may be redefined as g (x) = {0 if x ≥ 0 -2 x if x < 0

Therefore, gof gets defined as:

For x ≥ 0, (gof) (x) = g (f (x) = g (2x) = 0

and for x < 0, (gof ) (x) = g (f (x) = g (0) = 0.

Consequently, we have (g o f) (x) = 0, ∀ x ∈ R.

Similarly, fog gets defined as:

For x ≥ 0, (fog)(x) = f(g(x)) = f (0) = 0, and for x < 0, (f o g ) (x) = f (g(x)) = f (–2 x) = – 4x.

i.e. (fog)(x) = {0, x > 0 - 4 x , x < 0}

Example 15. Let R be the set of real numbers and f : R → R be the function defined by f (x) = 4x + 5. Show that f is invertible and find f-1.

Ans: Here the function f : R → R is defined as f(x) = 4x + 5 = y (say).

Then, 4 x = y - 5 or x = $\dfrac{{{{y - 5}}}}{{{4}}}$

This leads to a function g: R → R defined as

g(y) = $\dfrac{{{{y}} - {{5}}}}{{{4}}}$

Therefore, (gof) (x) = g(f(x)) = g(4 x +5) = $\dfrac{{{{4x+5-5}}}}{{{4}}}{{=x}}$

Or, gof = ${{{I}}_{{R}}}$

Similarly (fog) (y) = f(g(y) = f$\left( {\dfrac{{{{y - 5}}}}{{{4}}}} \right)$ = 4$\left( {\dfrac{{{{y - 5}}}}{{{4}}}} \right)$ + 5 = y

Or, fog = ${{{I}}_{{R}}}$

Hence f is invertible and ${{{f}}^{ - 1}}$ = g which is given by

${{{f}}^{ - 1}}$(x) = $\dfrac{{{{x - 5}}}}{{{4}}}$.

Example 16. Let * be a binary operation defined on Q. Find which of the following binary operations are associative

(i) a * b = a – b for a, b ∈ Q.

Ans: * is not associative for if we take a = 1, b = 2 and c = 3, then (a * b) * c = (1 * 2) * 3 = (1 – 2) * 3 = – 1 – 3 = – 4 and a * (b * c) = 1 * (2 * 3) = 1 * (2 – 3) = 1 – ( – 1) = 2.

Thus (a * b) * c ≠ a * (b * c) and hence * is not associative.

(ii) a * b = $\dfrac{{{\mathbf{ab}}}}{4}$ for a, b ∈ Q.

Ans: * is associative since Q is associative with respect to multiplication.

(iii) a * b = a – b + ab for a, b ∈ Q.

Ans: * is not associative for if we take a = 2, b = 3 and c = 4, then

(a * b) * c = (2 * 3) * 4 = (2 – 3 + 6) * 4 = 5 * 4 = 5 – 4 + 20 = 21, and

a * (b * c) = 2 * (3 * 4) = 2 * (3 – 4 + 12) = 2 * 11 = 2 – 11 + 22 = 13

Thus, (a * b) * c ≠ a * (b * c) and hence * is not associative.

(iv) a * b = ab2 for a, b ∈ Q.

Ans: * is not associative for if we take a = 1, b = 2 and c = 3, then

(a * b) * c = (1 * 2) * 3 = 4 * 3 = 4 × 9 = 36 and

a * (b * c) = 1 * (2 * 3) = 1 * 18 = 1 × 182 = 324.

Thus (a * b) * c ≠ a * (b * c) and hence * is not associative.

### Objective Type Questions

### Choose the correct answer from the given four options in each of the Examples 17 to 25.

Example 17. Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is

(A) Reflexive and symmetric

(B) Transitive and symmetric

(C) Equivalence

(D) Reflexive, transitive but not symmetric

Ans: The correct choice is (D).

Since n divides n, ∀ n ∈ N, R is reflexive.

R is not symmetric since for 3, 6 ∈ N, 3 R 6 ≠ 6 R 3.

R is transitive since for n, m, r whenever $\dfrac{n}{m}$ and $\dfrac{m}{r}$

⇒ $\dfrac{n}{r}$, i.e., n divides m and m divides r, then n will divide r.

Example 18. Let L denote the set of all straight lines in a plane. Let a relation R be defined by lRm if and only if l is perpendicular to m ∀ l, m ∈ L. Then R is

(A) reflexive

(B) symmetric

(C) transitive

(D) none of these

Ans: Since Two perpendicular lines are perpendicular to each other.

If l⊥m⇒m⊥l

So, Given relation is transitive.

The correct choice is (B).

Example 19. Let N be the set of natural numbers and the function f : N → N be defined by f (n) = 2n + 3 ∀ n ∈ N. Then f is

(A) surjective

(B) injective

(C) bijective

(D) none of these

Ans: (B) is the correct option

Here for different values for n the f(n) gives a unique value, that’s why it is a one-one function.

So the given function is an injective function.

Example 20. Set A has 3 elements and set B has 4 elements. Then the number of injective mappings that can be defined from A to B is

(A) 144

(B) 12

(C) 24

(D) 64

Ans: The correct choice is (C).

The total number of injective mappings from the set containing 3 elements into the set containing 4 elements is $^{4}P_{3}$ = 4! = 24

Example 21. Let f: R → R be defined by f (x) = sin x and g: R → R be defined by g (x) = x2, then fog is

(A) x2 sin x

(B) ${\left( {{\mathbf{sinx}}} \right)^2}$

(C ) sin ${{\mathbf{x}}^2}$

(D) $\dfrac{{{\mathbf{sin}}{\mathbf{x}}}}{{{{\mathbf{x}}^2}}}$

Ans: (C) is the correct choice.

Given: f(x)=sin x and g(x)=x

Since, fog=f(g(x)=sin(g(x))=sin(x2)

Thus, option (C) is the correct answer.

Example 22. Let f : R → R be defined by f (x) = 3 x- 4. Then ${{\mathbf{f}}^{ - 1}}$(x) is given by

(A) $\dfrac{{{\mathbf{x}} + 4}}{3}$

(B) $\dfrac{{\mathbf{x}}}{3} - 4$

(C) 3x + 4

(D) None of these

Ans: (C) is the correct choice.

Let f(x)=y=3x-4

$\Rightarrow y=3x-4$

$\Rightarrow x =\dfrac{y+4}{3}$

$\Rightarrow f^{-1}{y} =\dfrac{y+4}{3}$

$\Rightarrow f^{-1}{x} =\dfrac{x+4}{3}$

Thus option (C) is the correct answer.

Example 23. Let f : R → R be defined by f (x) = ${{\mathbf{x}}^2}$ + 1. Then, pre-images of 17 and – 3, respectively, are

(A) φ, {4, – 4}

(B) {3, – 3}, φ

(C) {4, –4}, φ

(D) {4, – 4, {2, – 2}

Ans: (C) is the correct choice since for ${{{f}}^{ - 1}}$(17) = x

⇒ f (x) = 17 or ${{{x}}^2}$ + 1 = 17

⇒ x = ± 4 or ${{{f}}^{ - 1}}$(17) = {4, – 4} and for ${{{f}}^{ - 1}}$(–3) = x

⇒ f (x) = – 3 ⇒ ${{{x}}^2}$ + 1 = – 3

⇒ ${{{x}}^2}$ = – 4 and hence ${{{f}}^{ - 1}}$(– 3) = φ.

Example 24. For real numbers x and y, define xRy if and only if x – y + $\sqrt 2 $ is an irrational number. Then the relation R is

(A) reflexive

(B) symmetric

(C) transitive

(D) none of these

Ans: (A) is the correct choice.

