Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.2 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 2.2

Question 1.
For what values of natural number n, 4th can end with the digit 6?
Answer:
4n = (22)n = 22n
= 2n × 2n
2 is a factor of 4n
∴ 4n is always even.

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Question 2.
If m, n are natural numbers, for what values of m, does 2n × 5n ends in 5?
Solution:
2n × 5m
2n is always even for all values of n.
5m is always odd and ends with 5 for all values of m.
But 2n × 5m is always even and ends in 0.
∴ 2n × 5m cannot end with the digit 5 for any values of m. No value of m will satisfy 2n × 5m ends in 5.

Question 3.
Find the H.C.F. of 252525 and 363636.
Answer:
To find the HCF of 252525 and 363636 by using Euclid’s Division algorithm.
363636 = 252525 × 1 + 111111
The remainder 111111 ≠ 0
By division of Euclid’s algorithm
252525 = 111111 × 2 + 30303
The remainder 30303 ≠ 0
Again by division of Euclid’s algorithm
111111 = 30303 × 3 + 20202
The remainder 20202 ≠ 0
Again by division of Euclid’s algorithm.
30303 = 20202 + 10101
The remainder 10101 ≠ 0
Again by division of Euclid’s algorithm.
20202 = 10101 × 2 + 0
The remainder is 0
∴ The H.C.F. is 10101

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Question 4.
If 13824 = 2a × 3b then find a and b?
Answer:
Using factor tree method factorise 13824
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 2
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 1
13824 = 29 × 33
Given 13824 = 2a × 3b
Compare we get a = 9 and b = 3

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Aliter:
13824 = 29 × 33
Compare with
13824 = 2a × 3b
The value of a = 9 b = 3

Question 5.
If p1x1 × p2x2 × p3x3 × p4x4 = 113400 where p1 p2, p3, p4 are primes in ascending order and x1, x2, x3, x4, are integers, find the value of p1,p2,p3,p4 and x1,x2,x3,x4.
Answer:
Given 113400 = p1x1 × p2x2 × p3x3 × p4x4
Using tree method factorize 113400
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 3
113400 = 23 × 34 × 52 × 7
compare with
113400 = p1x1 × p2x2 × p3x3 × p4x4
P1 = 2, x1 = 3
P2 = 3, x2 = 4
P3 = 5, x3 = 2
P4 = 7, x4 = 1

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.
Answer:
Factorise 408 and 170 by factor tree method
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 4
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 5
408 = 23 × 3 × 17
170 = 2 × 5 × 17
To find L.C.M. list all prime factors of 408 and 170 of their greatest exponents.
L.C.M. = 23 × 3 × 5 × 17
= 2040
To find the H.C.F. list all common factors of 408 and 170.
H.C.F. = 2 × 17 = 34
L.C.M. = 2040 ; HCF = 34

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Question 7.
Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?
Answer:
The greatest number of 6 digits is 999999.
The greatest number must be divisible by L.C.M. of 24, 15 and 36
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 6
24 = 23 × 3
15 = 3 × 5
36 = 22 × 32
L.C.M = 23 × 32 × 5
= 360
To find the greatest number 999999 must be subtracted by the remainder when 999999 is divided by 360
The greatest number in 6 digits = 999999 – 279
= 999720
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 7

Question 8.
What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Solution:
35 = 5 × 7
56 = 2 × 2 × 2 × 7
91 = 7 × 13
LCM of 35, 56, 91 = 5 × 7 × 2 × 2 × 2 × 13 = 3640
∴ Required number = 3647 which leaves remainder 7 in each case.

Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2

Question 9.
Find the least number that is divisible by the first ten natural numbers?
Answer:
Find the L.C.M of first 10 natural numbers
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.2 8
The least number is 2520

Modular Arithmetic
Two integers “a” and “b” are congruence modulo n if they differ by an integer multiple of n. That b – a = kn for some integer k. This can be written as a = b (mod n).

