Bhagya

Students can download Maths Chapter 1 Relations and Functions Ex 1.6 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.6

Multiple Choice Questions

Question 1.
If n(A × B) = 6 and A= {1, 3} then n (B) is ………….
(1) 1
(2) 2
(3) 3
(4) 6
Answer:
(3) 3
Hint: n(A × B) = 6
n(A) = 2
n(A × B) = n(A) × n(B)
6 = 2 × n(B)
n(B) = \(\frac { 6 }{ 2 } \) = 3

Question 2.
A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} then n[(A ∪ C) × B] is
(1) 8
(2) 20
(3) 12
(4) 16
Answer:
(3) 12
Hint:
A = {a, b, p}, B = {2, 3}, C = {p, q, r, s}
n (A ∪ C) × B
A ∪ C = {a, b, p, q, r, s}
(A ∪ C) × B = {{a, 2), (a, 3), (b, 2), (b, 3), (p, 2), (p, 3), (q, 2), (q, 3), (r, 2), (r, 3), (s, 2), (s, 3)
n [(A ∪ C) × B] = 12

Question 3.
If A = {1,2}, B = {1,2, 3, 4}, C = {5,6} and D = {5, 6, 7, 8} then state which of the following statement is true ……………….
(1) (A × C) ⊂ (B × D)
(2) (B × D) ⊂ (A × C)
(3) (A × B) ⊂ (A × D)
(4) (D × A) ⊂ (B × A)
Answer:
(1) (A × C) ⊂ (B × D)
Hint: n(A × B) = 2 × 4 = 8
(A × C) = 2 × 2 = 4
n(B × C) = 4 × 2 = 8
n(C × D) = 2 × 4 = 8
n(A × C) = 2 × 2 = 4
n(A × D) = 2 × 4 = 8
n(B × D) = 4 × 4 = 16
∴ (A × C) ⊂ (B × D)

Question 4.
If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is
(1) 3
(2) 2
(3) 4
(4) 6
Answer:
(2) 2
Hint:
n(A) = 5
n(B) = x
n(A × B) = 1024 = 2¹⁰
2^5x = 2¹⁰
⇒ 5x = 10
⇒ x =2

Question 5.
The range of the relation R = {(x, x²) a prime number less than 13} is ……………………
(1) {2, 3, 5, 7}
(2) {2, 3, 5, 7, 11}
(3) {4, 9, 25, 49, 121}
(4) {1, 4, 9, 25, 49, 121}
Answer:
(3) {4, 9, 25, 49, 121}
Hint:
Prime number less than 13 = {2, 3, 5, 7, 11}
Range (R) = {(x, x²)}
Range = {4, 9, 25, 49, 121} (square of x)

Question 6.
If the ordered pairs (a + 2, 4) and (5, 2a + b)are equal then (a, b) is
(1) (2, -2)
(2) (5, 1)
(3) (2, 3)
(4) (3, -2)
Answer:
(4) (3, -2)
Hint:
(a + 2, 4), (5, 2a + b)
a + 2 = 5
a = 3
2a + b = 4
6 + b = 4
b = -2

Question 7.
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 ……………..
(1) mⁿ
(2) n^m
(3) 2^mn – 1
(4) 2^mn
Answer:
(4) 2^mn

Question 8.
If {(a, 8),(6, b)}represents an identity function, then the value of a and b are respectively
(1) (8, 6)
(2) (8, 8)
(3) (6, 8)
(4) (6, 6)
Answer:
(1) (8, 6)
Hint:
{{a, 8), (6, b)}
a = 8
b = 6

Question 9.
Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}.
A function f: A → B given by f = {(1, 4), (2, 8),(3,9),(4,10)} is a ……………
(1) Many-one function
(2) Identity function
(3) One-to-one function
(4) Into function
Answer:
(3) One-to-one function
Hint:

Different elements of A has different images in B.
∴ It is one-to-one function.

Question 10.
If f (x) = 2x² and g(x) = \(\frac { 1 }{ 3x } \), then fog is …………..
(1) \(\frac{3}{2 x^{2}}\)
(2) \(\frac{2}{3 x^{2}}\)
(3) \(\frac{2}{9 x^{2}}\)
(4) \(\frac{1}{6 x^{2}}\)
Answer:
(3) \(\frac{2}{9 x^{2}}\)
Hint:

Question 11.
If f: A → B is a bijective function and if n(B) = 7, then n(A) is equal to
(1) 7
(2) 49
(3) 1
(4) 14
Answer:
(1) 7
Hint:
In a bijective function, n(A) = n(B)
⇒ n(A) = 7

Question 12.
Let f and g be two functions given by
f = {(0,1),(2, 0),(3-4),(4,2),(5,7)}
g = {(0,2),(1,0),(2, 4),(-4,2),(7,0)}
then the range of f o g is …………………
(1) {0,2,3,4,5}
(2) {-4,1,0,2,7}
(3) {1,2,3,4,5}
(4) {0,1,2}
Answer:
(4) {0,1,2}
Hint: f = {(0, 1)(2, 0)(3, -4) (4, 2) (5, 7)}
g = {(0,2)(l,0)(2,4)(-4,2)(7,0)}

fog = f[g(x)]
f [g(0)] = f(2) = 0
f [g(1)] = f(0) = 1
f [g(2)] = f(4) = 2
f[g(-4)] = f(2) = 0
f[g(7)] = f(0) = 1
Range of fog = {0,1,2}

Question 13.
Let f(x) = \(\sqrt{1+x^{2}}\) then
(1) f(xy) = f(x),f(y)
(2) f(xy) ≥ f(x),f(y)
(3) f(xy) ≤ f(x).f(y)
(4) None of these
Answer:
(3) f(xy) ≤ f(x).f(y)
Hint:
\(\sqrt{1+x^{2} y^{2}} \leq \sqrt{\left(1+x^{2}\right)} \sqrt{\left(1+y^{2}\right)}\)
⇒ f(xy) ≤ f(x) . f(y)

Question 14.
If g= {(1,1),(2,3),(3,5),(4,7)} is a function given by g(x) = αx + β then the values of α and β are
(1) (-1,2)
(2) (2,-1)
(3) (-1,-2)
(4) (1,2)
Answer:
(2) (2, -1)
Hint: g (x) = αx + β
g(1) = α(1) + β
1 = α + β ….(1)
g (2) = α (2) + β
3 = 2α + β ….(2)
Solve the two equations we get
α = 2, β = -1

Question 15.
f(x) = (x + 1)³ – (x – 1)³ represents a function which is
(1) linear
(2) cubic
(3) reciprocal
(4) quadratic
Answer:
(4) quadratic
Hint:
f(x) = (x + 1)³ – (x – 1)³
= x³ + 3x² + 3x + 1 -[x³ – 3x² + 3x – 1]
= x³ + 3x² + 3x + 1 – x³ + 3x² – 3x + 1 = 6x² + 2
It is a quadratic function.

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.3

Question 1.
Find the least positive value of x such that

(i) 71 = x (mod 8)
Answer:
71 = 7 (mod 8)
∴ The value of x = 7

(ii) 78 + x = 3 (mod 5)
78 + x – 3 = 5n (n is any integer)
75 + x = 5n
(Let us take x = 5)
75 + 5 = 80 (80 is a multiple of 5)
∴ The least value of x is 5

(iii) 89 = (x + 3) (mod 4)
89 – (x + 3) = 4n
(n may be any integer)
89 – x – 3 = 4n
89 – x = 4n
86 – x is a multiple of 4
(84 is a multiple of 4)
86 – 2 = 4n
84 = 4n
The value of x is 2

(iv) 96 = \(\frac { x }{ 7 } \) (mod 5)
96 – \(\frac { x }{ 7 } \) = 5n (n may be any integer)
672 – x = 35n (multiple of 35 is 665)
672 – 7 = 665
∴ The value of x = 7

(v) 5x = 4 (mod 6)
5x – 4 = 6n (n may be any integer)
5x = 6n + 4
x = \(\frac { 6n+4 }{ 5 } \)
Substitute the value of n as 1, 6, 11, 16 …. as n values in x = \(\frac { 6n+4 }{ 5 } \) which is divisible by 5.
2, 8, 14, 20,…………
The least positive value is 2.