Here we will have to choose such values of x and y where we remain with only $\sqrt{2}$ which is an irrational number. So for that x=y.

Hence, The relation is a reflexive relation.

Example 25. Consider the set A = {1, 2, 3} and R be the smallest equivalence relation on A, then R = ________

Ans: R = {(1, 1), (2, 2), (3, 3)}.

Example 26. The domain of the function f: R$ \to $R defined by f(x) = $\sqrt {{{\mathbf{x}}^2} - 3{\mathbf{x}} + 2}$ is __________.

Ans: Here ${{{x}}^2} - {{3x + 2}}$ $ \geqslant $ 0

$ \Rightarrow $ (x- 1) (x- 2) $ \geqslant $ 0

$ \Rightarrow $ x $ \leqslant $ 1 or x $ \geqslant $ 2

Hence, the domain of f = (– ∞, 1] ∪ [2, ∞).

Example 27. Consider the set A containing n elements. Then, the total number of injective functions from A onto itself is ________.

Ans: n!

Example 28. Let Z be the set of integers and R be the relation defined in Z such that aRb if a – b is divisible by 3. Then R partitions the set Z into ________ pairwise disjoint subsets.

Ans: Three.

Example 29. Let R be the set of real numbers and * be the binary operation defined on R as a * b = a + b – ab ∀ a, b ∈ R. Then, the identity element with respect to the binary operation * is _______.

Ans: 0 is the identity element with respect to the binary operation *.

### State True or False for the statements in each of the Examples 30 to 34

Example 30. Consider the set A = {1, 2, 3} and the relation

R = {(1, 2), (1, 3)}. R is a transitive relation.

Ans: True.

Example 31. Let A be a finite set. Then, each injective function from A into itself is not subjective.

Ans: False.

Example 32. For sets A, B and C, let f : A → B, g : B → C be functions such that g o f is injective. Then both f and g are injective functions.

Ans: False.

Example 33. For sets A, B and C, let f : A → B, g : B → C be functions such that g o f is subjective. Then g is subjective

Ans: True.

Example 34. Let N be the set of natural numbers. Then, the binary operation * in N defined as a * b = a + b, ∀ a, b ∈ N has identity element.

Ans: False.

### Exercise

### Short Answer Questions

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.

Ans: Give relation, R={(a, a),(b, c),(a, b)}

To make R reflexive we must add (b, b) and (c, c) to R.

Also, to make R transitive we must add (a, c) to R.

So, the minimum number of ordered pairs to be added is 3.

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

Ans: Given real valued function f(x), such that f(x) = $\sqrt {{{25}} - {{{x}}^2}{{}}} $, since, f(x) is real valued.

We must have

${{25}} - {{{x}}^2}$ $ \geqslant {{0}}$

$ \Rightarrow {{{x}}^2} \leqslant {{25}}$

$ \Rightarrow {{ - 5}} \leqslant {{x}} \leqslant {{5}}$

The domain is D = [-5, 5]

3. Let f, g : R → R be defined by f (x) = 2x + 1 and g (x) = x2-2, ∀ x ∈ R, respectively. Then, find g o f.

Ans: Given that f(x) = 2x + 1and g (x) = x2-2, ∀ x ∈ R.

Then (gof)x = g{f(x)}= g(2x + 1) = (2x + 1)2-2

= 4${{{x}}^2}$ + 4x +1-2

= 4${{{x}}^2}$ + 4x - 1

4. Let f : R → R be the function defined by f (x) = 2x – 3 ∀ x ∈ R. write f-1.

Ans: Given f(x) = 2 x - 3 ∀ x ∈ R

Now, lwt a, b ∈ R such that

f(a) = f(b)

$ \Rightarrow {{2a - 3}} = {{2b - 3}}$

$ \Rightarrow {{2a = 2b}}$

$ \Rightarrow $ f(x) is one- one.

Also, if x, y ∈ R , such that,

f(x) = y $ \Rightarrow $ 2x - 3 = y

$ \Rightarrow $ x = $\dfrac{{{{y}} + {{3}}}}{{{2}}}$= g(y) ∀ y ∈ R

$ \Rightarrow $ f(x) is onto and therefore is bijective implies f(x) has an inverse

f-1(x) = g(x) = $\dfrac{{{{x}} + {{3}}}}{{{2}}}$ ∀ x ∈ R

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

Ans: Let f:A $ \to $A , then inverse ${{{f}}^{ - 1}}$ is such that

f-1 = {(b, a), (d, b), (a, c), (c, d)}

f-1 : A $ \to $ A and f-1 = {(b, a), (d, b), (a, c), (c, d)}

6. If f: R $ \to $ R is defined by f(x) = x2-3x+2, write f(f(x)).

Ans: Given that f(x) = x2-3x+2

f(f(x)) = f(x2-3x+2)

= (x2-3x+2)2 - 3(x2-3x+2) + 2

= x4+9x2+4-6x3-12x+4x2-3x2+9x-6+2

= x4+10x2-6x3-3x

f(f(x)) = x4+10x2-6x3-3x

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

Ans: Yes g is a function since every element in the domain has a unique image.

Now, let g(x) = ${{\alpha }}$x + ${{\beta }}$

Then given,

g(1) = ${{\alpha }}$ + ${{\beta }}$ = 1

g(2) = 2${{\alpha }}$ + ${{\beta }}$ = 3

Subtracting g(1) from g(2) gives,

(2${{\alpha }}$ + ${{\beta }}$) - (${{\alpha }}$ + ${{\beta }}$) = ${{\alpha }}$ = 2 and substituting it into g(1)

We have ${{\beta }}$ = -1.

So, values of ${{\alpha \text{ and }\beta }}$ are 2 and -1.

8. Are the following set of ordered pair’s functions? If so, examine whether the mapping is injective or subjective.

(i) {(x, y) : x is a person, y is the mother of x}

Ans: Given set of ordered pairs is:

{(x, y) : x is a person, y is the mother of x}.

It represents a function. Here the image of distinct elements of x under f are not distinct, so it's not injective but it is subjective.

(ii) {(a, b) : a is a person , b is the ancestor of a}

Ans: Given set of ordered pairs is:

{(a, b): a is a person, b is the ancestor of a}

Here, each element of the domain does not have a unique image.

So, it does not represent function.

9. If the mapping f and g are given by f = {(1, 2), (3, 5), (4, 1)} and g ={ (2,3), (5,1), (1,3)}, write fog.

Ans: Given that,

f = {(1, 2), (3, 5), (4, 1)}

And g={(2, 3), (5, 1), (1, 3)}

Now, fog(2) = f{g(2)} = f(3) = 5

fog(5) = f{g(5)} = f(1) = 2

fog(1) = f{g(1)} = f(3) = 5

fog = {(2,5),(5,2),(1,5)}.

10. Let C be the set of complex numbers. Prove that the mapping f: C $ \to $R given by f(z) = $\left| {\mathbf{z}} \right|$,∀ z ∈ C, is neither one-one nor onto.

Ans: The mapping f : C $ \to $ R

Given, f(z) = $\left| {{z}} \right|$,∀ z ∈ C

f(1) = |1| = 1

f(-1) = |-1| = 1

f (1) = f(-1)

But, 1 $ \ne $ -1

So, f(z) is not one-one. Also, f(z) is not onto as there is no pre-image for any negative element of R under the mapping f(z).