Euclid’s Division Lemma and Modular Arithmetic

Let m and n be integers, where m is positive. By Euclid’s division Lemma we can write n = mq + r where 0 < r < m and q is an integer.
This n = r (mod m)

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1

Students can download Maths Chapter 1 Relations and Functions Unit Exercise 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 Unit Exercise 1

Question 1.
If the ordered pairs (x2 – 3x, y2 + 4y) and (-2, 5) are equal, then find x and y.
Answer:
(x2 – 3x, y2 + 4y) = (-2, 5)
x2 – 3x = -2
x2 – 3x + 2 = 0
(x – 2) (x – 1) = 0
x – 2 = 0 or x – 1 = 0
x = 2 or 1
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1 1
y2 + 4y = 5
y2 + 4y – 5 = 0
(y + 5) (y – 1) = 0
y + 5 = 0 or y – 1 = 0
y = -5 or y = 1
The value of x = 2, 1
and 7 = -5, 1
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1 2

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1

Question 2.
The Cartesian product A × A has 9 elements among which (-1, 0) and (0, 1) are found. Find the set A and the remaining elements of A × A.
Solution:
A = {-1, 0, 1}, B = {1, 0, -1}
A × B = {(-1, 1), (-1, 0), (-1, -1), (0, 1), (0, 0), (0, -1), (1, 1), (1, 0), (1, -1)}

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1

Question 3.
Given that f(x) = \(\left\{\begin{array}{rl}
{\sqrt{x-1}} & {x \geq 1} \\
{4} & {x<1}
\end{array}\right.\).
Find
(i) f(0) (ii)f (3) (iii) f(a + 1) in terms of a.(Given that a > 0)
Answer:
f(x) = \(\sqrt { x-1 }\) ; f(x) = 4
(i) f(0) = 4
(ii) f(3) = \(\sqrt { 3-1 }\) = \(\sqrt { 2 }\)
(iii) f(a + 1) = \(\sqrt { a+1-1 }\) = \(\sqrt { a }\)

Question 4.
Let A = {9, 10, 11, 12, 13, 14, 15, 16, 17} and let f: A → N be defined by f(n) = the highest prime factor of n ∈ A. Write f as a set of ordered pairs and find the range of f.
Solution:
A = {9, 10, 11, 12, 13, 14, 15, 16, 17}
f: A → N
f(n) = the highest prime factor of n ∈ A
f = {(9, 3), (10, 5), (11, 11), (12, 3), (13, 13), (14, 7), (15, 5), (16, 2), (17, 17)}
Range = {3, 5, 11, 13, 7, 2, 17}
= {2, 3, 5, 7, 11, 13, 17}

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1

Question 5.
Find the domain of the function
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1 3
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1 4
Domain of f(x) = {-1, 0, 1}

Question 6.
If f(x)= x2, g(x) = 3x and h(x) = x – 2 Prove that (fog)oh = fo(goh).
Solution:
f(x) = x2
g(x) = 3x
h(x) = x – 2
(fog)oh = x – 2
LHS = fo(goh)
fog = f(g(x)) = f(3x) = (3x)2 = 9x2
(fog)oh = (fog) h(x) = (fog) (x – 2)
= 9(x – 2)2 = 9(x2 – 4x + 4)
= 9x2 – 36x + 36 ……………. (1)
RHS = fo(goh)
(goh) = g(h(x)) = g(x – 2)
= 3(x – 2) = 3x – 6
fo(goh) = f(3x – 6) = (3x – 6)2
= 9x2 – 36x + 36 ………….. (2)
(1) = (2)
LHS = RHS
(fog)oh = fo(goh) is proved.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1

Question 7.
Let A= {1,2} and B = {1,2,3,4}, C = {5,6} and D = {5,6,7,8}. Verify whether A × C is a subset of B × D?
Answer:
Given A = {1, 2}
B = {1, 2, 3, 4}
C = {5,6}
D = {5,6, 7,8}
A × C = {1,2} × {5,6}
= {(1,5) (1,6) (2, 5) (2, 6)}
B × D = {1,2, 3, 4} × {5, 6, 7, 8}
= {(1,5) (1,6) (1,7) (1,8)
(2, 5) (2, 6) (2,7) (2, 8)
(3, 5) (3, 6) (3, 7) (3, 8)
(4, 5) (4, 6) (4, 7) (4, 8)}
∴ A × C ⊂ B × D
Hence it is verified

Question 8.
If f(x) = \(\frac{x-1}{x+1}, x \neq 1\) Show that
f(f(x)) = – \(\frac { 1 }{ x } \), Provided x ≠ 0.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1 5
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1 6