Question 2.
If x is congruent to 13 modulo 17 then 7x -3 is congruent to which number modulo 17?
Solution:
x ≡ 13 (mod 17)
Let p be the required number …………. (1)
7x – 3 ≡ p (mod 17) ………….. (2)
From (1),
x – 13 = 17n for some integer M.
x – 13 is a multiple of 17.
x must be 30.
∴ 30 – 13 = 17
which is a multiple of 17.
From (2),
7 × 30 – 3 ≡ p (mod 17)
210 – 3 ≡ p (mod 17)
207 ≡ p (mod 17)
207 ≡ 3 (mod 17)
∴ P ≡ 3

Question 3.
Solve 5x ≡ 4 (mod 6)
5x – 4 = 6n (n may be any integer)
5x = 6n + 4
x = \(\frac{6 n+4}{5}\)
The value of n 1, 6, 11, 16 ……..
∴ The value of x is 2, 8, 14, 20 …………..

Question 4.
Solve 3x – 2 = 0 (mod 11)
Answer:
Given 3x – 2 = 0(mod 11)
3x – 2 = 11n (n may be any integer)
3x = 2 + 11n
x = \(\frac { 11n+2 }{ 3 } \)
Substitute the value of n = 2, 5, 8, 11 ….
When n ≡ 2 ⇒ x = \(\frac { 22+2 }{ 3 } \) = \(\frac { 24 }{ 3 } \) = 8
When n = 5 ⇒ x = \(\frac { 55+2 }{ 3 } \) = \(\frac { 57 }{ 3 } \) = 19
When n = 8 ⇒ x = \(\frac { 88+2 }{ 3 } \) = \(\frac { 90 }{ 3 } \) = 30
When n = 11 ⇒ x = \(\frac { 121+2 }{ 3 } \) = \(\frac { 123 }{ 3 } \) = 41
∴ The value of x is 8, 19, 30,41

Question 5.
What is the time 100 hours after 7 a.m.?
Answer:
100 ≡ x (mod 12) Note: In a clock every 12 hours
100 ≡ 4 (mod 12) the numbers repeats.

The time repeat after 7 am is 7 + 4 = 11 o’ clock (or) 11 am.

Question 6.
What is time 15 hours before 11 p.m.?
Solution:
15 ≡ x (mod 12)
15 – x = 12n
15 – x is a multiple of 12 x must be 3.

∴ The time 15 hrs before 11 O’clock is 11 – 3 = 8 O’ clock i.e. 8 p.m

Question 7.
Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?
Answer:
Number of days in a week = 7
45 ≡ x (mod 7)
45 ≡ 3 (mod 7)

The value of x must be 3.
Three days after tuesday is friday uncle will come on friday.

Question 8.
Prove that 2ⁿ + 6 × 9ⁿ is always divisible by 7 for any positive integer n.
Solution:
2¹ + 6 × 9¹ = 2 + 54 = 56 is divisible by 7
When n = k,
2^k + 6 × 9^k = 7 m [where m is a scalar]
⇒ 6 × 9^k = 7 m – 2^k …………. (1)
Let us prove for n = k + 1
Consider 2^k+1 + 6 × 9^k+1 = 2^k+1 + 6 × 9^k × 9
= 2^k+1 + (7m – 2^k)9 (using (1))
= 2^k+1 + 63m – 9.2^k = 63m + 2^k.2¹ – 9.2^k
= 63m – 2^k (9 – 2) = 63m – 7.2^k
= 7 (9m – 2^k) which is divisible by 7
∴ 2ⁿ + 6 × 9ⁿ is divisible by 7 for any positive integer n

Question 9.
Find the remainder when 2⁸¹ is divided by 17?
Answer:
2⁸¹ ≡ x(mod 17)
2⁴⁰ × 2⁴⁰ × 2¹ ≡ x(mod 17)
(2⁴)¹⁰ × (2⁴)¹⁰ × 2¹ ≡ x(mod 17)
(16)¹⁰ × (16)¹⁰ × 2¹ ≡ x(mod 17)
(16²)⁵ × (16²)⁵ × 2¹ ≡ x(mod 17)
= 1 × 1 × 2 (mod 17)
[(16)² = 256 = 1 (mod 17)]
= 2 (mod 17)
2⁸¹ = 2(mod 17)
∴ x = 2
The remainder is 2

Question 10.
The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is four and half hours ahead to that of London’s time, then find the time at London, when will the flight lands at London Airport?
Answer:
Duration of the flight time = 11 hours
(Chennai to London)
Starting time on Sunday = 23 : 30 hour

Time difference is 4 \(\frac { 1 }{ 2 } \) horns ahead to london
The time to reach London airport = (10.30 – 4.30)
= 6 am
The first reach the london airport next day (monday) at 6 am

Students can download Maths Chapter 2 Real Numbers Ex 2.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.5

Question 1.
Write the following in the form of \(5^n\):
(i) 625
(ii) \(\frac{1}{5}\)
(iii) \(\sqrt{5}\)
(iv) \(\sqrt{125}\)
Solution:
(i) 625 = 5⁴

(ii) \(\frac{1}{5}\) = 5^-1
(iii) \(\sqrt{5}\) = \(5^\frac{1}{2}\)
(iv) \(\sqrt{125}\) = \(\sqrt{5^3}\) = \((5^3)^\frac{1}{2} = 5^\frac{3}{2}\)

Question 2.
Write the following in the form of \(4^n\):
(i) 16
(ii) 8
(iii) 32
Solution:
(i) 16
= 4 × 4
= 4²

(ii) 8
= 4 × 2
= 4 × \(\left(2^{2}\right)^{\frac{1}{2}} \)
= 4 \(\times 4^{\frac{1}{2}} \)
= 4\(^{1+\frac{1}{2}} \)
= 4\(^{\frac{2+1}{2}} \)
= 4\(^{3 / 2}\)

(iii) 32
= 4 × 4 × 2
= 4² × \(\left(2^{2}\right)^{\frac{1}{2}} \)
= 4\(^{2} \times 4^{\frac{1}{2}} \)
= 4\(^{2+\frac{1}{2}} \)
= 4\(^{\frac{4+1}{2}} \)
= 4\(^{\frac{5}{2}} \)

Question 3.
Find the value of
(i) (49)\(^\frac{1}{2}\)
(ii) (243)\(^\frac{2}{5}\)
(iii) (9)\(^\frac{-3}{2}\)
(iv) \((\frac{64}{125})^\frac{-2}{3}\)
Solution:
(i) 49\(^\frac{1}{2}\) = \((7^2)^\frac{1}{2}\) = 7\(^{2 × \frac{1}{2}}\) = 7
(ii) (243)\(^\frac{2}{5}\) = \((3^5)^\frac{2}{5}\) = 3\(^{5 × \frac{2}{5}}\) = 3² = 9

(iii) \(9^{\frac{-3}{2}}=\left(3^{2}\right)^{\frac{-3}{2}}=3^{2 \times \frac{-3}{2}}=3^{-3}=\frac{1}{3^{3}}=\frac{1}{27}\)
(iv) \(\left(\frac{64}{125}\right)^{\frac{-2}{3}}=\left(\frac{4^{3}}{5^{3}}\right)^{\frac{-2}{3}}=\left[\left(\frac{4}{5}\right)^{3}\right]^{\frac{-2}{3}}=\left(\frac{4}{5}\right)^{3 \times \frac{-2}{3}}=\left(\frac{4}{5}\right)^{-2}=\frac{4^{-2}}{5^{-2}}=\frac{5^{2}}{4^{2}}=\frac{25}{16} \)