11. Let the function f : R $ \to $ R be defined by f(x) = cos x , ∀ x ∈ R. Show that f is neither one-one nor onto.

Ans: Given function f(x) = cos x, ∀ x ∈ R

Now, f\[\left( {\dfrac{\pi }{2}} \right)\] = cos\[\left( {\dfrac{\pi }{2}} \right)\] = 0

f\[\left( {\dfrac{{ - \,\pi }}{2}} \right)\] = cos\[\left( {\dfrac{\pi }{2}} \right)\] = 0

f\[\left( {\dfrac{\pi }{2}} \right)\] = f\[\left( {\dfrac{{ - \,\pi }}{2}} \right)\]

f\[\left( {\dfrac{{ - \,\pi }}{2}} \right)\] = f\[\left( {\dfrac{\pi }{2}} \right)\]

But, \[\dfrac{\pi }{2} \ne \dfrac{{ - \,\pi }}{2} = 0\]

So, f(x) is not one-one.

Now, f(x) = cos x, ∀ x ∈ R is not onto as there is no pre- image for any real number, which does not belong to the intervals [-1, 1], the range of cos x.

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

(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)}

Ans: f is not a function because f has no unique image.

(ii) g = {(1, 4), (2, 4), (3, 4)}

Ans: g is a function as each element of the domain has a unique image.

(iii) h = {(1,4), (2, 5), (3, 5)}

Ans: h is a function

(iv) k = {(1,4), (2, 5)}.

Ans: k is not a function as 3 does not have any image under the mapping.

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

Ans: Given that f: A $ \to $B and g: B$ \to $A satisfy gof= ${{{I}}_{{A}}}$

Since, gof = IA,

$ \Rightarrow $ gof { f(${{{x}}_1})\} $ = gof {f(${{{x}}_2})\} $

$ \Rightarrow $ g(${{{x}}_1}){{}}$= g(${{{x}}_2}){{}}$[ Since gof = IA]

$ \Rightarrow $ ${{{x}}_1}$ = ${{{x}}_2}$

Hence, f is one-one and g is onto.

14. Let f:R $ \to $R be the function defined f(x) = $\dfrac{1}{{2 - {\mathbf{cos}}{\mathbf{x}}}},$∀ x ∈ R. Then find the range of f.

Ans: Given function, f(x) = $\dfrac{{{1}}}{{{{2}} - {{cosx}}}},$∀ x ∈ R

Let y = $\dfrac{{{1}}}{{{{2}} - {{cosx}}}}$

$ \Rightarrow $ 2y - y cos x = 1

$ \Rightarrow $ y cos x = 2y - 1

$ \Rightarrow $ cos x = $\dfrac{{{{2y - 1}}}}{{{y}}} = {{2}} - \dfrac{{{1}}}{{{y}}}$

$ \Rightarrow $ cos x = ${{2}} - \dfrac{{{1}}}{{{y}}}$

$ \Rightarrow $ -1 $ \leqslant $ cos x $ \leqslant $ 1

$ \Rightarrow $ -1 $ \leqslant $ ${{2}} - \dfrac{{{1}}}{{{y}}} \leqslant $ 1

$ \Rightarrow $ -3 $ \leqslant $ $ - \dfrac{{{1}}}{{{y}}} \leqslant $ -1

$ \Rightarrow $ 3 $ \geqslant \dfrac{{{1}}}{{{y}}} \geqslant $ 1

$ \Rightarrow $ $\dfrac{{{1}}}{{{3}}}$ $ \leqslant $ ${{y}} \leqslant $ 1

So, the range of y, that is f(x) is $\left[ {\dfrac{{{1}}}{{{3}}},\,{{1}}} \right]$.

15. Let n be a fixed positive integer. Define a relation R in Z as follows: ∀ a, b ∈ Z, aRb if and only if a – b is divisible by n. Show that R is an equivalence relation.

Ans: Given that,

∀ a, b ∈ Z, aRb if and only if a – b is divisible by n.

Now,

Reflexive

aRb $ \Rightarrow $ (a-a) is divisible by n , which is true for any integer ‘a’ as ‘0’ is divisible by n. Hence, r is reflexive.

Symmetric

aRb

$ \Rightarrow $ a - b is divisible by n.

$ \Rightarrow $ - b+ a is divisible by n.

$ \Rightarrow $ - (b- a) is divisible by n.

$ \Rightarrow $ (b- a) is divisible by n.

bRa

Hence, R is symmetric.

Transitive

Let aRb and bRc

$ \Rightarrow $ (a - b) is divisible by n and (b-c) is divisible by n.

$ \Rightarrow $ (a - b) + (b-c) is divisible by n.

$ \Rightarrow $ (a - c) is divisible by n.

aRc

Hence, R is transitive.

So, R is an equivalence relation.

### Long Answers:

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

(a) reflexive, transitive but not symmetric

Ans: Given that: A = {1, 2, 3, 4}

Let R1 = {(1, 1), (1, 2), (2, 3), (2, 2), (1, 3), (3, 3)}

R1 is reflexive, since (1, 1), (2, 2), (3, 3) lies in R1

Now, (1, 2) R1, (2, 3) R1 \[ \Rightarrow \](1, 3) R1

Hence, R1 is also transitive but (1, 2) R1 \[ \Rightarrow \] (2, 1) $ \notin $ R1

So, it is not symmetric.

(b) symmetric but neither reflexive nor transitive

Ans: Given that: A = {1, 2, 3, 4}

Let R2 = {(1, 2), (2, 1)}

Now, (1,2) , (2,1)

So, it is symmetric.

(c) reflexive, symmetric and transitive.

Ans: Given that: A = {1, 2, 3, 4}

Let ${{{R}}_3}$= {(1, 2), (2, 1), (1, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 3)}

Hence, ${{{R}}_3}$ is reflexive, symmetric and transitive.

17. Let R be relation defined on the set of natural numbers N as follows:

R = {(x, y): x ∈N, y ∈N, 2x + y = 41}.

Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive.

Ans: Given that: R = {(x, y): x ∈N, y ∈N, 2x + y = 41}.

Domain = {1, 2, 3 …..20}

Range = {(1, 39), (2, 37), (3, 35) ...... (19, 3), (20, 1)}

R is not reflexive as (2, 2) $ \notin {{R}}$

2 x 2 + 2 $ \ne $ 41

So, R is not symmetric.

As, (1, 39) ∈ R but (39, 1) $ \notin $ R

So, R is not transitive

As (11, 19) ∈ R, (19, 3) ∈ R but (11, 3) $ \notin $ R

Hence, R is neither reflexive, nor symmetric nor transitive.

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

Ans: Given that, A = {2, 3, 4}, B = {2, 5, 6, 7}.

Let f : A $ \to $ B denote a mapping

f = {(x, y) : y = x + 3}

I.e f = {(2, 5), (3, -6), (4, 7)}

Which is an injective mapping.

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

Ans: Given that A = {2, 3, 4}, B = {2, 5, 6, 7}.

Let g : A $ \to $ B denote a mapping

Such that g = {(2, 2), (3, 5), (4, 7)} which is not an injective mapping.

(c) a mapping from B to A.

Ans: Given that A = {2, 3, 4}, B = {2, 5, 6, 7}.

Let h : B $ \to $ A denote a mapping

Such that h = { (2,2),(5,3),(6,4),(7,4)} which is a mapping from B$ \to {{A}}$.

19. Give an example of a map

(i) which is one-one but not onto

Ans: Let f : N $ \to $ N, be a mapping defined by f(x) = 2x which is one-one

For f(x1) = f(x2)

$ \Rightarrow {{2}}{{{x}}_1} = {{2}}{{{x}}_2}$

$ \Rightarrow {{}}{{{x}}_1} = {{}}{{{x}}_2}$

Further f is not onto, as for 1$ \in $ N, there does not exist any x in N such that f(x) = 2x + 1.