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1

Question 9.
The functions f and g are defined by f{x) = 6x + 8; g(x) = \(\frac { x-2 }{ 3 } \)
(i) Calculate the value of gg [latex]\frac { 1 }{ 2 } [/latex]
(a) Write an expression for gf (x) in its simplest form.
Answer:
f(x) = 6x + 8 ; g(x) = \(\frac { x-2 }{ 3 } \)
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1 7
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1 8
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1 88

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1

Question 10.
Write the domain of the following real functions
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Unit Exercise 1 9
Answer:
(i) f (x) = \(\frac { 2x+1 }{ x-9 } \)
If the denominator vanishes when x = 9
So f(x) is not defined at x = 9
∴ Domain is x ∈ [R – {9}]

(ii) if p(x) = \(=\frac{-5}{4 x^{2}+1}\)
p(x) is defined for all values of x. So domain is x ∈ R.

(iii) g(x) = \(\sqrt { x-2 }\)
When x < 2 g(x) becomes complex. But given “g” is real valued function.
So x > 2
Domain x ∈ (2, α)

(iv) h (x) = x + 6
For all values of x, h(x) is defined. Hence domain is x ∈ R.

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

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

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

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)
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.6

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
25x = 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)

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

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) nm
(3) 2mn – 1
(4) 2mn
Answer:
(4) 2mn

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

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

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:
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.6 2
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:
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.6 3

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

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)}
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.6 4
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}

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

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

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

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.

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

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
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.3 1

(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

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

(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

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

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.
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.3 11
The time repeat after 7 am is 7 + 4 = 11 o’ clock (or) 11 am.

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

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.
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.3 33
∴ 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)
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.3 3
The value of x must be 3.
Three days after tuesday is friday uncle will come on friday.

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

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,
2k + 6 × 9k = 7 m [where m is a scalar]
⇒ 6 × 9k = 7 m – 2k …………. (1)
Let us prove for n = k + 1
Consider 2k+1 + 6 × 9k+1 = 2k+1 + 6 × 9k × 9
= 2k+1 + (7m – 2k)9 (using (1))
= 2k+1 + 63m – 9.2k = 63m + 2k.21 – 9.2k
= 63m – 2k (9 – 2) = 63m – 7.2k
= 7 (9m – 2k) 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

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

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
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.3 4
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

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

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)}

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

(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)}

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

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.

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

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.

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

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.

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

(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.

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

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)

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

(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)

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

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}

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

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,…

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

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

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

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.

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

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.

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

(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.

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

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

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

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

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

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.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Students can download Maths Chapter 1 Relations and Functions Additional 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 Additional Questions

I. Multiple Choice Questions.

Question 1.
If n(A × B) = 15 and B = {1, 3, 7} then n(A) is ……………
(1) 3
(2) 5
(3) 1
(4) 15
Answer:
(2) 5
Hint: B(A × B) = 15
n(A) × n(B) = 15
n(A) × 3 = 15
n(A) = \(\frac { 15 }{ 3 } \) = 5

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 2.
If A = {a, b,c) B = {b, d, e}
C = {a, e, i, o, u} then n [A ∩ C] × B] is
(1) 18
(2) 36
(3) 9
(4) 3
Answer:
(4) 3
Hint:
A ∩ C = {a,b,c} ∩ {a, e, i, o, u}
= {a}
n(A ∩ C) = 1
n[(A ∩ C) × B] = n(A ∩ C) × n(B)
= 1 × 3
= 3

Question 3.
If there are 28 relation from a set A = {2,4, 6, 8} to a set B, then the number of elements in B is ………………
(1) 7
(2) 14
(3) 5
(4) 4
Answer:
(1) 7
Hint: n(A) = 4
n(A × B) = 28
n(A) × n(B) = 28
4 × n(B) = 28
n(B) = \(\frac { 28 }{ 4 } \) = 7

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 4.
The ordered pairs (a + 1, 4) (3, 4a + b) are equal then (a, b) is ………………..
(1) (4, 20)
(2) (20, 4)
(3) (-4, 20)
(4) (20, -4)
Answer:
(3) (-4, 20)
Hint: (a + 7, 4) = (3, 4a + b)
a + 7 = 3
a = 3 – 7
= – 4
4a + b = 4
4(-4) + b = 4
-16 + b = 4
b = 4 + 16 = 20
The pair (a, 6) is (-4, 20)