Question 4.
Use a fractional index to write:
(i) \(\sqrt{5}\)
(ii) \(\sqrt[2]{7}\)
(iii) (\(\sqrt[3]{49})^{5}\)
(iv) \((\frac{1}{\sqrt[3]{100}})^{7}\)
Solution:
(i) \(\sqrt{5}\) = (5)\(^\frac{1}{2}\)
(ii) \(\sqrt[2]{7}\) = 7\(^\frac{1}{2}\)
(iii) \((\sqrt[3]{49})^{5}=\left[(49)^{\frac{1}{3}}\right]^{5}=\left[\left(7^{2}\right)^{\frac{1}{3}}\right]^{5}=\left(7^{\frac{2}{3}}\right)^{5}=7^{\frac{2}{3} \times 5}=7^{\frac{10}{3}}\)
(iv) \(\left(\frac{1}{\sqrt[3]{100}}\right)^{7}=\left[\frac{1}{\sqrt[3]{10^{2}}}\right]^{7}=\left[\frac{1}{\left(10^{2}\right)^{1 / 3}}\right]^{7}=\left[\frac{1}{10^{2 / 3}}\right]^{7}=\left(10^{\frac{-2}{3}}\right)^{7}=10^{\frac{-2}{3} \times 7}=10^{\frac{-14}{3}}\)

Question 5.
Find the 5th root of:
(i) 32
(ii) 243
(iii) 100000
(iv) \(\frac{1024}{3125}\)
Solution:
(i) \(\sqrt[5]{32}=(32)^{\frac{1}{5}}=\left(2^{5}\right)^{\frac{1}{5}}=2^{5 \times \frac{1}{5}} \) = 2

(ii) \(\sqrt[5]{243}=(243)^{\frac{1}{5}}=\left(3^{5}\right)^{\frac{1}{5}}=3^{5 \times \frac{1}{5}}\) = 3

(iii) \(\sqrt[5]{100000}=(100000)^{\frac{1}{5}}=\left(10^{5}\right)^{\frac{1}{5}}\)
= \(10^{5}\times{\frac{1}{5}}\)

Students can download Maths Chapter 2 Real Numbers Ex 2.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.4

Question 1.
Represent the following numbers on the number line.
(i) 5.348
Solution:
5.348 lies between 5 and 6.

Steps of construction:
1. Divide the distance between 5 and 6 into 10 equal intervals.
2. Mark the point 5.3 which is the sixth from the left of 6 and 3 from the right of 5.
3. 5.34 lies between 5.3 and 5.4. Divide the distance into 10 equal intervals.
4. Mark the point 5.34 which is sixth from the left of 5.40
5. 5.348 lies between 5.34 and 5.35. Divide the distance into 10 equal intervals.
6. Mark a point 5.348 which is second from the left of 5.350 and seventh form the right of 5.340

(ii) 6.\(\overline {4}\) upto 3 decimal places.
Solution:
6.\(\overline {4}\) = 6.4444
6.\(\overline {4}\) = 6.444 (correct to 3 decimal places)
The number lies between 6 and 7.

Steps of construction:
1. Divide the distance between 6 and 7 into 10 equal intervals.
2. Mark the point 6.4 which is the sixth from the left of 7 and fourth from the right of 6.
3. 6.44 lies between 6.44 and 6.45. Divide the distance into 10 equal intervals.
4. Mark the point 6.44 which is sixth from the left of 6.5 and fourth from the right of 6.40.
5. Mark the point 6.444 which is sixth from the left of 6.450 and fourth from the right of 6.440.

(ii) 4.\(\overline {73}\) upto 4 decimal places.
Solution:
4.\(\overline {73}\) = 4.737373……..
= 4.737374 (correct to 4 decimal places 4.7374 lies between 4 and 5)

Steps of construction:
1. Divide the distance between 4 and 5 into 10 equal parts.
2. Mark the point 4.7 which is third from the left of 5 and seventh from the right of 4.
3. 4.73 lies between 4.7 and 4.8. Divide the distance into 10 equal intervals.
4. Mark the point 4.73 which is seventh from the left of 4.80 and third from the left of 4.70.
5. 4.737 lies between 4.73 and 4.74. Divide the distance into 10 equal intervals.
6. Mark the point 4.737 which is third from the left of 4.740 and seventh from the right of 4.730.
7. 4.7374 lies between 4.737 and 4.738. Divide the distance into 10 equal intervals.
8. Mark the point 4.7374 which is sixth from the left of 4.7380 and fourth from the right of 4.7370.

Students can download Maths Chapter 1 Set Language Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Additional Questions

Choose the correct answer

Question 1.
A = {set of odd natural numbers}, B = {set of even natural numbers}, then A and B are……….
(a) equal set
(b) equivalent sets
(c) overlapping sets
(d) disjoint sets
Solution:
(d) disjoint sets

Question 2.
Number of subsets in set A = {1, 2, 3} is
(a) 3
(b) 6
(c) 8
(d) 9
Solution:
(c) 8

Question 3.
The set does not have a proper subset is
(a) Finite set
(b) Infinite set
(c) Null set
(d) Singleton set
Solution:
(c) Null set

Question 4.
Sets having the same number of elements are called
(a) overlapping sets
(b) disjoints sets
(c) equivalent sets
(d) equal sets
Solution:
(c) equivalent sets

Question 5.
The set (A – B) ∪ (B – A) is
(a) AΔB
(b) A∪B
(c) A∩B
(d) A’∪B’
Solution:
(a) AΔB

Question 6.
The set of (A∪B) – (A∩B) is
(a) (A∪B)’
(b) AΔB
(c) (A∩B)’
(d) A’∪B’
Solution:
(b) AΔB

Question 7.
The set {x : x ∈ A, x ∈ B, x ∉ A∩B} is
(a) A∩B
(b) A∪B
(c) A – B
(d) AΔB
Solution:
(d) AΔB

Question 8.
The number of elements of the set {x : x ∈ Z , x² = 1} is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(c) 2

Question 9.
If A is a proper subset of B, then A∩B =…………..
(a) A
(b) B
(c) 0
(d)A∪B
Solution:
(a) A

Question 10.
The shade region with adjoint diagram represents ……….

(a) A – B
(b) B – A
(c) A’
(d) B’
Solution:
(c) A’

Question 11.
From the given venn diagram (A∪B)’ is ………..

(a) {5, 6}
(b) {1, 2, 3, 4, 7}
(c) {1, 2, 3, 4, 5, 6, 7}
(d) {8, 9}
Solution:
(d) {8, 9}

Question 12.
If n(A∪B∪C) = 73, n(A) = 2x, n(B) = 3x, n(C) = 5x, n(A∩B) = 10, n(B∩C) = 15, n(A∩C) = 5 and n(A∩B∩C) = 3, then the value of x is ………
(a) 9
(b) 10
(c) 5
(d) 18
Solution:
(b) 10

Question 13.
For any three sets, n(A∪B∪C) = 60, n(A) = 25, n(B) = 20, n(C) = 15, n(A∩B) = 10, n(B∩C) = 7, n(A∩C) = 3, then n(A∩B∩C) is……….
(a) 10
(b) 15
(c) 20
(d) 25
Solution:
(c) 20

Question 14.
If n(U) = 70, n(A) = 25, n(B) = 30, n(A∩B) = 5, then n(A∪B)’ is……….
(a) 5
(b) 10
(c) 15
(d) 20
Solution:
(d) 20

Question 15.
Which of the following is not correct?
(a) A – (B∪C) = (A – B) ∩ (A – C)
(b) A – (B∩C) = (A – B) ∪ (A – C)
(c) (A∪B)’ = A’∩B’
(d) A’∪B’ = (A – B)’
Solution:
(d) A’∪B’ = (A – B)’

Answer the following questions.