(ii) which is not one-one but onto

Ans: Let f: N$ \to {{N}}$ , given by f(1) = f(2) = 1 and f(x) = x- 1 for every x> 2 is onto but not one-one. f is not one-one as f(1)= f(2) = 1. But f is onto.

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

Ans: The mapping f: R$ \to {{R}}$ defined by f(x) = ${{{x}}^2}{{}},$ is neither one -one nor onto.

20. Let A = R – {3}, B = R – {1}. Let f : A → B be defined by f (x) = $\dfrac{{{\mathbf{x}} - 2}}{{{\mathbf{x}} - 3}}$ ∀ x ∈ A . Then show that f is bijective.

Ans: Given that, A = R- {3}, B = R -{1}.

f : A $ \to $ B is defined by f(x) = $\dfrac{{{{x - 2}}}}{{{{x - 3}}}}$ ∀ x ∈ A.

For injectivity

Let f(x1) = f(x2)

$ \Rightarrow $ $\dfrac{{{{{x}}_1} - {{2}}}}{{{{{x}}_1} - {{3}}}}$ = $\dfrac{{{{{x}}_2} - {{2}}}}{{{{{x}}_2}{{}} - {{3}}}}$

$ \Rightarrow $ (${{{x}}_1} - {{2}}$)(${{{x}}_2}{{}} - {{3}}$) = (${{{x}}_2} - {{2}}$)(${{{x}}_1}{{}} - {{3}}$)

$ \Rightarrow $ ${{{x}}_1}{{{x}}_2}$ - 3${{{x}}_1}$- 2${{{x}}_2}$+ 6 = ${{{x}}_1}{{{x}}_2}$- 3 ${{{x}}_2}$- 2 ${{{x}}_1}$ + 6

$ \Rightarrow $ - 3${{{x}}_1}$- 2${{{x}}_2}$ = -3${{{x}}_2}$- 2 ${{{x}}_1}$

$ \Rightarrow $ - ${{{x}}_1}$= -${{{x}}_2}$

$ \Rightarrow $ ${{{x}}_1}$= ${{{x}}_2}$

So, f(x) is an injective function.

For subjectivity,

Let y = $\dfrac{{{{x - 2}}}}{{{{x - 3}}}}$

$ \Rightarrow $ x - 2 = xy - 3y

$ \Rightarrow $ x (1-y) = 2 - 3y

$ \Rightarrow $ x = $\dfrac{{{{2 - 3y}}}}{{{{1}} - {{y}}}}$

$ \Rightarrow $ x = $\dfrac{{{{2}} - {{3y}}}}{{{{1}} - {{y}}}}$ $ \in $ A ∀ y ∈ B (codomain)

So, f(x) is a subjective function.

Hence, f(x) is a bijective function.

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

(i) f(x) = $\dfrac{{\mathbf{x}}}{2}$

Ans: Given : A = [-1,1]

f(x) = $\dfrac{{{x}}}{{{2}}}$

Let f(${{{x}}_1}$) = f(${{{x}}_2}$)

$ \Rightarrow $ $\dfrac{{{{{x}}_1}}}{{{2}}}$= $\dfrac{{{{{x}}_2}}}{{{2}}}$

$ \Rightarrow $ ${{{x}}_1}$ = ${{{x}}_2}$

So, f(x) is one-one.

$ \Rightarrow $ x = 2y $ \notin $ A, ∀ y ∈ A

As for y = 1 ∈ A, x = 2 $ \notin {{A}}$

So, f(x) is not onto.

Also, f(x) is not bijective as it is not onto.

(ii) g(x) = |x|

Ans: Let g(x1) = g (x2)

$ \Rightarrow $ |x1| = |x2|

$ \Rightarrow $ x1 = $ \pm $ x2

So, g(x) is not one-one.

Now, y = $\left| {{x}} \right| \Rightarrow $ x = $ \pm $ y $ \notin $ A, ∀ y ∈ A

So, g(x) is not onto also, g(x) is not bijective.

(iii) h (x) = x |x|

Ans: h (x) = x|x|

$ \Rightarrow $ x1|x1|=x2|x2|

$ \Rightarrow $ x1 = x2

So, h(x) is one-one.

Now, let y = x|x|

$ \Rightarrow $ y = ${{{x}}^{2{{}}}}$∈ A, ∀ x ∈A

So, h(x) is onto also, h(x) is bijective.

(iv) k(x) = x2

Let k(${{{x}}_1}$) = k(${{{x}}_2}$)

$ \Rightarrow $ ${{{x}}_1}^{2{{}}}$ = ${{{x}}_2}^{2{{}}}$

$ \Rightarrow $ ${{{x}}_1}$ = $ \pm {{}}{{{x}}_2}$

Thus, k(x) is not one-one.

Now, let y = ${{{x}}^2}$

$ \Rightarrow $ x = $\sqrt {{y}} $ $ \in $ A, ∀ y ∈A

$ \Rightarrow $ x = $\sqrt {{y}} $ $ \notin $ A, ∀ y ∈A

As for y = -1, x = $\sqrt { - {{1}}} $ $ \notin $ A

Hence, k(x) is neither one-one nor onto.

22. Each of the following defines a relation on N. Determine which of the above relations are reflexive, symmetric, and transitive.

(i) x is greater than y, x, y ∈ N

Ans: x is greater than y, x, y ∈ N

(x, x) ∈ R

For xRx, x > x is not true for any x ∈ N

Therefore, R is not reflexive.

Let (x, y) ∈ R $ \Rightarrow $ xRy

x > y

But y > x is not true for any x, y ∈ N

Thus, R is not symmetric.

Let xRy and yRz

x > y and y > z $ \Rightarrow $ x > z

$ \Rightarrow $ x R z

So, r is transitive.

(ii) x + y = 10, x, y ∈ N

Ans: x + y = 10, x, y ∈ N

R = {(x, y); x+y = 10x, y ∈ N}

R = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)} (1, 1) $ \notin $ R

So, R is not reflexive.

(x, y) ∈ R $ \Rightarrow $ (y,x) ∈ R

Therefore, R is symmetric.

(1, 9) ∈ R, (9, 1) ∈ R $ \Rightarrow $ (1, 1)$ \notin $ R

Hence, R is not transitive.

(iii) x y is the square of an integer x, y ∈ N

Ans: Given xy is the square of the integer. x, y ∈ N

$ \Rightarrow $ R = {(x,y): xy is a square of an integer x, y ∈ N}

(x, x) ∈ R, ∀ x ∈ N

As x2 is the square of an integer for any x ∈ N

Hence, R is reflexive.

If (x, y) ∈ R $ \Rightarrow $ (y, x) ∈ R

Therefore, R is symmetric.

If (x, y) ∈ R, (y, z) ∈ R

So, xy is square of an integer and yz is square of an integer.

Let xy = ${{{m}}^2}$ and yz = ${{{n}}^2}$ for some m, n ∈ Z

x = $\dfrac{{{{{m}}^2}}}{{{Y}}}$ and z = $\dfrac{{{{{n}}^2}}}{{{Y}}}$

xz = $\dfrac{{{{{m}}^2}{{{n}}^2}}}{{{Y}}}$ , which is square of an integer.

So, R is transitive.

(iv) x + 4y = 10 x, y ∈ N.

Ans: R = {(x, y): x + 4y = 10, x, y ∈ N}

R = {(2, 2), (6, 1)}

(1, 1), (3, 3) ……. $ \notin $ R

Thus, R is not reflexive.