Question 5.
The range of the relation R = {(x, x3) / x} is a prime number less than 13} is …………………
(1) (2,3,5,7,11)
(2) (4,9,25,49,121)
(3) (8,27, 125,343, 1331)
(4) (1,8,27, 125,343, 1331)
Answer:
(3) (8, 27, 125,343, 1331)
Hint: x = {2, 3, 5, 7, 11}
Range (x3) = {8, 27, 125, 343, 1331}

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 6.
If {( x, 2), (4, y) } represents an identity function, then (x, y) is
(1) (2, 4)
(2) (4, 2)
(3) (2, 2)
(4) (4, 4)
Answer:
(1) (2, 4)
Hint: In an identity function each element is associated with itself.

Question 7.
If {(7, 11), (5, a)} represents a constant
function, then the value of ‘a’ is
(1) 7
(2) 11
(3) 5
(4) 9
Answer:
(2) 11
Hint: All the images are same in a constant function.

Question 8.
Given f(x) = (- 1)x is a function from N to Z. Then the range of f is
(1) {1}
(2) N
(3) { 1,- 1 }
(4) Z
Answer:
(3) {1, – 1}
Hint: f(x) = (- 1)x = ± 1

Question 9.
If f = { (6, 3), (8, 9), (5, 3), (-1, 6) }, then the pre-images of 3 are
(1) 5 and-1
(2) 6 and 8
(3) 8 and-1
(4) 6 and 5
Answer:
(4) 6 and 5.
Hint: The Pre images of 3 are 6 and 5

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 10.
Let A= { 1, 3, 4, 7, 11 }, B = {-1, 1, 2, 5, 7, 9 } and f : A → B be given by
f = {(1, -1), (3,2), (4, 1), (7, 5), (11, 9)}.
Then f is ………………….
(1) one-one
(2) onto
(3) bijective
(4) not a function
Answer:
(1) one – one
Hint:
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 1

Question 11.
The given diagram represents
(1) an onto function
(2) a constant function
(3) an one-one function
(4) not a function
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 2
Answer:
(4) not a function
Hint: 2 has two images 4 and 2.
It is not a function.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 12.
If A = { 5, 6, 7 }, B = { 1, 2, 3, 4, 5 }and f: A → B is defined by f(x) = x – 2, then the range of f is …………….
(1) {1,4, 5}
(2) {1,2, 3, 4, 5}
(3) { 2, 3, 4 }
(4) { 3, 4, 5 }
Answer:
(4) {3, 4, 5}
Hint: f(x) = x – 2
f(5) = 5 – 2 = 3
f(6) = 6 – 2 = 4
f(7) = 7 – 2 = 5
Range of f = {3, 4, 5}

Question 13.
If f(x) = x2 + 5, then f(-4) = ………
(1) 26
(2) 21
(3) 20
(4) – 20
Answer:
(2) 21
Hint: f(x) = x2 + 5
f(- 4) = (-4)2 + 5 = 16 + 5 = 21

Question 14.
If the range of a function is a singleton set, then it is ……………..
(1) a constant function
(2) an identity function
(3) a bijective function
(4) an one-one function
Answer:
(1) a constant function
Hint: Every element of the first set has same image in the second set.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 15.
If f : A → B is a bijective function and if n(A) = 5 , then n(B) is equal to ………………
(1) 10
(2) 4
(3) 5
(4) 25
Answer:
(3) 5
Hint: If A and B are Bijective (one-one and onto) function then n (A) = n (B)

Question 16.
If f: R → R defined by f(x) = 3x – 6 and g : R → R defined by g(x) = 3x + k if fog – gof then the value of k is …………………..
(1) – 5
(2) 5
(3) 6
(4) -6
Answer:
(4) – 6
Hint: f(x) = 3x – 6 ;g(x) = 3x + k
fog = f[g(x)]
= f(3x + k)
= 3 (3x + k) – 6
= 9x + 3k – 6
g o f = g[f(x)]
= g(3x – 6)
= 3(3x – 6 ) + k
= 9x – 18 + k
But fog = gof
9x + 3k – 6 = 9x – 18 + k
3k – k = -18 + 6
2k = -12
k = \(\frac { -12 }{ 2 } \) = -6