Question 1.
Write the following in “Roster” form?
(a) A = set of the months having 31 days.
(b) B = {x : x is a natural number of 2 digits divisible by 13}
(c) C = {set of vowels in the word “father”}
(d) D = {x : 5 < x ≤ 10; x ∈ N}
(e) E = {x : x is a square natural number less than 16}
Solution:
(a) A = {Jan, March, May, July, Aug, Oct, Dec}
(b) B = {13, 26, 39, 52, 65, 78, 91}
(c) C = {a, e}
(d) D = {6, 7, 8, 9, 10}
(e) E = {1, 4, 9}

Question 2.
Given that A = {1, 3, 5, 7} B = {1, 2, 4, 6, 8}. Find
(i) AΔB and
(ii) BΔA
Solution:
(i) A = {1, 3, 5, 7}; B = {1, 2, 4, 6, 8}
A – B = {1, 3, 5, 7} – {1, 2, 4, 6, 8}
= {3, 5, 7}
B – A = {1, 2, 4, 6, 8} – {1, 3, 5, 7}
= {2, 4, 6, 8}
AΔB = (A – B) ∪ (B – A)
= {3, 5, 7} ∪ {2, 4, 6, 8}
= {2, 3, 4, 5, 6, 7, 8}
(ii) BΔA = (B – A) ∪ (A – B)
= {2, 4, 6, 8} ∪ {3, 5, 7}
= {2, 3, 4, 5, 6, 7, 8}

Question 3.
From the venn-diagram, list the following:

(i) A
(ii) B
(iii) A∩B
(iv)A∪B
(v) A – B
(vi) B – A
(vii) (A – B) ∩ (B – A)
Solution:
(i) A = {1, 2, 5, 6, 7}
(ii) B = {3, 4, 5, 6}
(iii) A∩B = {5, 6}
(iv) A∪B = {1, 2, 3, 4, 5, 6, 7}
(v) A – B = {1, 2, 7}
(vi) B – A = {3, 4}
(vii) (A – B) ∩ (B – A) = { }

Question 4.
In a class there are 40 students. 26 have opted for Mathematics and 24 have opted for Science. How many student have opted for Mathematics and Science.
Solution:
Let M be the set of students opting for Mathematics.
Let S be the set of students opting for Science.
n (M∪S) = 40, n (M) = 26, n(S) = 24
n(M∪S) = n (M) + n (S)- n(M∩S)
40 = 26 + 24 – n(M∩S)
n (M∩S) = 26 + 24 – 40 = 50 – 40 = 10
∴ Number of students opted for Mathematics and Science = 10.
Another Method:
Let “x” be the number of students opted for Mathematics and Science.
Let M and S represent students opting Mathematics and Science.
n(M∪S) = 40, n(M) = 26, n(S) = 24

By venn-diagram, number of students in a class = 26 – x + x + 24 – x
40 = 50 – x
x = 50 – 40 = 10
x = 10
∴ Number of students opted for Mathematics and Science = 10.

Question 5.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {4, 5, 7, 9}, B = {1, 3, 5, 7, 8}, Verify De Morgan’s Laws for complementation.
De Morgan’s Laws (i) (A∪B)’ = A’∩B’ (ii) (A∩B)’ = A’∪B’
Solution:
(i) A∪B = {4, 5, 7, 9} ∪ {1, 3, 5, 7, 8}
= {1, 3, 4, 5, 7, 8, 9}
(A∪B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 4, 5, 7, 8, 9}
= {2, 6}……….(1)
A’= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {4, 5, 7, 9}
= {1, 2, 3, 6, 8}
B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 5, 7, 8}
= {2, 4, 6, 9}
A’∩B’ = {1, 2, 3, 6, 8} ∩ {2, 4, 6, 9}
= {2, 6}………(2)
From (1) and (2) we get (A∪B)’ = A’∩B’.

(ii) A∩B = {4, 5, 7, 9} ∩ {1, 3, 5, 7, 8}
= {5, 7}
(A∩B)’= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 7}
= {1, 2, 3, 4, 6, 8, 9}………(1)
A’ = {1, 2, 3, 6, 8}
B’ = {2, 4, 6, 9}
A’∪B’ = {1, 2, 3, 6, 8} ∪ {2, 4, 6, 9}
= {1, 2, 3, 4, 6, 8, 9}………(2)
From (1) and (2) we get (A∩B)’ = A’∪B’.

Students can download Maths Chapter 1 Set Language Ex 1.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.3

Question 1.
Using the given venn diagram, write the elements of

(i) A
(ii) B
(iii) A∪B
(iv) A∩B
(v) A – B
(vi) B – A
(vii) A’
(viii) B’
(ix) U
Solution:
(i) A = {2, 4, 7, 8, 10}
(ii) B = {3, 4, 6, 7, 9, 11}
(iii) A∪B = {2, 3, 4, 6, 7, 8, 9, 10, 11}
(iv) A∩B = {4, 7}
(v) A – B = {2, 8, 10}
(vi) B – A = {3, 6, 9, 11}
(vii) A’ = {1, 3, 6, 9, 11, 12}
(viii) B’ = {1,2, 8, 10, 12}
(ix) U = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}

Question 2.
Find A∪B, A∩B, A – B and B – A for the following sets
(i) A = {2, 6, 10, 14} and B = {2, 5, 14, 16}
Solution:
A∪B = {2, 6, 10, 14} ∪ {2, 5, 14, 16}
= {2, 5, 6, 10, 14, 16}
A∩B = {2, 6, 10, 14} ∩ {2, 5, 14, 16}
= {2, 14}
A – B = {2, 6, 10, 14} – {2, 5, 14, 16}
= {6, 10}
B – A = {2, 5, 14, 16} – {2, 6, 10, 14}
= {5, 16}

(ii) A = {a, b, c, e, u} and B = {a, e, i, o, u}
Solution:
A∪B = {a, b, c, e, u} ∪ {a, e, i, o, u}
= {a, b, c, e, i, o, u}
A∩B = {a, b, c, e, u} ∩ {a, e, i, o, u}
= {a, e, u}
A – B = {a, b, c, e, u} – {a, e, i, o, u}
= {b, c}
B – A = {a, e, i, o, u} – {a, b, c, e, u}
= {i, o}

(iii) A = {x : x ∈ N, x ≤ 10} and B = {x : x ∈ W, x < 6}
Solution:
A = {1, 2, 3, 4, 5, 6,7, 8, 9, 10} and B = {0, 1, 2, 3, 4, 5}
A∪B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∪ {0, 1, 2, 3, 4, 5}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A∩B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 3, 4, 5}
= {1, 2, 3, 4, 5}
A – B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {0, 1, 2, 3, 4, 5}
= {6, 7, 8, 9, 10}
B – A = {0, 1, 2, 3, 4, 5} – {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
= {0}

(iv) A = Set of all letters in the word “mathematics” and B = Set of all letters in the word “geometry”
Solution:
A = {m, a, t, h, e, i, c, s}
B = {g, e, o, m, t, r, y}
A∪B = {m, a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y}
= {a, c, e, g, h, i, m, o, r, s, t, y}
A∩B= {m, a, t, h, e, i, c, s} ∩ {g, e, o, m, t, r, y}
= {e, m, t}
A – B = {m, a, t, h, e, i, c, 5} – {g, e, o, m, t, r, y}
= {a, c, h, i, s}
B -A = {g, e, o, m, t, r, y} – {m, a, t, h, e, i, c, s}
= {g, o, r, y}

Question 3.
If U = {a, b, c, d, e, f, g, h), A = {b, d, f, h} and B = {a, d, e, h}, find the following sets.
(i) A’
Solution:
A’ = U – A
= {a, b, c, d, e, f, g, h} – {b, d, f, h)
= {a, c, e, g}

(ii) B’
Solution:
B’ = U – B
= {a, b, c, d, e, f, g, h} – {a, d, e, h}
= {b, c, f, g}

(iii) A’∪B’
Solution:
A’∪B’ = {a, c, e, g} ∪ {b, c,f, g}
= {a, b, c, e, f, g}

(iv) A’∩B’
Solution:
A’∩B’ = {a, c, e, g} ∩ {b, c, f, g}
= {c, g}

(v) (A∪B)’
Solution:
A∪B = {b, d, f, h} ∪ {a, d, e, h}
= {a, b, d, e, f, h}
(A∪B)’ = U – (A∪B)
= {a, b, c, d, e, f, g, h} – {a, b, d, e, f, h}
= {c, g}

(vi) (A∩B)’
Solution:
(A∩B) = {b, d, f, h) ∩ {a, d, e, h)
= {d, h}
(A∩B)’ = U – (A∩B)
= {a, b, c, d, e, f, g, h} – {d, h}
= {a, b, c, e, f, g}

(vii) (A’)’
Solution:
A’ = {a, c, e, g}
(A’)’ = U – A’
= {a, b, c, d, e, f, g, h} – {a, c, e, g}
= {b, d, f, h}

