THE INDIAN INTERNATIONAL SCHOOL, DSO

CHAPTER 02 RELATIONS AND FUNCTIONS (ANSWERS)

SUBJECT: MATHEMATICS MAX. MARKS : 40

CLASS : XI DURATION : 1½ hrs
General Instructions:
(i). All questions are compulsory.
(ii). This question paper contains 20 questions divided into five Sections A, B, C, D and E.
(iii). Section A comprises of 10 MCQs of 1 mark each. Section B comprises of 4 questions of 2 marks
each. Section C comprises of 3 questions of 3 marks each. Section D comprises of 1 question of 5
marks each and Section E comprises of 2 Case Study Based Questions of 4 marks each.
(iv). There is no overall choice.
(v). Use of Calculators is not permitted

SECTION – A
Questions 1 to 10 carry 1 mark each.
1. If A × B = {(a, 1), (b, 3), (a, 3), (b, 1), (a, 2), (b, 2)}, find A and B, then set B is
(a) {a} (b) {a, b} (c) {1, 2} (d) {1, 2, 3}
Ans: (d) {1, 2, 3}
First entry ∈ set A and second entry ∈ set B
∴ A = {a, b}, B = {1, 2, 3}
x
2. Range of the function f (x) = is
x2
(a) R (b) R – {2} (c) R – {1} (d) R – {–2}
Ans: (c) R – {1}

3. If n(A) = 3, n(B) = 2, then number of non empty relations from set A to set B are
(a) 8 (b) 4 (c) 64 (d) 63
Ans: (d) 63, as n(A × B) = 6
Total relations = 26 = 64
Total non-empty relations = 64 – 1 = 63
x4
4. Range of the function f (x) = is
| x4|
(a) {4} (b) {–4} (c) {–1, 1} (d) any real number
Ans: (c) {–1, 1}

5. If [x]2 – 5[x] + 6 = 0, where [ ] denote the greatest integer function, then

(a) x ∈ [3, 4) (b) x ∈ [2, 3) (c) x ∈ [2, 3) (d) x ∈ [2, 4)
Ans: (d) x ∈ [2, 4), we have [x]2 – 5[x] + 6 = 0
⇒ [x]2 – 3[x] – 2[x] + 6 = 0
⇒ [x] ([x] – 3) – 2([x] – 3) = 0
⇒ ([x] – 2) ([x] – 3) = 0 ⇒ [x] – 2 = 0 or [x] – 3 = 0
⇒ [x] = 2 or [x] = 3 ⇒ x ∈ [2, 3) or x ∈ [3, 4) ⇒ x ∈ [2, 4)

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6. Domain of a 2  x 2 (a > 0) is
(a) (–a, a) (b) [–a, a] (c) [0, a] (d) (–a, 0]
2 2
Ans: (b) [–a, a], let y = a  x the function y is defined if
a2 – x2 ≥ 0 ⇒ x2 – a2 ≤ 0 or x2 ≤ a2
–a ≤ x ≤ a
So, domain of y = [–a, a]

7. Given set A = {1, 2, 3, ....., 10}. Relation R is defined in set A as R = {(a, b) ∈ A × A : a = 2b}.
Then range of relation R is
(a) {2, 4, 6, 8, 10} (b) {1, 3, 5, 7, 9}
(c) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5)} (d) {1, 2, 3, 4, 5}
Ans: (d) {1, 2, 3, 4, 5}, as R = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5)}

8. Let n(A) = m and n(B) = n. Then the total number of non-empty relations that can be defined
from A to B is
(a) mn (b) nm – 1 (c) mn – 1 (d) 2mn – 1
Ans: (d) 2mn – 1, as n(A) = m, n(B) = n ⇒ n(A × B) = mn
So, number of relations = 2mn including void relation f.
Number of non-empty relations = 2nm – 1

For Q9 and Q10, a statement of assertion (A) is followed by a statement of reason (R). Choose
the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

9. Assertion (A): Let A = {1, 2} and B = {3, 4}. Then, number of relations from A to B is 16.
Reason (R): If n(A) = p and n(B) = q, then number of relations is 2pq.
Ans: (a) Both A and R are true and R is the correct explanation of A.

10. Assertion (A): The domain of the relation R = {(x + 2, x + 4) : x ∈ N, x < 8} is {3, 4, 5, 6, 7, 8,
9).
Reason (R): The range of the relation R = {(x + 2, x + 4) : x ∈ N, x < 8} is {1, 2, 3, 4, 5, 6, 7}.
Ans: (c) A is true but R is false.