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 17.
If f(x) = x2 – x then f (x – 1) – f(x + 1) is ……………….
(1) 4x
(2) 4x + 2
(3) 2 – 4x
(4) 4x – 2
Answer:
(3) 2 – 4x
Hint: f(x – 1) = (x – 1)2 – (x – 1)
= x2 – 2x + 1 – x + 1
= x2 – 3x + 2
f(x + 1) = (x + 1)2 – (x + 1)
= x2 + 2x + 1 – x – 1
= x2 + x
f(x – 1) – f(x + 1) = x2 – 3x + 2 – (x2 + x)
= x2 – 3x + 2 – x2 – x
= -4x + 2 = 2 – 4x

Question 18.
If K(x) = 3x – 9 then L (x) = 7x – 10 then LOK is ……………..
(1) 21x + 73
(2) – 21x + 73
(3) 21x – 73
(4) 22x – 73
Answer:
(3) 21x – 73
Hint: K (x) = 3x – 9 ; L(x) = 7x – 10
LOK = L[K(x)]
= L (3x – 9)
= 7(3x – 9) – 10
= 21x – 63 – 10
= 21x – 73

Question 19.
Composition of function is ……………..
(1) commutative
(2) associative
(3) commutative and associative
(4) not associative
Answer:
(2) associative

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 20.
A comet is heading for Jupiter with acceleration a = 50 kms-2. The velocity of the comet at time ”t” is given by f(t) = at2 – at + 1. Then the velocity at time t = 5 seconds is …………..
(1) 900kms-1
(2) 1001 kms-1
(3) 2001 kms-1
(4) 50 kms-1
Answer:
(2) 1001 kms-1
Hint: f(t) = at2 – at + 1
m = 50(5)2 – 50(5) + 1
= 1250 – 250 + 1
= 1001 kms-1

II. Answer the following questions.

Question 1.
f(x) = (1 + x)
f(x) = (2x – 1)
Show that fo(g(x)) = gof(x)
Solution:
f(x) = 1 + x
g(x) = (2x – 1)
fog(x) = f(g(x)) = f(2x – 1)
= 1 + 2x – 1 = 2x ………….. (1)
gof(x) = g(f(x)) = g(1 + x) = 2(1 + x) = 1
= 2 + 2x – 1
= 2x + 1 ……………. (2)
(1) ≠ (2)
∴ fog(x) + gof(x) It is verified.

Question 2.
If A × B = {(a, x) (a, y) (b, x) (b, y) (c, x) (c, y)} then find A and B
Answer:
A × B = {(a, x) (a, y) (b, x) (b, y) (c, x) (c, y)}
A = {a, b, c}
B = {x,y}

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 3.
Let A = {x ∈ w/3 < x < 7},
B = {x ∈ N/0 < x < 3}, C = {x ∈ w/x < 2}
verify A × (B ∩ C) = (A × B) ∩ (A × C)
Answer:
A = {4,5,6} ; B = {1,2} C = {0, 1}
B ∩ C = {1,2} ∩ {0, 1}
= {1}
A × (B ∩ C) = {4,5,6} × {1}
= {(4, 1) (5, 1) (6, 1)} …. (1)
A × B = {4,5,6} × {1,2}
= {(4, 1) (4, 2) (5, 1) (5, 2) (6, 1) (6, 2)}
A × C = {4,5,6} x {0, 1}
= {(4,0) (4,1) (5,0)
(5, 1) (6, 0) (6, 1)}
(A × B) ∩ (A × C) = {(4, 1) (5, 1) (6, 1)}…. (2)
From (1) and (2) we get
A × (B ∩ C) = (A × B) ∩ (A × C)

Question 4.
Let A = {10, 11, 12, 13, 14}; B = {0, 1, 2, 3, 5} and fi: A → B, i = 1, 2, 3. State the type of function for the following (give reason):
(i) f1 = {(10,1), (11,2), (12,3), (13,5), (14,3)}
Answer:
The element 12 and 14 in A have same image 3 in B.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 4
∴ It is not one-one function.
The element ‘0’ in B has no preimage in A
∴ It is not onto function
So the given function is neither one – one nor onto function.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