(viii) (B’)’
Solution:
B’ = {b, c, f, g}
(B’)’ = U – B’
= {a, b, c, d, e, f, g, h) – {b, c, f, g)
= {a, d, e, h}

Question 4.
Let U = {0, 1, 2, 3, 4, 5, 6, 7} A = {1, 3, 5, 7} and B = {0, 2, 3, 5, 7}, find the following sets.
(i) A’
Solution:
A’ = U – A
= {0, 1, 2, 3, 4, 5, 6, 7} – {1, 3, 5, 7}
= {0, 2, 4, 6}

(ii) B’ = U – B
Solution:
= {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 3, 5, 7}
= {1, 4, 6}

(iii) A’∪B’
Solution:
A’∪B’ = {0, 2, 4, 6,}∪{1, 4, 6}
{0, 1, 2, 4, 6}

(iv) A’∩B’
Solution:
A’∩B’ = {0, 2, 4, 6,}∩{1, 4, 6}
{4, 6}

(v) (A∪B)’
Solution:
A∪B = {1, 3, 5, 7}∪{0, 2, 3, 5, 7}
= {0, 1, 2, 3, 5, 7}
(A∪B)’ = U – (A∪B)
{0, 1, 2, 3, 4, 5, 6, 7} – {0, 1, 2, 3, 5, 7}
{4, 6}

(vi) (A∩B)’
Solution:
(A∩B)= {1, 3, 5, 7}∩{0, 2, 3, 5, 7}
= {3, 5, 7}
(A∩B)’ = U – (A∩B)
= {0, 1, 2, 3, 4, 5, 6, 7} – {3, 5, 7}
= {0, 1, 2, 4, 6}

(vii) (A’)’
A’ = {0, 2, 4, 6}
(A’)’ = U – A’
= {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 4, 6}
= {1, 3, 5, 7}

(viii) (B’)’
B’ = {1, 4, 6}
(B’)’ = {0, 1, 2, 3, 4, 5, 6, 7} – {1, 4, 6}
= {0, 2, 3, 5, 7}

Question 5.
Find the symmetric difference between the following sets.
(i) P = {2, 3, 5, 7, 11} and Q = {1, 3, 5, 11}
Solution:
Use anyone of the formula to find A & B
AΔB = (A – B)∪(B – A) or AΔB = (A∪B) – (A∩B)
P∪Q = {2, 3, 5, 7, 11} ∪ {1,3,5, 11}
= {1, 2, 3, 5, 7, 11}
P∩Q = {2, 3, 5, 7, 11}∩{1, 3, 5,11}
= {3, 5, 11}
PΔQ = (P∪Q) – (P∩Q)
= {1, 2, 3, 5, 7, 11} – {3, 5, 11}
= {1, 2, 7}
(OR)
P – Q = {2, 3, 5, 7, 11} – {1, 3, 5, 11}
= {2,7}
Q – P = {1, 3, 5, 11} – {2, 3, 5, 7, 11}
= {1}
PΔQ = (P – Q)∪(Q – P)
= {2, 7} ∪ {1}
= {1, 2, 7}

(ii) R = {l, m, n, o, p} and S = {j, l, n, q}
Solution:
R- S = {l, m, n, o, p} – {j, l, n, q}
= {m, o, p}
S – R = {j, l, n, q} – {l, m, n, o, p}
= {j, q}
RΔS = (R – S)∪(S – R)
= {m, o, p} – {j, q} = {j, m, o, p, q)
(OR)
R∪S = {l, m, n, o, p} ∪ {j,l,n,q}
= {l, m, n, o, p, j, q}
R∩S = {l, m, n, o,p} ∩ {j, l, n, q}
= {l, n}
RΔS = (R∪S) – (R∩S)
= {l, m, n, o, p, j, q} – { l, n}
= {m, o, p, j, q}

(iii) X = {5, 6, 7} and Y = {5, 7, 9, 10}
Solution:
X∪Y = {5, 6, 7} ∪ {5, 7, 9, 10}
= {5 ,6, 7, 9, 10}
X∩Y = {5, 6, 7} ∩ {5, 7, 9, 10}
= {5, 7}
XΔY = (X∪Y) – (X∩Y)
= {5, 6, 7, 9, 10} – {5, 7}
= {6, 9, 10}
OR
X – Y = {5, 6, 7} – {5, 7, 9, 10} = {6}
Y – X = {5, 7, 9, 10} – {5, 6, 7} = {9, 10}
XΔY = (X – Y) ∪ (Y – X)
= {6}∪{9, 10}
= {6, 9, 10}

Question 6.
Using the set symbols, write down the expressions for the shaded region in the following

Solution:

Question 7.
Let A and B be two overlapping sets and the universal set U. Draw appropriate Venn diagram for each of the following,
(i) A∪B
(ii) A∩B
(iii) (A∩B)’
(iv) (B – A)’
(v) A’∪B’
(Vi) A’∩B’
(vii) What do you observe from the Venn diagram (iii) and (v)?
Solution:
(i) A∪B

(ii) A∩B

(iii) (A∩B)’

(iv) (B – A)’

(v) A’∪B’

(Vi) A’∩B’

(vii) What do you observe from the diagram (iii) and (v)?
From the diagram (iii) and (v) we get (A∩B)’ = A’∪B’

Students can download Maths Chapter 1 Set Language Ex 1.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.7

I. Multiple Choice Questions.

Question 1.
Which of the following is correct?
(a) {7} ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(b) 1 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(c) 7 ∉ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(d) {7} ⊄ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Solution:
(b) 1 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Question 2.
The set P = {x | x ∈ Z, -1 < x < 1} is a ……….
(a) Singleton set
(b) Power set
(c) Null set
(d) Subset
Solution:
(a) Singleton set
Hint: P = {0}

Question 3.
If U = {x | x ∈ N, x < 10} and A = {x | x ∈ N, 2 ≤ x < 6} then (A’)’ is………..
{a) {1, 6, 7, 8, 9}
(b) {1, 2, 3, 4}
(c) {2, 3, 4, 5}
(d) { }
Solution:
(c) {2, 3, 4, 5}
Hint:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A= {2, 3, 4, 5}; A’ = {1, 6, 7, 8, 9}
(A’)’ = U – A
(A’)’ = {2, 3, 4, 5}

Question 4.
If B⊆A then n(A∩B) is………..
(a) n(A – B)
(b) n(B)
(c) n(B – A)
(d) n(A)
Solution:
(b) n(B)

Question 5.
If A = {x, y, z} then the number of non- empty subsets of A is…………..
(a) 8
(b) 5
(c) 6
(d) 7
Solution:
(d) 7
Hint:
n(A) = 3; P(A) = 2^m = 2³ = 8
Non-empty subsets of A = 8 – 1 = 7

Question 6.
Which of the following is correct?
(a) Ø ⊆ {a, b}
(b) Ø ∈ {a, b}
(c) {a} ∈ {a, b}
(d) a ⊆ {a, b}
Solution:
(a) Ø ⊆ {a, b}
Hint:
‘Q’ is a subset of every set.

Question 7.
If A∪B = A∩B then ……………
(a) A ≠ B
(b) A = B
(c) A ⊂ B
(d) B ⊂ A
Solution:
(b) A = B

Question 8.
If B – A is B, then A∩B is………….
(a) A
(b) B
(c) U
(d) Ø
Solution:
(d) Ø

Question 9.
From the adjacent diagram n[P(AΔB)] is………….

