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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Choose the correct or the most suitable answer.

Question 1.
If A = {(x, y) : y = ex ; x ∈ R } and B = {(x, y) : y = e-x, x ∈ R } then n(A ∩ B)
(a) Infinity
(b) 0
(c) 1
(d) 2
Solution:
(c) 1
Hint.
A∩B = (0, 1)
n(A∩B) = 1
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 1

Question 2.
IfA {(x, y) : y = sin x, x ∈ R) and 8= (x, y) : y = cos x, x ∈ R) then A∩B contains ……..
(a) no element
(b) infinitely many elements
(c) only one element
(d) cannot be determined.
Solution:
(b) infinitely many elements
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5
Question 3.
The relation R defined on a set A = {0, -1, 1, 2} by xRy if |x2 +y2| ≤ 2, then which one of the following is true?
(a) R = {(0, 0), (0, -1), (0, 1), (-1, 0), (-1, 1), (1, 2), (1, 0)}
(b) R = {(0, 0), (0, -1), (0, 1), (-1, 0), (1, 0)
(c) Domain of R is {0, -1, 1, 2}
Solution:
(a) Range of R is {0, -1, 1}
Hint.
A= {0, -1, 1, 2}
|x2 + y2| ≤ 2
The values of x and y can be 0, -1 or 1
So range = {0, -1, 1}

Question 4.
If f(x) = |x – 2| + |x + 2|, x ∈ R, then
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 2
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 3
Hint.
f(x) = |x – 2| + |x + 2|
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 4

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 5.
Let R be the set of all real numbers. Consider the following subsets of the plane R x R: S = {(x, y): y = x + 1 and 0 < x < 2} and T = {(x, y) : x – y is an integer} Then which of the following is true?
(a) T is an equivalence relation but S is not an equivalence relation.
(b) Neither S nor T is an equivalence relation
(c) Both S and T are equivalence relation
(d) S is an equivalence relation but T is not an equivalence relation.
Solution:
(a) T is an equivalence relation but S is not an equivalence relation.
Hint.
(0, 1), (1, 2) it is not an equivalence relation
T is an equivalence relation

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 6.
Let A and B be subsets of the universal set N, the set of natural numbers. Then
A’ ∪ [(A ∩ B) ∪ B’] is ………
(a) A
(b) A’
(c) B
(d) N
Solution:
(d) N
Hint.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 9

Question 7.
The number of students who take both the subjects Mathematics and Chemistry is 70. This represents 10% of the enrollment in Mathematics and 14% of the enrollment in Chemistry. How many students take at least one of these two subjects?
(a) 1120
(b) 1130
(c) 1100
(d) insufficient data
Solution:
(b) 1130
Hint.
n(M ∪ C) = n(M) + n(C) – n(M ∩ C)
= 700 + 500 – 70 = 1130

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 8.
If n[(A × B) ∩ (A × C)] = 8 and n(B ∩ C) = 2 , then n(A) is
(a) 6
(b) 4
(c) 8
(d) 16
Solution:
(b) 4

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 9.
If n(A) = 2 and n(B ∪ C) = 3, then n[(A × B) ∪ (A × C)] is …….
(a) 23
(b) 32
(c) 6
(d) 5
Solution:
(c) 6
Hint.
n[(A × B) ∪ (A × C)] = n[ A × (B ∪ C)] = n(A) × n(B ∪ C) = 2 × 3 = 6

Question 10.
If two sets A and B have 17 elements in common, then the number of elements common to the set A × B and B × A is
(a) 217
(b) 172
(c) 34
(d) insufficient data
Solution:
(b) 172
Hint.
n (A ∩ B) = 17
So n [(A × B) ∩ (B × A)]
= n(A ∩ B) × n(B ∩ A) = 17 × 17 = 172

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 11.
For non-empty sets A and B, if A ⊂ B then (A × B) ∩ (B × A) is equal to ……….
(a) A ∩ B
(b) A × A
(c) B × B
(d) None of these
Solution:
(b) A × A
Hint.

When A ⊂ B, (A × B) ∩ (B × A) = A × A

Question 12.
The number of relations on a set containing 3 elements is
(a) 9
(b) 81
(c) 512
(d) 1024
Solution:
(c) 512
Hint.
Number of relations = 2n2 = 232 = 29 = 512

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 13.
Let R be the universal relation on a set X with more than one element. Then R is
(a) Not reflexive
(b) Not symmetric
(c) Transitive
(d) None of the above
Solution:
(c) Transitive

Question 14.
Let X = {1, 2, 3, 4} and R = {(1, 1), (1, 2), (1, 3), (2, 2), (3, 3), (2, 1), (3, 1), (1,4), (4, 1)}. Then R is ……..
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) Equivalence
Solution:
(b) Symmetric
Hint.
(4, 4} ∉ R ⇒ R is not reflexive
(1, 4), (4, 1) ∈ R ⇒ R is symmetric
(1, 4), (4, 1) ∈ R but (4, 4) ∉ R
So R is not transitive

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 15.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 20
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 21

Question 16.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 22
Solution:
(c) [0, 1)
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 23

Question 17.
The rule f(x) = x2 is a bijection if the domain and the co-domain are given by …..
(a) R,R
(b) R, (0, ∞)
(c) (0, ∞), R
(d) [0, ∞), [0, ∞)
Solution:
(d) [0, ∞), [0, ∞)

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 18.
The number of constant functions from a set containing m elements to a set containing n elements is
(a) mn
(b) m
(c) n
(d) m + n
Solution:
(c) n
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 25

Question 19.
The function f: [0, 2π] ➝ [-1, 1] defined by f(x) = sin x is
(a) One to one
(b) Onto
(c) Bijection
(d) Cannot be defined
Solution:
(b) Onto
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 26
So it is not one-to-one
So it is an onto function

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 20.
If the function f : [-3, 3] ➝ S defined by f(x) = x2 is onto, then S is ………
(a)[-9, 9]
(b) R
(c) [-3, 3]
(d) [0, 9]
Solution:
(d) [0, 9]

Question 21.
Let X = {1, 2, 3, 4}, Y = {a, b, c, d) and f = {(1, a), (4, b), (2, c), (3, d) (2, d)}. Then f is ………
(a) An one-to-one function
(b) An onto function
(c) A function which is not one-to-one
(d) Not a function
Solution:
(d) Not a function
Hint.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 266
Since the element 2 has two images, it is not a function

Question 22.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 28
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 29
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 30
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 45

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 23.
Let f : R ➝ R be defined by f(x) = 1 – |x|. Then the range of f is ………
(a) R
(b) (1, ∞)
(c) (-1, ∞)
(d) (-∞, 1]
Solution:
(d) (-∞, 1]
Hint.
f: R ➝ R defined by
f(x) = 1 – |x|
For example,
f(1) = 1 – 1 = 0
f(8) = 1 – 8 = -1
f(-9) = 1 – 9 = -8
f(-0.2) = 1 – 0.2 = 0.8
so range = (-∞, 1]

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 24.
The function f : R ➝ R is defined by f(x) = sin x + cos x is ……
(a) An odd function
(b) Neither an odd function nor an even function
(c) An even function
(d) Both odd function and even function
Solution:
(b) Neither an odd function nor an even function

Question 25.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 32
(a) An odd function
(b) Neither an odd function nor an even function
(c) An even function
(d) Both odd function arid even function
Solution:
(c) An even function
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 33
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 34