(ii) f2 = {(10,1), (11,1), (12,1), (13,1), (14,1)}
Answer:
f2 is a constant function
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 3
(iii) f3 = {(10,0), (11,1), (12,2), (13,3), (14,5)}
Answer:
f3 is one-one and onto function (or) bijective function.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 44

Question 5.
If X = {1, 2, 3, 4, 5}, Y = {1, 3, 5, 7, 9} determine which of the following relations from X to Y are functions? Give reason for your answer. If it is a function, state its type.
(i) R1 = {(x,y)| y = x + 2,x ∈ X,y ∈ Y}
Answer:
Given y = x + 2
When x = 1 ; y = 1 + 2 = 3
When x = 2 ; y = 2 + 2 = 4
When x = 3 ; y = 3 + 2 = 5
When x = 4 ; y = 4 + 2 = 6
When x = 5 ; y = 5 + 2 = 7
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 5
R1 = {1,3), (2,4), (3, 5), (4, 6), (5,7)}
R1 is not a function ; 2 and 4 has no image in Y.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

(ii) R2 = {(1,1), (2,1), (3,3), (4,3), (5,5)}
Answer:
R2 is a function.
Every element of X has unique image in Y.
1 and 2 have same image 1
3 and 4 have same image 3
It is not one – one function …. (1)
7 and 9 has no pre image in X
It is not an onto function …. (2)
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 6
From (1) and (2) we know that, it is
neither one – one nor onto function.

(iii) R3 = {(1,1), (1,3), (3,5), (3,7), (5,7)}
Answer:
R3 is not a function.
1 has two images 1 and 3
3 has two images 5 and 7
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 7

(iv) R4 = {(1,3), (2,5), (4,7), (5,9), (3,1)}
Answer:
Every element of X has unique image in
Y. Range = Co-domain
R4 is a function.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 8
It is an one-one and onto function (or) bijective function

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 6.
A= {-2,-1, 1, 2} and f = {(x,\(\frac { 1 }{ x } \)) ; x ∈ A}
Write down the range of f. Is f a function from A to A?
Answer:
Given, f = (x,\(\frac { 1 }{ x } \)) ; So f(x) = \(\frac { 1 }{ x } \)
f (-2) = \(\frac { 1 }{ -2 } \) = – \(\frac { 1 }{ 2 } \) ; f(-1) = \(\frac { 1 }{ -1 } \) = -1
f(1) = \(\frac { 1 }{ 1 } \) = 1 ; f(2) = \(\frac { 1 }{ 2 } \) = \(\frac { 1 }{ 2 } \)
Range of f = {\(\frac { -1 }{ 2 } \), -1, 1, \(\frac { 1 }{ 2 } \)}
It is not a function from A to A since – \(\frac { 1 }{ 2 } \) ,\(\frac { 1 }{ 2 } \) ∈ A

Question 7.
Let A = {1, 2, 3, 4, 5}, B = N and f: A → B be defined by f(x) = x2. Find the range of f. Identify the type of function.
Solution:
A = {1, 2, 3, 4, 5}
B = {1, 2, 3, 4 ….}
f: A → B, f(x) = x2
∴ f(1) = 12 = 1
f(2) = 22 = 4
f(3) = 32 = 9
f(4) = 42 = 16
f(5) = 52 = 25
∴ Range of f = {1, 4, 9, 16, 25)
Elements in A have been different elements in B. Therefore it is one-one function. But not all the elements in B have preimages in A. Therefore it is not on-to function.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 8.
Let A = { 1, 2, 3, 4, 5 }, B = N and f: A → B be defined by f(x) = x2.
Find the range of f. Identify the type of function.
Answer:
Now, A = { 1, 2, 3, 4, 5 };
B = { 1, 2, 3, 4, … }
Given f: A → B and f(x) = x2
f(1) = 12 = 1;
f(2) = 4;
f(3) = 9;
f(4) = 16;
f(5) = 25.
Range of f = {1, 4, 9, 16, 25}
Since distinct elements are mapped into distinct images, it is a one-one function.
However, the function is not onto, since 3 ∈ B but there is no x ∈ A
such that
f(x) = x2 = 3.