(6, 1) ∈ R but (1, 6) $ \notin $ R

Hence, R is not symmetric.

(x, y) ∈ R $ \Rightarrow $ x + 4y = 10 but (y, z) ∈ R

y + 4 z = 10 $ \Rightarrow $ (x, z) ∈ R

So, R is transitive.

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

Ans: Given that A = {1, 2, 3, ... 9}

(a, b) R (c, d) if a + d = b + c for (a, b), (c, d) ∈ A × A

So, (a, b) ∈ A × A

(c, d ) ∈ A × A

Let (a, b) R (a, b)

$ \Rightarrow {{}}$a + b = b + a, ∀ a, b ∈A which is true for any a, b ∈ A.

Hence, R is reflexive.

Let (a, b) R (c, d)

a + b = b + c

c + b = d + a

$ \Rightarrow $ (c, d) R (a, b)

So, R is symmetric.

Let (a, b) R (c, d) and (c, d) R (e, f)

$ \Rightarrow $ a + d = b + c and c + f = d+ e

$ \Rightarrow $ a + d = b + c and d+ e = c + f

$ \Rightarrow $ (a + d) - (d + e) = (b + c) - (c + f)

$ \Rightarrow $ (a - e) = b- f

$ \Rightarrow $ a + f = b + e

$ \Rightarrow $ (a, b) R (e, f)

Now, equivalence class containing

[(2,5) is { (1,4) ,(2,5),(3,6),(4,7),(5,8),(6,9)}].

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

Ans: A function f : x $ \to $ y is defined to be invertible, if there exist a function

g = y $ \to $ x such that gof = ${{{I}}_{{x}}}{{}}$and fog = ${{{I}}_{{y}}}$. The function is called the inverse of f and is denoted by ${{{f}}^{ - 1}}.$

A function f = x $ \to $ y is invertible if f is a bijective function.

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

(i) f o g

Ans: Given that f(x) = ${{{x}}^2}$ + 3x + 1, g(x) = 2 x - 3

fog = f{g(x)} = f (2 x- 3)

= ${\left( {2{{x}} - {{}}3} \right)^2}{{}} + {{}}3\left( {2{{x}} - 3} \right){{}} + 1{{}}$

= 4 ${{{x}}^2}$ + 9 - 12 x + 6x -9x + 1

= 4 ${{{x}}^2}$- 6 x + 1

(ii) g o f

Ans: Given that f(x) = ${{{x}}^2}$ + 3x + 1, g(x) = 2 x - 3

gof = g { f(x) } = g ( ${{{x}}^2}$ + 3x + 1)

= 2( ${{{x}}^2}$ + 3x + 1 )- 3

= 2 ${{{x}}^2}$ + 6 x + 2 - 3

= 2 ${{{x}}^2}$ + 6 x - 1

(iii) f o f

Ans: Given that f(x) = ${{{x}}^2}$ + 3x + 1, g(x) = 2 x - 3

fof = f{f(x)} = f( ${{{x}}^2}$ + 3x + 1)

= ${\left( {{{{x}}^2}{{}} + {{}}3{{x}} + {{}}1} \right)^2}{{}} + {{}}3\left( {{{{x}}^2}{{}} + {{}}3{{x}} + {{}}1} \right){{}} + 1{{}}$

= ${{{x}}^4} + 9{{}}{{{x}}^2}$ + 1 + 6 ${{{x}}^3}$ + 6x + 2 ${{{x}}^2}$ + 3 ${{{x}}^2}$+ 9 x + 3+ 1

= ${{{x}}^4}$+ 6 ${{{x}}^3}$+ 14 ${{{x}}^2}$+ 15 x + 5

(iv) g o g

Ans: Given that f(x) = ${{{x}}^2}$ + 3x + 1, g(x) = 2 x - 3

gog = g{g(x)} = g( 2 x- 3)

= 2(2 x -3)-3

= 4 x - 6 - 3

= 4 x - 9

26. Let * be the binary operation defined on Q. Find which of the following binary operations are commutative

(i) a * b = a – b ∀ a, b ∈Q

Ans: a * b = a - b ∀ a, b ∈ Q and b * a = b- a

So, a * b $ \ne $ b * a $\left[Since, b-a \ne a - b\right]$

Hence, * is not commutative.

(ii) a * b = ${{\mathbf{a}}^2} + {{\mathbf{b}}^2}$ ∀ a, b ∈ Q

Ans: a * b = ${{{a}}^2} + {{{b}}^2}{{}}$ ∀ a, b ∈ Q

b * a = ${{{b}}^2} + {{{a}}^2}{{}}$

Hence, * is commutative. (Since, + is on rational is commutative).

(iii) a * b = a + ab ∀ a, b ∈ Q

Ans: a * b = a + ab ∀ a, b ∈ Q

b * a = b + ab

Clearly, a + ab $ \ne $ b + ab

Hence, * is not commutative.

(iv) a * b = ${\left( {{\mathbf{a}} - {\mathbf{b}}} \right)^2}$ ∀ a, b ∈ Q

Ans: a * b = (a-b)2 ∀ a, b ∈ Q

b * a = (b-a)2

Since, (a-b)2=(b-a)2

Hence, * is commutative.

27. Let * be binary operation defined on R by a * b = 1 + ab, ∀ a, b ∈ 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

Ans: Option (i) is the correct answer.

Given that a * b = 1 + ab, ∀ a, b ∈ R

a * b = ab + 1 = b * a

So, * is a commutative binary operation

Also, a* (b* c) = a * (1+ bc) = 1 + a (1+ bc)

a* (b* c) = 1 + a + abc ………. (1)

(a* b*) c = (1 + ab) * c

= 1 + (1+ab)c = 1 + c + abc ……….. (2)

From (1) and (2)

a* (b* c) $ \ne $ (a* b)* c

So, * is not associative

Hence, * is commutative but not associative.

### Objective Type Questions

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 ∀ a, b ∈ T. Then R is

(A) reflexive but not transitive

(B) transitive but not symmetric

(C) equivalence

(D) none of these

Ans: Option (C) is the correct answer.

Consider that aRb, if a is congruent to b, ∀ a, b ∈ T.

Then, aRa $ \Rightarrow $ a $ \simeq $ a,

Which is true for all a ∈ T.

So, R is reflexive ----------- (i)

Let aRb $ \Rightarrow $ a $ \simeq $b,

$ \Rightarrow $ b $ \simeq $a

$ \Rightarrow $ bRa

So, R is symmetric --------- (ii)

Let aRb and bRc

$ \Rightarrow $ a $ \simeq $ b and b $ \simeq $ c

$ \Rightarrow $ a $ \simeq $ c

$ \Rightarrow $ aRc

So, R is transitive ----------- (iii)

Hence, R is an equivalence relation.

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

(A) symmetric but not transitive

(B) transitive but not symmetric

(C) neither symmetric nor transitive

(D) both symmetric and transitive

Ans: Option (B) is the correct answer.

Given: aRb $ \Rightarrow $ a is the brother of b.

Therefore aRa $ \Rightarrow $ a is the brother of a, which is not true.

So, R is not reflexive.

aRb $ \Rightarrow $ a is the brother of b.

This doesn't mean b is also a brother of an as b can be the sister of a.

Hence, R is not symmetric.

aRb $ \Rightarrow $ a is the brother of b.

And, bRc $ \Rightarrow $ b is the brother of c.

So, a is the brother of c

Hence, R is transitive.

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

(A) 1

(B) 2

(C) 3

(D) 5

Ans: Option (D) is the correct answer.