(a) 8
(b) 16
(c) 32
(d) 64
Solution:
(c) 32
Hint:
A – B = {60, 85, 75}
B – A = {90, 70}
AΔB = (A – B) ∪ (B – A)
= {60, 70, 75, 85, 90}
n[P(AΔB)] = 2⁵ = 32

Question 10.
If n(A) = 10 and n(B) = 15 then the minimum and maximum number of elements in A∩B is …………
(a) (10, 15)
(b) (15, 10)
(c) (10, 0)
(d) (0, 10)
Solution:
(d) (0, 10)

Question 11.
Let A = {Ø} and B = P(A) then A∩B is………..
(a) {Ø, {Ø}}
(b) {Ø}
(c) Ø
{d) {0}
Solution:
(b) {Ø}
Hint:
A = {Ø}, B = {{ }, {Ø}}
A∩B = {Ø}

Question 12.
In a class of 50 boys, 35 boys play carrom and 20 boys play chess then the number of boys play both games is……..
(a) 5
(b) 30
(c) 15
(d) 10
Solution:
(a) 5
Hint:
n(A∪B) = 50, n(A) = 35, n(B) = 20
n(A∩B) = n(A) + n(B) – n(A∪B)
= 35 + 20 – 50 = 5

Question 13.
If U = {x : x ∈ N and x < 10}, A = {1, 2, 3, 5, 8} and B = (2, 5, 6, 7, 9}, then n [(A∪B)’] is
(a) 1
(b) 2
(c) 4
(d) 8
Solution:
(a) 1
Hint:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A∪B ={1, 2, 3, 5, 8} ∪ {2, 5, 6, 7, 9}
= {1, 2, 3, 5, 6, 7, 8, 9}
(A∪B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 5, 6, 7, 8, 9}
= {4}
n(A∪B)’ = 1

Question 14.
For any three sets P, Q and R, P – (Q∩R) is ………
(a) P – (Q∪R)
(i)(P∩Q) – R
(c) (P – Q) ∪ (P – R)
(d) (P – Q) ∩ (P – R)
Solution:
(c)(P – Q) ∪ (P – R)

Question 15.
Which of the following is true?
(a) A – B = A∩B
(b) A – B = B – A
(c) (A∪B)’ = A’∪B’
(d) (A∩B)’ = A’∪B’
Solution:
(d) (A∩B)’ = A’∪B’

Question 16.
If n(A∪B∪C) = 100, n(A) = 4x, n(B) = 6x, n(C) = 5x, n(A∩B) = 20, n(B∩C) = 15 and n(A∩C) = 25, then n(A∩B∩C) = 10, then the value of x is……….
(a) 10
(b) 15
(c) 25
(d) 30
Solution:
(b) 15
Hint:
n(A∪B∪C) = n( A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)
100 = 4x + 6x + 5x – 20 – 15 – 25 + 10
100 = 15x – 50 ⇒ 150 = 15x ⇒ x = \(\frac{150}{15}\) = 10

Question 17.
For any three sets A, B and C, (A – B) ∩ (B – C) is equal to
(a) A only
(b) B only
(c) C only
(d) \(\phi \)
Solution:
(d) \(\phi \)

Question 18.
If J = Set of three sided shapes, K = Set of shapes with two equal sides and L = Set of shapes with right angle, then J∩K∩L is………
(a) Set of isosceles triangles
(b) Set of equilateral triangles
(c) Set of isosceles right triangles
(d) Set of right angled triangles
Solution:
(c) Set of isosceles right triangles
J = Set of three sided shapes; It is a triangle
K = Set of shapes with two equal sides (Isosceles triangle)
L = Set of shapes with right angle (One angle is right angle)
∴ J∩K∩L = Set of isosceles right triangles

Question 19.
The shaded region in the Venn diagram is

(a) Z – (X∪Y)
(b) (X∪Y) ∩ Z
(c) Z – (X∩Y)
(d) Z ∪ (X∩Y)
Solution:
(c) Z – (X∩Y)

Question 20.
In a city, 40% people like only one fruit, 35% people like only two fruits, 20% people like all the three fruits. How many percentage of people do not like any one of the above three fruits?
(a) 5
(b) 8
(c) 10
(d) 15
Solution:
(a) 5

Students can download Maths Chapter 1 Relations and Functions Ex 1.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.1

1. Find A × B, A × A and B × A
(i) A = {2, -2, 3} and B = {1, -4}
(ii) A = B = {p, q}
(iii) A – {m, n} ; B = Φ
Answer:
(i) A = {2, -2, 3} and B = {1, -4}
A × B = {2,-2, 3} × {1,-4}
= {(2, 1), (2, -4)(-2, 1) (-2, -4) (3, 1) (3,-4)}
A × A = {2,-2, 3} × {2,-2, 3}
= {(2, 2)(2, -2)(2, 3)(-2, 2)
(-2, -2)(-2, 3X3,2) (3,-2) (3,3)}
B × A = {1,-4} × {2,-2, 3}
= {(1, 2)(1, -2)( 1, 3)(-4, 2) (-4,-2)(-4, 3)}

(ii) A = B = {p, q}
A × B = {p, q) × {p, q}
= {(p,p),(p,q)(q,p)(q,q)}
A × A = {p,q) × (p,q)
= {(p,p)(p,q)(q,p)(q,q)
B × A = {p,q} × {p,q}
= {(p,p)(p,q)(q,p)(q,q)

(iii) A = {m, n} × B = Φ
Note: B = Φ or {}
A × B = {m, n) × { }
= { )
A × A = {m, n) × (m, n)}
= {(m, m)(w, w)(n, m)(n, n)}
B × A = { } × {w, n}
= { }

Question 2.
Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A × B and B × A.
Solution:
A = {1, 2, 3}, B = {2, 3, 5, 7}
A × B = {(1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3) , (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7)}
B × A = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3) , (5, 1), (5, 2), (5, 3), (7, 1), (7, 2) , (7, 3)}

Question 3.
If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4), (3,3) ,(3, 4)} find A and B.
Answer:
B × A = {(-2, 3)(-2, 4) (0, 3) (0, 4) (3, 3) (3,4)}
A = {3,4}
B = {-2,0,3}

Question 4.
If A ={5, 6}, B = {4, 5, 6} , C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C).
Solution:
A = {5,6}, B = {4, 5, 6},C = {5, 6, 7}
A × A = {(5, 5), (5, 6), (6, 5), (6, 6)} ……….. (1)
B × B = {(4, 4), (4, 5), (4, 6), (5, 4),
(5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} …(2)
C × C = {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6),
(6, 7), (7, 5), (7, 6), (7, 7)} …(3)
(B × B) ∩ (C × C) = {(5, 5), (5,6), (6, 5), (6,6)} …(4)
(1) = (4)
A × A = (B × B) ∩ (C × C)
It is proved.

Question 5.
Given A = {1,2,3}, B = {2,3,5}, C = {3,4} and D = {1,3,5}, check if
(A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D) is true?
Answer:
A = {1,2, 3}, B = {2, 3, 5}, C = {3,4} D = {1,3,5}
A ∩ c = {1,2,3} ∩ {3,4}
= (3}
B ∩ D = {2,3, 5} ∩ {1,3,5}
= {3,5}
(A ∩ C) × (B ∩ D) = {3} × {3,5}
= {(3, 3)(3, 5)} ….(1)
A × B = {1,2,3} × {2,3,5}
= {(1,2) (1,3) (1,5) (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5)}
C × D = {3,4} × {1,3,5}
= {(3,1) (3, 3) (3, 5) (4,1) (4, 3) (4, 5)}
(A × B) ∩ (C × D) = {(3, 3) (3, 5)} ….(2)
From (1) and (2) we get
(A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D)
This is true.

Question 6.
Let A = {x ∈ W | x < 2}, B = {x ∈ N |1 < x < 4} and C = {3, 5}. Verify that
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
(iv) A × (B ∪ C) = (A × B) ∪ (A × C)
Solution:
A = {x ∈ W|x < 2} = {0,1}
B = {x ∈ N |1 < x < 4} = {2,3,4}
C = {3,5}
LHS =A × (B ∪ C)
B ∪ C = {2, 3, 4} ∪ {3, 5}
= {2, 3, 4, 5}
A × (B ∪ C) = {(0, 2), (0, 3), (0,4), (0, 5), (1, 2) , (1, 3), (1, 4),(1, 5)} ………. (1)
RHS = (A × B) ∪ (A × C)
(A × B) = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
(A × B) ∪ (A × C)= {(0, 2), (0, 3), (0,4), (1, 2), (1, 3), (1, 4), (0, 5), (1, 5)} ….(2)
(1) = (2), LHS = RHS
Hence it is proved.