SECTION – B
Questions 11 to 14 carry 2 marks each.
11. Let f : R → R be given by f(x) = x2 + 3 Find (i) {x : f(x) = 28} (ii) The pre-images of 39 and 2
under ‘f ’.
Ans: (i) 28 = x2 + 3 ⇒ x2 = 25 ⇒ x = ± 5
(ii) 39 = x2 + 3 ⇒ x2 = 36 ⇒ x = ± 6 ;
2 = x2 + 3 ⇒ x2 = – 1, not possible

12. Determine the domain and range of the relation R defined by R = {(x + 1, x + 5) : x ∈ (0, 1, 2, 3,
4, 5)}
Ans: Relation R is {(1, 5), (2, 6), (3, 7), (4, 8), (5, 9), (6, 10)}
Domain of R = {1, 2, 3, 4, 5, 6} ; Range of R = {5, 6, 7, 8, 9, 10}

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 x3  x  3
13. Find the domain of each of the following functions given by : f ( x) 
x2 1
x3  x  3
Ans: f ( x)  ,
x2 1
For domain, x2 – 1 ≠ 0 ⇒ x ≠ ±1
∴ Domain = R – {–1, 1}

| x 4|
14. Find the range of the following functions given by : f ( x ) 
x4
( x  4) ( x  4)
Ans: y  or , i.e. 1 or –1
( x  4) ( x  4)
∴ range {–1, 1}

SECTION – C
Questions 15 to 17 carry 3 marks each.

15. Find the domain and the range of the function : f (x) = x2  4
Ans: Given, f (x) = x 2  4 ; For Df., f (x) must be a real number.
⇒ x 2  4 must be a real number. ⇒ x2 – 4 ≥ 0 ⇒ (x + 2) (x – 2) ≥ 0
⇒ Either x ≤ – 2 or x ≥ 2. ⇒ Df = (– ∞, – 2] ∪ [2, ∞).
For Rf, let y = x 2  4 ... (i)
As square root of a real number is always non-negative, y ≥ 0.
On squaring (i), we get y2 = x2 – 4 ⇒ x2 = y2 + 4 but x2 ≥ 0 ∀ x ∈ Df.
⇒ y2 + 4 ≥ 0 ⇒ y2 ≥ – 4, which is true ∀ y ∈ R,
Also, y ≥ 0. ⇒ Rf = [0, ∞).

16. Find the domain and range of the real function f (x) = 9  x2
Ans: Given function is f (x) = 9  x 2
For domain of ‘f ’, 9 – x2 ≥ 0
⇒ 9 ≥ x2 ⇒ x2 ≤ 9 ⇒ –3 ≤ x ≤ 3
∴ Domain is {x ∈ R | –3 ≤ x ≤ 3}, i.e. [–3, 3]
For range : f (x) = 9  x2  y  9  x2
9  x 2 is always +ve
⇒ y is always +ve.
⇒ y2 = 9 – x2 ⇒ x2 = 9 – y2
⇒ x = 9  y2
For x to exist 9 – y2 ≥ 0 ⇒ y2 ≤ 9 ⇒ –3 ≤ y ≤ 3
As y ≥ 0
∴ Range = [0, 3]

17. If A = {x : x ∈ W, x < 2}, B = {x : x ∈ N, 1 < x < 5}, C = {3, 5} find

(i) A × (B ∩ C) (ii) A × (B ∪ C)
Ans: A = {x : x ∈ W, x < 2} = {0, 1},
B = {x : x ∈ N, 1 < x < 5} = {2, 3, 4};
C = {3, 5}
(i) A × (B ∩ C) = {0, 1} × {3} = {(0, 3), (1,3)}
(ii) A × (B ∪ C) = {0, 1} × {2, 3, 4, 5}
= {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}

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 SECTION – D
Questions 18 carry 5 marks.
18. (a) Relations R1 and R2 are defined on the set Z of integers as follows :
(x, y) ∈ R1 ⇔ x2 + y2 = 25 ; (y, x) ∈ R2 ⇔ x2 + y2 = 25
Express R1 and R2 as the sets of ordered pairs and hence find their respective domains.
(b) A relation R is defined from a set A = {2, 3, 4, 5} to a set B = {3, 6, 7, 10} as follows : (x, y)
∈ R ⇔ x divides y. Express R as a set of ordered pairs and determine the domain and range of R.
Ans: (a) (x, y) ∈ R ⇔ x2 + y2 = 25
⇔ y = ± 25  x 2
We observe that : x = 0 ⇒ y = ± 5
x = ± 3 ⇒ y = 25  9 = ± 4,
x = ± 4,y = 25  16 = ± 3
x = ± 5, ⇒ y = 25  25 = 0.
R1 = {(0, 5), (0, – 5), (3, 4), (– 3, 4), (3, – 4), (– 3, – 4), (4, 3), (– 4, 3), (4, – 3), (– 4, – 3), (5, 0),
(– 5, 0)}
R2 = {(5, 0), (– 5, 0), (4, 3), (4, – 3), (– 4, 3), (– 4, – 3), (3, 4), (3, – 4), (– 3, 4), (– 3, – 4), (0, 5),
(0, – 5)}
Domain (R1) = {0, 3, – 3, 4, – 4, 5, – 5} = Domain (R2)
(b) a|b stands for ‘a divides b’. For the elements of the given sets A and B, we find that 2 |6,
2|10, 3|3, 3|6 and 5|10.
∴ (2, 6) ∈ R, (2, 10) ∈ R, (3, 3) ∈ R, (3, 6) ∈ R and (5, 10) ∈ R.
Thus, R = {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)}
Clearly, Domain (R) = {2, 3, 5} and range (R) = {3, 6, 10}.