Given that, A= {1, 2, 3}

Now, number of equivalence relations are as follows:

${{{R}}_1} = {{}}\left\{ {{{}}\left( {{{1,1}}} \right){{,}}\left( {{{2,2}}} \right){{,}}\left( {{{3,3}}} \right)} \right\}$

${{{R}}_2} = {{}}\left\{ {{{}}\left( {{{1,1}}} \right){{,}}\left( {{{2,2}}} \right){{,}}\left( {{{3,3}}} \right){{,}}\left( {{{1,2}}} \right){{,}}\left( {{{2,1}}} \right)} \right\}$

${{{R}}_3} = {{}}\left\{ {{{}}\left( {{{1,1}}} \right){{,}}\left( {{{2,2}}} \right){{,}}\left( {{{3,3}}} \right){{,}}\left( {{{1,3}}} \right){{,}}\left( {{{3,1}}} \right)} \right\}$

${{{R}}_{{4}}}{{ = }}\left\{ {{{}}\left( {{{1,1}}} \right){{,}}\left( {{{2,2}}} \right){{,}}\left( {{{3,3}}} \right){{,}}\left( {{{2,3}}} \right){{,}}\left( {{{3,2}}} \right)} \right\}$

${{{R}}_5} = {{}}\left\{ {{{}}\left( {{{1,2,3}}} \right) \Leftrightarrow {{AXA}} = {{}}{{{A}}^2}} \right\}$

Maximum number of equivalence relation on the set A = {1, 2, 3} = 5.

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

(A) reflexive

(B) transitive

(C) symmetric

(D) none of these

Ans: The correct answer is option (B).

R on the set {1, 2, 3} be defined by R = {(1, 2)}

It is clear that R is transitive.

32. Let us define a relation R in R as aRb if a ≥ b. Then R is

(A) an equivalence relation

(B) reflexive, transitive but not symmetric

(C) symmetric, transitive but

(D) neither transitive nor reflexive, not reflexive but symmetric.

Ans: The correct answer is option (B).

Given that, aRb if a$ \geqslant $b

$ \Rightarrow $ aRa $ \Rightarrow $ a$ \geqslant $a , which is true.

Let aRb, a$ \geqslant $b, then b $ \geqslant $a which is not true R is not symmetric.

But aRb and bRc

$ \Rightarrow $ a$ \geqslant $b and b$ \geqslant $c

$ \Rightarrow $ a$ \geqslant $c

Hence, R is transitive.

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

(A) reflexive but not symmetric

(B) reflexive but not transitive

(C) symmetric and transitive

(D) neither symmetric, nor transitive.

Ans: The correct answer is option (A)

Given that A = {1, 2, 3}

And R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}.

Since, (1, 1), (2, 2), (3, 3) $ \in $ R

Hence, R is reflexive.

(1, 2) $ \in $ R but (2, 1) $ \notin $ R

Hence, R is not symmetric.

(1, 2) $ \in $ R and (2, 3) $ \in $ R

$ \Rightarrow $ (1, 3) $ \in $ R

Hence, R is transitive.

34. The identity element for the binary operation * defined on Q ~ {0} as a * b = $\dfrac{{{\mathbf{ab}}}}{2}$, ∀ a, b ∈ Q ~ {0} is

(A) 1

(B) 0

(C) 2

(D) none of these

Ans: Option (C) is the correct answer.

Given that a * b = $\dfrac{{{{ab}}}}{{{2}}}$ , ∀ a, b ∈ Q ~ {0}

Let e be the identity element for *

Therefore, a * e = $\dfrac{{{{ae}}}}{{{2}}}$ (a* e = e* a= a)

$ \Rightarrow $ a = $\dfrac{{{{ae}}}}{{{2}}}$

$ \Rightarrow $ e = 2

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

(A) 720

(B) 120

(C) 0

(D) none of these

Ans: Option (C) is the correct answer.

We know that, if A and B are two non- empty finite set containing m and n elements respectively, then the number of one-one and onto mapping from A to B is

n! If m = n

0, if m$ \ne $ n

Given that m = 5 and n = 6

Therefore, m $ \ne $ n

Number of mapping = 0

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

(A) \[{}^{n}{P_2}\]

(B) 2n-2

(C) 2n – 1

(D) None of these

Ans: Option (A) is the correct answer.

Given that:

A = {1, 2, 3, ...n} and B = {a, b}

We know that, if A and B are two non- empty finite set containing

m and n elements respectively, then the number of surjections from A to B is

\[^n{C_m} \times m!\]

(if n $ \geqslant {{m}}$ ),

0, if m < n.

Here, m = 2.

\[^n{C_m} \times 2!\]

$= {{}}\dfrac{{{{n}}!}}{{2!{{}}\left( {{{n}} - 2} \right)!}}$ \[ \times \] 2!

$= $ \[^n{C_2}\]

$= {{}}\dfrac{{{{n}}\left( {{{n}} - {{1}}} \right)\left( {{{n}} - {{2}}} \right)!}}{{{{2}} \times {{1}}\left( {{{n}} - {{2}}} \right)!}}$ \[ \times \] 2!

$=$ n2 – n

37. Let f : R → R be defined by f (x) = $\dfrac{1}{{\mathbf{x}}}$, ∀ x ∈ R. Then f is

(A) one-one

(B) onto

(C) bijective

(D) f is not defined

Ans: Option (D) is the correct answer.

Given that, f (x) = $\dfrac{{{1}}}{{{x}}}$, ∀ x ∈ R.

For x = 0, f(x) is not defined.

Hence, f(x) is a not defined function.

38. Let f : R $ \to $ R be defined by f(x) = 3x2-5 and g : R $ \to $ R by g(x) = $\dfrac{{\mathbf{x}}}{{{{\mathbf{x}}^2} + 1}}.$ Then gof is:

(A) $\dfrac{{3{{\mathbf{x}}^2} - 5}}{{9{{\mathbf{x}}^4} - 30{{\mathbf{x}}^2} + 26}}$

(B) $\dfrac{{3{{\mathbf{x}}^2} - 5}}{{9{{\mathbf{x}}^4} - 6{{\mathbf{x}}^2} + 26}}$

(C) $\dfrac{{3{{\mathbf{x}}^2}}}{{{{\mathbf{x}}^4} + 2{{\mathbf{x}}^2} - 4}}$

(D) $\dfrac{{3{{\mathbf{x}}^2}}}{{9{{\mathbf{x}}^4} + 30{{\mathbf{x}}^2} - 2}}$

Ans: Option (A) is the correct answer.

Given that f(x) = 3${{{x}}^2} - {{5}}$ and g(x) = $\dfrac{{{x}}}{{{{{x}}^2}{{}} + {{1}}}}$

gof = g{f(x)}= g (3${{{x}}^2} - {{5}}$ )

= $\dfrac{{{{3}}{{{x}}^{{2}}}{{ - 5}}}}{{{{\left( {{{3}}{{{x}}^{{2}}}{{ - 5}}} \right)}^{{2}}}{{ + 1}}}}$

= $\dfrac{{{{3}}{{{x}}^{{2}}}{{ - 5}}}}{{{{9}}{{{x}}^{{4}}}{{ - 30}}{{{x}}^{{2}}}{{ + 25 + 1}}}}$

= $\dfrac{{{{3}}{{{x}}^{{2}}}{{ - 5}}}}{{{{9}}{{{x}}^{{4}}}{{ - 6}}{{{x}}^{{2}}}{{ + 26}}}}$

40. Let f: R$ \to $R be the functions defined by f(x) = ${{\mathbf{x}}^3} + 5$. Then ${{\mathbf{f}}^{ - 1}}\left( {\mathbf{x}} \right)$ is:

(A) ${({\mathbf{x}} + 5)^{\dfrac{1}{3}}}$

(B) ${({\mathbf{x}} - 5)^{\dfrac{1}{3}}}$

(C ) ${(5 - {\mathbf{x}})^{\dfrac{1}{3}}}$

(D) 5 - x

Ans: Option (B) is the correct answer.