Question 9.
Identify the type of function.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 9
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 10
Answer:
(i) Many – one into
(ii) One – one onto
(iii) Constant function
(iv) One – one into

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 10.
Find the domain and range of the following
(i) f = {(1, 2), (2, 3), (3, 4), (4, 5) (5, 6)}
(ii) R = {(-2, 4), (-1,1), (2,4), (1,1) (-3, 9)}
Answer:
(i) f = {( 1,2), (2, 3), (3, 4), (4, 5) (5, 6)}
Domain = {1,2, 3,4, 5}
Range = {2, 3, 4, 5, 6}

(ii) R = {(-2,4), (-1, 1),(2,4), (1,1) (-3,9)}
Domain = {-2, -1,2, 1,-3} (or)
= {-3,-2,-1, 1,2}
Range = {4, 1, 9} (or) {1, 4, 9}

Question 11.
Given P ={-2,-1, 0,1}
Q = {1,-2, 6,-3}
R = {x,y/y = x2 – 3 x ∈ P,y ∈ Q}
(i) List the elements of R
(ii) Is the relation a function? If so identity the function
Answer:
P = {-2, -1, 0, 1}; Q = {1, -2,6, -3}
y = x2- 3 x ∈ P, y ∈ Q
When x = -2 ⇒ y = (-2)2 – 3 = 4 – 3 = 1
When x = -1 ⇒ y = (-1 )2 – 3 = 1 – 3 = -2
When x = 0 ⇒ y = (0)2 – 3 = 0 -3 = -3
When x = 1 ⇒ y = 12 – 3 = 1 – 3 = -2
(i) R = {(-2,1), (-1,-2), (0,-3), (1,-2)}
(ii) Yes the relation is a function many – one into function.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 12.
Given f(x) = 3x – 2; g(x) = 2x2 find
(i) fog and
(ii) gof what do you find
Answer:
f(x) = 3x – 2 ; g(x) = 2x2
(i) f o g = f[g(x)]
= f(2x2)
= 3(2x2) – 2
= 6x2 – 2

(ii) g o f = g [f(x)]
= g (3x – 2)
= 2(3x – 2)2
= 2(9x2 + 4 – 12x)
= 18x2 – 24x + 8
we find that fog ≠ gof
Composition of function is not commutative.

Question 13.
If f: R → R is defined by f(x) = ax + 3 and g: R → R is defined by g (x) = 4x – 3 find a so that fog = gof
Answer:
f(x) = ax + 3 ; g(x) = 4x -3
fog = f[g(x)]
= f(4x – 3)
= a (4x – 3) + 3
= 4ax – 3a + 3
gof = g [f(x)]
= g (ax + 3)
= 4 (ax + 3) – 3
= 4 ax + 12 – 3
= 4ax + 9
But fog = gof
4ax – 3a + 3 = 4ax + 9
-3a + 3 = 9
– 3a = 6
a = – 2
The value of a = – 2

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 14.
Given f(x) = 3 + x ; g(x) = x2 ;
h(x) = \(\frac { 1 }{ x } \) find fo (goh)
Answer:
f(x) = 3 + x ; g (x) = x2, h(x) = \(\frac { 1 }{ x } \)
goh = g[h(x)]
= g (\(\frac { 1 }{ x } \))
= (\(\frac { 1 }{ x } \))2
goh = \(\frac{1}{x^{2}}\)
fo(goh) = f (\(\frac{1}{x^{2}}\))
= 3 + \(\frac{1}{x^{2}}\)

Question 15.
If f(x) = x + 3 where A = {4, 6, 8,10} B = {7, 9,11,13} and f: A → B
(i) Draw the arrow diagram
(ii) Why type of function is f.
Answer:
A= {4, 6, 8, 10}
f(x) = x + 3
f(4) = 4 + 3 = 7
f(6) = 6 + 3 = 9
f(8) = 8 + 3 = 11
f(10) = 10 + 3 = 13
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 11
(ii) one – one onto function