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
LHS = A × (B ∩ C)
(B ∩ C) = {3}
A × (B ∩ C) = {(0, 3), (1, 3)} …(1)
RHS = (A × B) ∩ (A × C)
(A × B) = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
(A × B) ∩ (A × C) = {(0, 3), (1, 3)} ……….. (2)
(1) = (2) ⇒ LHS = RHS.
Hence it is verified.

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
LHS = (A ∪ B) × C
A ∪ B = {0, 1, 2, 3, 4}
(A ∪ B) × C = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} …………. (1)
RHS = (A × C) ∪ (B × C)
(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
(B × C) = {(2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)}
(A × C) ∪ (B × C) = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} ………… (2)
(1) = (2)
∴ LHS = RHS. Hence it is verified.

Question 7.
Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that
(i) (A ∩ B) × C = (A × c) ∩ (B × C)
(ii) A × (B – C) = (A × B) – (A × C)
Answer:
A = {1,2, 3, 4, 5,6, 7}
B = {2, 3, 5,7}
C = {2}

(i) (A ∩ B) × C = (A × C) ∩ (B × C)
A ∩ B = {1, 2, 3, 4, 5, 6, 7} ∩ {2, 3, 5, 7}
= {2, 3, 5, 7}
(A ∩ B) × C = {2, 3, 5, 7} × {2}
= {(2, 2) (3, 2) (5, 2) (7, 2)} ….(1)
A × C = {1,2, 3, 4, 5, 6, 7} × {2}
= {(1,2) (2, 2) (3, 2) (4, 2)
(5.2) (6, 2) (7, 2)}
B × C = {2, 3, 5, 7} × {2}
= {(2, 2) (3, 2) (5, 2) (7, 2)}
(A × C) ∩ (B × C) = {(2, 2) (3, 2) (5, 2) (7, 2)} ….(2)
From (1) and (2) we get
(A ∩ B) × C = (A × C) ∩ (B × C)

(ii) A × (B – C) = (A × B) – (A × C)
B – C = {2, 3, 5, 7} – {2}
= {3,5,7}
A × (B – C) = {1,2, 3, 4, 5, 6,7} × {3,5,7}
= {(1, 3) (1, 5) (1, 7) (2, 3) (2, 5)
(2, 7) (3, 3) (3, 5) (3, 7) (4, 3)
(4, 5) (4, 7) (5, 3) (5, 5) (5, 7)
(6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} ………….(1)
A × B = {1,2, 3, 4, 5, 6, 7} × {2, 3, 5,7}
= {(1, 2) (1, 3) (1, 5) (1, 7) (2, 2) (2, 3)
(2, 5) (2, 7) (3, 2) (3, 3) (3, 5) (3, 7)
(4, 2) (4, 3) (4, 5) (4, 7) (5, 2) (5, 3) (5, 5)
(5, 7) (6, 2) (6, 3) (6, 5) (6, 7)
(7, 2) (7, 3) (7, 5) (7, 7)}
A × C = {1,2, 3,4, 5, 6, 7} × {2}
= {(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6.2) (7,2)}
(A × B) – (A × C) = {(1, 3) (1, 5) (1, 7)
(2, 3) (2, 5) (2, 7) (3, 3) (3, 5)
(3, 7) (4, 3) (4, 5) (4, 7) (5, 3) (5, 5)
(5, 7) (6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} ….(2)
From (1) and (2) we get
A × (B – C) = (A × B) – (A × C)

Relations
Let A and B be any two non-empty sets. A “relation” R from A to B is a subset of A × B satisfying some specified conditions.

Note:

1. The domain of the relations R = {x ∈ A/xRy, for some y ∈ B}
2. The co-domain of the relation R is B
3. The range of the ralation

R = (y ∈ B/xRy for some x ∈ A}

Students can download Maths Chapter 2 Real Numbers Ex 2.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.1

Question 1.
Which arrow best shows the position of \(\frac{11}{3}\) on the number line?

Solution:
D represent \(\frac{11}{3}\) on the number line.

Question 2.
Find any three rational numbers between \(\frac{-7}{11}\) and \(\frac{2}{11}\)
Solution:
Three rational numbers between \(\frac{-7}{11}\) and \(\frac{2}{11}\)
\(\frac{-6}{11}\), \(\frac{-5}{11}\), \(\frac{-4}{11}\), ……… \(\frac{1}{11}\)

Question 3.
Find any five rational numbers between
(i) \(\frac{1}{4}\) and \(\frac{1}{5}\)
Solution:
Converting the given rational numbers with the same denominators.
\(\frac{1}{4}\) = \(\frac{1×30}{4×30}\) = \(\frac{30}{120}\)
\(\frac{1}{5}\) = \(\frac{1×24}{5×24}\) = \(\frac{24}{120}\)
Five rational numbers between \(\frac{30}{120}\) and \(\frac{24}{120}\) are \(\frac{25}{120}\), \(\frac{26}{120}\), \(\frac{27}{120}\), \(\frac{28}{120}\) and \(\frac{29}{120}\)
Five rational numbers between \(\frac{1}{4}\) and \(\frac{1}{5}\) are \(\frac{25}{120}\), \(\frac{26}{120}\), \(\frac{27}{120}\), \(\frac{28}{120}\) and \(\frac{29}{120}\)
Other Method:
A rational numbers between \(\frac{1}{4}\) and \(\frac{1}{5}\) = \(\frac{1}{2}\)(\(\frac{1}{4}\)+\(\frac{1}{5}\)) = \(\frac{1}{2}\)(\(\frac{5+4}{20}\)) = \(\frac{1}{2}\) × \(\frac{9}{20}\) = \(\frac{9}{40}\)
A rational numbers between \(\frac{1}{4}\) and \(\frac{9}{40}\) = \(\frac{1}{2}\)(\(\frac{1}{4}\)+\(\frac{9}{40}\)) = \(\frac{1}{2}\)(\(\frac{10+9}{40}\)) = \(\frac{19}{80}\)
A rational numbers between \(\frac{1}{4}\) and \(\frac{19}{80}\) = \(\frac{1}{2}\)(\(\frac{1}{4}\)+\(\frac{19}{20}\)) = \(\frac{1}{2}\)(\(\frac{20+19}{80}\)) = \(\frac{39}{160}\)
A rational numbers between \(\frac{1}{4}\) and \(\frac{39}{160}\) = \(\frac{1}{2}\)(\(\frac{1}{4}\)+\(\frac{39}{160}\)) = \(\frac{1}{2}\)(\(\frac{40+39}{160}\)) = \(\frac{79}{320}\)
A rational numbers between \(\frac{1}{4}\) and \(\frac{79}{320}\) = \(\frac{1}{2}\)(\(\frac{1}{4}\)+\(\frac{79}{320}\)) = \(\frac{1}{2}\)(\(\frac{80+79}{320}\)) = \(\frac{159}{640}\)
∴ Five rational numbers are between \(\frac{9}{40}\), \(\frac{19}{80}\), \(\frac{39}{160}\), \(\frac{79}{320}\) and \(\frac{159}{640}\)

(ii) 0.1 and 0.11
Solution:
\(\frac{1×100}{10×100}\) = \(\frac{100}{1000}\)
\(\frac{11×10}{100×10}\) = \(\frac{110}{1000}\)
The five rational numbers are \(\frac{101}{1000}\), \(\frac{102}{1000}\), \(\frac{103}{1000}\), \(\frac{104}{1000}\), \(\frac{105}{1000}\), …….. (or)
The five rational numbers are 0.101, 0.102, 0.103, 0.104 and 0.105.