Given that, f(x) = ${{{x}}^3}{{}} + {{5}}$

Let y = ${{{x}}^3}{{}} + {{5}}$

$ \Rightarrow $ x3 = y-5

$ \Rightarrow $ ${{x}} = {({{y}} - {{5}})^{\dfrac{1}{3}}}$

$ \Rightarrow $ ${{{f}}^{ - 1}}\left( {{x}} \right)$ = ${({{x}} - {{5}})^{\dfrac{1}{3}}}$

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

(A) ${{\mathbf{f}}^{ - 1}}$o ${{\mathbf{g}}^{ - 1}}$

(B) fog

(C) ${{\mathbf{g}}^{ - 1}}$o ${{\mathbf{f}}^{ - 1}}$

(D) gof

Ans: Option (A) is the correct answer.

Given that f : A $ \to $ B and g : B $ \to $ C be the bijective functions.

${\left( {{{gof}}} \right)^{ - 1}}$ = ${{{f}}^{ - 1}}$o ${{{g}}^{ - 1}}$

42. Let \[{{f:R - }}\left\{ {\dfrac{{{3}}}{{{5}}}} \right\} \to {{R}}\] be defined by f(x) = $\dfrac{{3{\mathbf{x}} + 2}}{{5{\mathbf{x}} - 3}}.$ Then

(A) ${{\mathbf{f}}^{ - 1}}$(x) = f(x)

(B) ${{\mathbf{f}}^{ - 1}}$(x) = -f(x)

(C) (fof)x = -x

(D) ${{\mathbf{f}}^{ - 1}}$ (x) = $\dfrac{1}{{19}}{\mathbf{f}}\left( {\mathbf{x}} \right)$

Ans: Option (A) is the correct answer.

Given that, f(x) = $\dfrac{{{{3x + 2}}}}{{{{5x - 3}}}}$

Let y = $\dfrac{{{{3x + 2}}}}{{{{5x - 3}}}}$

3 x + 2 = 5 x y - 3 y

$ \Rightarrow $ x (3-5y) = -3y - 2

x = $\dfrac{{{{3x + 2}}}}{{{{5x - 3}}}}$

$ \Rightarrow $ ${{{f}}^{ - 1}}$(x) = $\dfrac{{{{3x + 2}}}}{{{{5x - 3}}}}$

$ \Rightarrow $ ${{{f}}^{ - 1}}$(x) = f(x)

43. Let f : [0, 1] \[ \to \] [0, 1] be defined by

$f\left( x \right)= \left\{ \begin{align} & x,\text{if x is a rational number} \\ & 1-x,\text{if x is irrational}\\ \end{align} \right.$

Then (fof)x is

(A) Constant

(B) 1 + x

(C) x

(D) none of these

Ans: Option (C) is the correct answer.

Given that f : [0, 1] $ \to $ [0, 1] be defined by

$f\left( x \right)= \left\{ \begin{align} & x,\text{if x is a rational number} \\ & 1-x,\text{if x is irrational}\\ \end{align} \right.$

Therefore,

(fof)x = f(f(x)) = x

44. Let f : [2, ∞) → R be the function defined by f (x) = x2-4x+5, then the range of f is

(A) R

(B) [1, ∞)

(C) [4, ∞)

(D) [5, ∞)

Ans: Option (B) is the correct answer.

Given that, f (x) = ${{{x}}^2}$ – 4x + 5,

y = ${{{x}}^2}$ – 4x + 5

$ \Rightarrow $ y = x2 – 4x + 4 + 1 = (x-2)2 +1

$ \Rightarrow $ (x-2)2 = y - 1

$ \Rightarrow $ (x - 2) = $\sqrt {{{y}} - {{1}}} $

$ \Rightarrow $ x = 2 + $\sqrt {{{y}} - {{1}}} $

Therefore, (y -1) $ \geqslant $ 0, y $ \geqslant $ 1.

Range = [1, ∞)

45. Let f : N → R be the function defined by f (x) = $\dfrac{{2{\mathbf{x}} - 1}}{2}$ and g : Q → R be another function defined by g (x) = x + 2. Then (g o f) $\dfrac{3}{2}$ is

(A) 1

(B) 3

(C) $\dfrac{7}{2}$

(D) none of these

Ans: Option (B) is the correct answer.

Given that f (x) = $\dfrac{{{{2x - 1}}}}{{{2}}}$ and g(x) = x + 2

\[{(gof)^{\dfrac{3}{2}}} = g\left[ {f\left( {\dfrac{3}{2}} \right)} \right] = g\left( {\dfrac{{2 \times \dfrac{3}{2} - 1}}{2}} \right)\]

= g(1) = 1+ 2 = 3

46. Let f : R → R be defined by

$f\left( x \right)=\left\{ \begin{align} & 2x,\text{ }x > 3 \\ & {x}^{2},\text{ }1 < x \leqslant 3 \\ & 3x,\text{}1 < x \leqslant 3 \end{align} \right.$

Then f (– 1) + f (2) + f (4) is

(A) 9

(B) 14

(C) 5

(D) none of these

Ans: Option (A) is the correct answer.

Given that,

$f\left( x \right)=\left\{ \begin{align} & 2x,\text{ }x > 3 \\ & {x}^{2},\text{ }1 < x \leqslant 3 \\ & 3x,\text{}1 < x \leqslant 3 \end{align} \right.$

f(–1) + f(2) + f(4)

= 3(-1) + 22 + (2 \[ \times \] 4)

= -3 + 4 + 8

= 9

47. Let f : R → R be given by f (x) = tan x. Then f-1(1) is

(A) 4

(B) \[\left\{ {n\pi + \dfrac{\pi }{4}:n \in Z} \right\}\]

(C) does not exist

(D) none of these

Ans: Option (A) is the correct answer.

Given that: f (x) = tan x

Let y = tan x

$ \Rightarrow $ x = ${{ta}}{{{n}}^{ - 1}}$y

$ \Rightarrow $ ${{{f}}^{ - 1}}$ (x) = ${{ta}}{{{n}}^{ - 1}}$x

$ \Rightarrow $ ${{{f}}^{ - 1}}$ (1) = ${{ta}}{{{n}}^{ - 1}}$1

$ \Rightarrow $ ${{ta}}{{{n}}^{ - 1}}$ tan $\dfrac{{{\pi }}}{{{4}}}$ = $\dfrac{{{\pi }}}{{{4}}}$ (Since tan $\dfrac{{{\pi }}}{{{4}}}$ = 1)

48. Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = ______.

Ans: Given that, 2a + 3b = 30

3 b = 30 - 2a

b = $\dfrac{{{{30 - 2a}}}}{{{3}}}$

For a = 3, b = 8

a = 6, b = 6

a = 9, b = 4

a = 12, b = 2

R = {(3, 8), (6, 6), (9, 4), (12, 2)}

49. Let the relation R be defined on the set A = {1, 2, 3, 4, 5} by R = {(a, b) : |a2-b2| < 8}. Then R is given by _______.

Ans: Given A = {1, 2, 3, 4, 5}

R = {(a, b) : |a2-b2| < 8}

R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (4, 3), (3, 4), (4, 4), (5, 5)}.