SECTION – E (Case Study Based Questions)

Questions 19 to 20 carry 4 marks each.
19. Maths teacher started the lesson Relations and Functions in Class XI. He explained the following
topics:
Ordered Pairs: The ordered pair of two elements a and b is denoted by (a, b) : a is first element
(or first component) and b is second element (or second component).
Two ordered pairs are equal if their corresponding elements are equal. i.e., (a, b) = (c, d) ⇒ a = c
and b = d
Cartesian Product of Two Sets: For two non-empty sets A and B, the cartesian product A x B
is the set of all ordered pairs of elements from sets A and B.
In symbolic form, it can be written as A x B= {(a, b) : a ∈ A, b ∈ B}
Based on the above topics, answer the following questions.
(i) If (a – 3, b + 7) = (3, 7), then find the value of a and b
(ii) If (x + 6, y – 2) = (0, 6), then find the value of x and y
(iii) If (x + 2, 4) = (5, 2x + y), then find the value of x and y
(iv) Find x and y, if (x + 3, 5) = (6, 2x + y).
Ans:
(i) We know that, two ordered pairs are equal, if their corresponding elements are equal.
(a – 3, b + 7) = (3, 7)
⇒ a – 3 = 3 and b + 7 = 7 [equating corresponding elements]
⇒ a = 3 + 3 and b = 7 – 7 ⇒ a= 6 and b = 0
(ii) (x + 6, y – 2) = (0, 6)
⇒ x + 6 = 0 ⇒ x = -6 and y – 2 = 6 ⇒ y = 6 + 2 = 8
(iii) (x + 2, 4) = (5, 2x + y)
⇒ x + 2 =5 ⇒ x = 5 – 2 = 3 and 4 = 2x + y ⇒ 4 = 2 x 3 + y ⇒ y = 4 – 6 = -2
(iv) x + 3 = 6, 2x + y = 5 ⇒ x = 3, y = 1

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20. Maths teacher explained the topics:
Method to Find the Sets When Cartesian Product is Given
For finding these two sets, we write first element of each ordered pair in first set say A and
corresponding second element in second set B (say).
Number of Elements in Cartesian Product of Two Sets
If there are p elements in set A and q elements in set B, then there will be pq n(A) = p and n(B) =
q, then n(A x B) = pq
Based on the above two topic, answer the following questions.
(i) If A x B = {(a, 1), (b, 3), (a, 3), (b, 1), (a, 2), (b, 2)}. Then, find A and B
(ii) If the set A has 3 elements and set B has 4 elements, then find the number of elements in A x
B
(iii) A and B are two sets given in such a way that A x B contains 6 elements. If three elements
of A x B are (1, 3), (2, 5) and (3, 3), then find A, B
(iv) The cartesian product P x P has 16 elements among which are found (a, 1) and (b, 2). Then,
find the set P
Ans: (i) Here, first element of each ordered pair of A x B gives the elements of set A and
corresponding second element gives the elements of set B.
∴ A = {a, b} and B = {1,3,2}
(ii) Given, n(A) = 3 and n(B) = 4.
∴ The number of elements in A x B is n(A x B) = n(A) x n(B) = 3 × 4 = 12
(iii) A = {1, 2, 3} and B = {3, 5}
∵ A x B = {1, 2, 3} x {3, 5} = {(1,3), (1,5), (2,3), (2,5), (3,3), (3,5)}
(iv) Given n( P  P)  16
 n( P).n( P)  16  n( P)  4
Now, as (a, 1) ∈ P
⇒ a ∈ P and 1 ∈ P
and as (b, 2) ∈ P
⇒ b ∈ P and 2 ∈ P
⇒ a, b, 1, 2 ∈ P
Hence P has exactly four elements.

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