(iii) -1 and -2
Solution:
Converting to rational numbers, – 1 = \(-\frac{10}{11}\) and – 2 = \(-\frac{20}{10}\)
So five rational numbers between -2 and -1 are \(-\frac{11}{10}\), \(-\frac{12}{10}\), \(-\frac{13}{10}\), \(-\frac{14}{10}\), \(-\frac{15}{10}\).
Other Method:
A rational number between -1 and -2 = \(\frac{1}{2}\)[-1-2] = \(\frac{1}{2}\)[-3] = \(-\frac{3}{2}\)
A rational number between -1 and \(-\frac{3}{2}\) = \(\frac{1}{2}\)[-1 – \(\frac{3}{2}\)] = \(\frac{1}{2}\)(\(\frac{-2-3}{2}\)) = \(-\frac{5}{4}\)
A rational number between -1 and \(-\frac{5}{4}\) = \(\frac{1}{2}\)[-1 – \(\frac{5}{4}\)] = \(\frac{1}{2}\)(\(\frac{-4-5}{4}\)) = \(-\frac{9}{8}\)
A rational number between -1 and \(-\frac{9}{8}\) = \(\frac{1}{2}\)[-1 – \(\frac{9}{8}\)] = \(\frac{1}{2}\)(\(\frac{-8-9}{8}\)) = \(-\frac{17}{16}\)
A rational number between -1 and \(-\frac{17}{16}\) = \(\frac{1}{2}\)[1 – \(\frac{17}{16}\)] = \(\frac{1}{2}\)(\(\frac{-16-17}{16}\)) = \(\frac{1}{2}\) (\(\frac{-33}{16}\)) = \(\frac{-33}{32}\)
The five rational numbers are \(-\frac{3}{2}\), \(-\frac{5}{4}\), \(-\frac{9}{8}\), \(-\frac{17}{16}\), and \(\frac{-33}{32}\)

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.1

Question 1.
Find all positive integers which when divided by 3 leaves remainder 2.
Answer:
All the positive integers when divided by 3 leaves remainder 2
By Euclid’s division lemma
a = bq + r, 0 < r < b
a = 3q + r where 0 < q < 3
a leaves remainder 2 when divided by 3
∴ The positive integers are 2, 5, 8, 11,…

Question 2.
A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over.
Solution:
Using Euclid’s division algorithm,
a = 21q + r, we get 532 = 21 × 25 + 7.
The remainder is 7.
No. of completed rows = 25, left over flower pots = 7 pots.

Question 3.
Prove that the product of two consecutive positive integers is divisible by 2.
Answer:
Let n – 1 and n be two consecutive positive integers, then the product is n (n – 1)
n(n – 1) = n² – n
We know that any positive integer is of the form 2q or 2q + 1 for same integer q

Case 1:
when n = 2 q
n² – n = (2q)² – 2q
= 4q² – 2q
= 2q (2q – 1)
= 2 [q (2q – 1)]
n² – n = 2 r
r = q(2q – 1)
Hence n² – n. divisible by 2 for every positive integer.

Case 2:
when n = 2q + 1
n² – n = (2q + 1 )² – (2q + 1 )
= (2q + 1) [2q + 1 – 1]
= 2q (2q + 1)
n² – n = 2r
r = q (2q + 1)
n² – n divisible by 2 for every positive integer.

Question 4.
When the positive integers be a, b and c are divided by 13, the respective remainders are 9, 7 and 10. Show that a + b + c is divisible by 13.
Solution:
Let the positive integers be a, b, and c.
a = 13 q + 9
b = 13q + 1
c = 13 q + 10
a + b + c = 13q + 9 + 13q + 7 + 13q + 10
= 39q + 26
= 13 (3q + 2)
which is divisible by 13.

Question 5.
Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.
Answer:
Let the integer be ” x ”
The square of its integer is “x² ”
Let x be an even integer
x = 2q + 0
x2 = 4q²
When x is an odd integer
x = 2k + 1
x² = (2k + 1)²
= 4k² + 4k + 1
= 4k (k + 1) + 1
= 4q + 1 [q = k(k + 1)]
It is divisible by 4
Hence it is proved

Question 6.
Use Euclid’s Division Algorithm to find the Highest Common Factor (H.C.F) of
(i) 340 and 412
(ii) 867 and 255
(iii) 10224 and 9648
(iv) 84, 90 and 120
Solution:
To find the H.C.F. of 340 and 412. Using Euclid’s division algorithm.
We get 412 = 340 × 1 + 72
The remainder 72 ≠ 0
Again applying Euclid’s division algorithm
340 = 72 × 4 + 52
The remainder 52 ≠ 0.
Again applying Euclid’s division algorithm
72 = 52 × 1 + 20
The remainder 20 ≠ 0.
Again applying Euclid’s division algorithm,
52 = 20 × 2 + 12
The remainder 12 ≠ 0.
Again applying Euclid’s division algorithm.
20 = 12 × 1 + 8
The remainder 8 ≠ 0.
Again applying Euclid’s division algorithm
12 = 8 × 1 + 4
The remainder 4 ≠ 0.
Again applying Euclid’s division algorithm
8 = 4 × 2 + 0
The remainder is zero.
Therefore H.C.F. of 340 and 412 is 4.

(ii) To find the H.C.F. of 867 and 255, using Euclid’s division algorithm.
867 = 255 × 3 + 102
The remainder 102 ≠ 0.
Again using Euclid’s division algorithm
255 = 102 × 2 + 51
The remainder 51 ≠ 0.
Again using Euclid’s division algorithm
102 = 51 × 2 + 0
The remainder is zero.
Therefore the H.C.F. of 867 and 255 is 51.

(iii) To find H.C.F. 10224 and 9648. Using Euclid’s division algorithm.
10224 = 9648 × 1 + 576
The remainder 576 ≠ 0.
Again using Euclid’s division algorithm
9648 = 576 × 16 + 432
Remainder 432 ≠ 0.
Again applying Euclid’s division algorithm
576 = 432 × 1 + 144
Remainder 144 ≠ 0.
Again using Euclid’s division algorithm
432 = 144 × 3 + 0
The remainder is zero.
There H.C.F. of 10224 and 9648 is 144.

(iv) To find H.C.F. of 84, 90 and 120.
Using Euclid’s division algorithm
90 = 84 × 1 + 6
The remainder 6 ≠ 0.
Again using Euclid’s division algorithm
84 = 6 × 14 + 0
The remainder is zero.
∴ The H.C.F. of 84 and 90 is 6. To find the H.C.F. of 6 and 120 using Euclid’s division algorithm.
120 = 6 × 20 + 0
The remainder is zero.
Therefore H.C.F. of 120 and 6 is 6
∴ H.C.F. of 84, 90 and 120 is 6.

Question 7.
Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.
Answer:
Find the HCF of ( 1230 – 12) and (1926- 12)
i.e HCF of 1218 and 1914
By Euclid’s division algorithm
1914 = 1218 × 1 + 696
The remainder 696 ≠ 0
By Euclid’s division algorithm
1218 = 696 × 1 + 522
The remainder 522 ≠ 0
Again by Euclid’s division algorithm
696 = 522 × 1 + 174
The remainder 174 ≠ 0 Again by Euclid’s division algorithm
522 = 174 × 3 + 0
The remainder is zero
∴ HCF of 1218 and 1914 is 174
The largest value is 174

Question 8.
If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.
Solution:
Applying Euclid’s divison lemma to 32 and 60, we get
60 = 32 × 1 + 28 ……………. (i)
The remainder is 28 ≠ 0.
Again applying division lemma
32 = 28 × 1 + 4 ……………. (ii)
The remainder 4 ≠ 0.
Again applying division lemma
28 = 4 × 7 + 0 ………….. (iii)
The remainder zero.
∴ H.C.F. of 32 and 60 is 4.
From (ii), we get
32 = 28 × 1 + 4
⇒ 4 = 32 – 28 × 1
⇒ 4 = 32 – (60 – 32 × 1) × 1
⇒ 4 = 32 – 60 + 32
⇒ 4 = 32 × 2+(-1) × 60
∴ x = 2 and y = -1

Question 9.
A positive integer, when divided by 88, gives the remainder 61. What will be the remainder when the same number is divided by 11?
Answer:
Let the positive integer be “x”
x = 88 × y + 61 (a = pq + r)
since 88 is a multiple of 11
61 = 11 × 5 + 6
∴ The remainder is 6

Question 10.
Prove that two consecutive positive integers are always coprime.
Solution:
Let the numbers be I, I + 1:
They are co-prime if only +ve integer that divides both is 1.
I is given to be +ve integer.
So I = 1, 2, 3, ….
∴ One is odd and the other one is even. Hence H.C.F. of the two consecutive numbers is 1. Hence the result.