50. Let f = {(1, 2), (3, 5), (4, 1) and g = {(2, 3), (5, 1), (1, 3)}. Then g o f = ______ and f o g = ________.

Ans: Given that

f = {(1, 2), (3, 5), (4, 1) and g = {(2, 3), (5, 1), (1, 3)}.

gof(1) = g{f(1)} = g(2) = 3

gof(3) = g{f(3)} = g(5) = 1

gof(4) = g{f(4)} = g(1) = 3

gof = {(1,3),(3,1),(4, 3)}

fog(x) = f{ g(x)}

fog(2) = f {g(2)} = f(3) = 5

fog(5) = f {g(5)} = f(1) = 2

fog(1) = f {g(1)} = f(3) = 5

fog={(2,5),(5,2),(1,5)}

51. Let f : R → R be defined by f(x) = $\dfrac{{\mathbf{x}}}{{\sqrt {1 + {{\mathbf{x}}^2}} }}$, then ( f o f o f ) (x) = _______.

Ans: Given that, f(x) = $\dfrac{{{x}}}{{\sqrt {{{1}} + {{}}{{{x}}^2}} }}$

(fofof)(x) = \[f\left[f{f(x)}\right]\]

= \[f\left[ {f\left( {\dfrac{x}{{\sqrt {1 + {x^2}} }}} \right)} \right]\]

= \[f\left( {\dfrac{{\dfrac{x}{{\sqrt {1 + {x^2}} }}}}{{\sqrt {1 + \dfrac{{{x^2}}}{{1 + {x^2}}}} }}} \right)\]

= \[f\left( {\dfrac{{x\sqrt {1 + {x^2}} }}{{\sqrt {1 + {x^2}\left( {\sqrt {2{x^2}} + 1} \right)} }}} \right)\]

= \[f\left( {\dfrac{x}{{\sqrt {1 + 2{x^2}} }}} \right)f\left( {\dfrac{{\dfrac{x}{{\sqrt {1 + 2{x^2}} }}}}{{\dfrac{{\sqrt {1 + {x^2} + {x^2}} }}{{\sqrt {1 + {x^2}} }}}}} \right)\]

= \[f\left( {\dfrac{x}{{\sqrt {1 + 2{x^2}} }}} \right)\]

= \[\dfrac{{\dfrac{x}{{\sqrt {1 + 2{x^2}} }}}}{{\sqrt {1 + \dfrac{{{x^2}}}{{1 + {x^2}}}} }}\]

= \[\dfrac{{x\sqrt {1 + 2{x^2}} }}{{\sqrt {1 + 2{x^2}} \sqrt {1 + 3{x^2}} }}\]

= \[\dfrac{x}{{\sqrt {1 + 3{x^2}} }}\]

52. If f (x) = [4 – (x – 7)3], then f-1(x) = _______.

Ans: Given that, f (x) = $\left[4 - {\left( {{{x}} - 7} \right)^3}\right]$

Let y =$\left[4 - {\left( {{{x}} - 7} \right)^3}\right]$

${\left( {{{x}} - 7} \right)^3}$ = 4 - y

$ \Rightarrow $ (x - 7) = ${\left( {{{4}} - {{y}}} \right)^{1/3}}$

$ \Rightarrow $ x = ${{7}} + {\left( {{{4}} - {{y}}} \right)^{1/3}}$

$ \Rightarrow $ f-1(x) = \[7 + {(4 - x)^{\dfrac{1}{3}}}\]

### True/False:

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.

Ans: The given statement is false.

Given that, R = {(3, 1), (1, 3), (3, 3)} be defined on the set A = {1, 2, 3}

(1, 1) $ \notin $ R

So, R is not reflexive.

(3, 1) $ \notin $ R, (1, 3) $ \in $ R

Hence, R is symmetric.

Since, (3, 1) $ \in $ R, (1, 3) $ \in $ R

But (1, 1) $ \notin $ R

Hence, R is not transitive.

54. Let f: R → R be the function defined by f (x) = sin (3x+2) ∀ x ∈ R. Then f is invertible.

Ans: The given statement is false.

Given f(x) = sin(3x + 2) ∀ x ∈ R is not one-one function for all x $ \in $ R.

So, f is not invertible.

55. Every relation which is symmetric and transitive is also reflexive.

Ans: The given statement is false.

Let R be a relation defined by:

R = { (1,2),(2,1),(1,1),(2,2) } on the set A = {1,2,3}

It is clear that (3, 3) $ \in $ R. So, it is not reflexive.

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

Ans: The given statement is false.

The given relation is reflexive and transitive but not symmetric.

57. Let A = {0, 1} and N be the set of natural numbers. Then the mapping f : N → A defined by f (2n–1) = 0, f (2n) = 1, ∀ n ∈ N, is onto.

Ans: The given statement is true.

Given, A = {0, 1}

F(2n - 1) = 0 , f(2n) = 1,∀ n ∈ N

So, the mapping f : N $ \to $ A is onto.

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.

Ans: The given statement is false.

Given that R = {(1, 1), (1, 2), (2, 1), (3, 3)}

(2, 2) $ \notin $ R

So, R is not reflexive.

59. The composition of functions is commutative.

Ans: The given statement is false.

Let f(x) = ${{{x}}^2}$

And g(x) = x + 1

fog(x) = f{g(x)} = f (x + 1)

= ${\left( {{{x}} + {{1}}} \right)^2} = {{}}{{{x}}^2}{{}} + {{2x + 1}}$

gof(x) = g{f(x)} = g (${{{x}}^2}$) = ${{{x}}^2}$ + 1

Therefore, fog(x) $ \ne $ gof(x).

60. The composition of functions is associative.

Ans: The given statement is true.

Let f(x) = x, g(x) = x + 1

And h(x) = 2 x - 1

Then, fo{goh(x)}

= $f\left[g{h(x)}\right]$

= f{g(2x - 1)}

= f(2x -1) + 1

= f(2x) = 2x

(fog)oh(x) = (fog){h(x)}

= (fog)(2x-1)

= f {g(2x-1)}

= f(2x - 1+ 1)

= f (2x) = 2x

61. Every function is invertible.

Ans: The given statement is false.

Only bijective functions are invertible.

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

Ans: The given statement is false.

‘+’ is a binary operation on the set N but it has no identity element.

### Introduction of Chapter 1 NCERT Exemplar

Chapter 1 of NCERT Exemplar Class 12 Maths Solutions is critical for laying a foundation for the full academic year. We will assist you in locating the answers to all NCERT questions from the Chapter, including complete steps and thorough answers. There are two parts to this Chapter of NCERT Class 12 Maths solutions: Relations and Functions. We have detailed explanations and solutions for questions from the parts. For students to enhance their understanding of the topics, NCERT Exemplar Class 12 Maths Solutions Chapter 1 PDF download is accessible.

This is the beginning of high school algebra, which serves as the foundation for higher learning Mathematics. Students are taught about Relations and Functions, which are a sort of relation, in NCERT Exemplar Class 12 Maths Chapter 1 solutions. The relation is nothing more than a relationship between two groups or an input-output relationship.

Although the function is a relation, it is used to get the output from a set of inputs. Studying the connections and functions in Calculus Mathematics from the beginning of 12th grade will aid in establishing one's roots in the subject for admission tests and higher education.

Several topics are discussed in Chapter 1 of Class 12 Maths NCERT Exemplar Solutions, which discusses the various forms of functions and connections. Universal relations, reflective relations, empty relations, symmetric relations, and other types of relations will be covered.

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