III. Answer the following Questions

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

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 2.
Study the relation given below an set- builder form. Represent each of them by
(a) an arrow diagram
(b) a graph
(c) a set in roster.
If {{x,y}/y = 2x + 1; x < 10 and y < 12 x ∈ N, y ∈ N}
Answer:
y = 2x + 1
when x = 1 ⇒ y = 2(1) + 1 = 2 + 1 = 3
when x = 2 ⇒ y = 2(2) + 1 = 4 + 1 = 5
when x = 3 ⇒ y = 2(3) + 1 = 6 + 1 = 7
when x = 4 ⇒ y = 2(4) + 1 = 8 + 1 = 9
when x = 5 ⇒ y = 2(5) + 1 = 10 + 1 = 11
f = {(1,3) (2, 5) (3, 7) (4, 9) (5, 11)}

(a) Arrow diagram
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 12
(b) A graph
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 13
(c) Roster form: R = {(1,3) (2,5) (3,7) (4,9) (5,11)}

Question 3.
State whether the following graphs represents a function. Give reason for your answer.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 14
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 15
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 16
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 17
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 18
Answer:
(i) The given graph represents a function. The vertical line cuts the graph at most one point R
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 19
(ii) The vertical line cuts the graph at most one point Q. The given graph represents a function.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 20
(iii) The vertical line cuts the graph at A and B. The given graph does not represents a function.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 21
(iv) The vertical line cuts the graph at A and B. The given graph does not represents a function.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 22
(v) The vertical line cuts the graph at most one point R. The given graph represents a function.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 23

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 4.
Let A = {6, 9,15,18, 21}; B = {1, 2, 4, 5, 6} and f: A → B be defined by f(x) = \(\frac { x-3 }{ 3 } \) Represent f by, (i) an arrow diagram, (ii) a set of ordered pairs, (iii) a table, (iv) a graph.
Given, A = {6, 9, 15, 18, 21}, B = {1, 2, 4, 5, 6}
f(x) = \(\frac { x-3 }{ 3 } \)
f(6) = \(\frac { 6-3 }{ 3 } \) = \(\frac { 3 }{ 3 } \) = 1
f(9) = \(\frac { 9-3 }{ 3 } \) = \(\frac { 6 }{ 3 } \) = 2
f(15) = \(\frac { 15-3 }{ 3 } \) = \(\frac { 12 }{ 3 } \) = 4
f(18) = \(\frac { 18-3 }{ 3 } \) = \(\frac { 12 }{ 3 } \) = 4
f(18) = \(\frac { 18-3 }{ 3 } \) = \(\frac { 15 }{ 3 } \) = 5
f(21) = \(\frac { 21-3 }{ 3 } \) = \(\frac { 18 }{ 3 } \) = 6

(i) an arrow diagram
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 24
(ii) a set of ordered pairs
f = {(6,1), (9, 2), (15, 4), (18, 5), (21,6)}

(iii) a table
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 25

(iv) a graph
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 26

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 5.
Let A = {4,6,8,10} and B = {3,4,5,6,7}. If f: A → B is defined by f(x) = \(\frac { 1 }{ 2 } \) x + 1 then represent f by (i) an arrow diagram, (ii) a set of ordered pairs and, (iii) a table.
Answer:
Given, A = {4, 6, 8, 10}
B = {3, 4, 5, 6, 7}
f(x) = \(\frac { x }{ 2 } \) + 1
f(4) = \(\frac { 4 }{ 2 } \) + 1 = 2 + 1 = 3
f(6) = \(\frac { 6 }{ 2 } \) + 1 = 3 + 1 = 4
f(8) = \(\frac { 8 }{ 2 } \) + 1 = 4 + 1 = 5
f(10) = \(\frac { 10 }{ 2 } \) + 1 = 5 + 1 = 6

(i) an arrow diagram
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 27
(ii) a set of ordered pairs
f = {(4, 3), (6, 4), (8, 5), (10, 6)}

(iii) a table
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 28

Question 6.
A function f[- 3, 7 ) → R is defined as follows f(x) =
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 29
Find (i) f(5) + f(6)
(ii) f(1) – f(-3)
(iii) f(-2) – f(4)
(iv) \(\frac{f(3)+f(-1)}{2 f(6)-f(1)}\)
Answer:
Given, f(x) = 4x2 – 1; x = {-3, -2, -1, 0, 1}
f(x) = 3x – 2; x = {2,3,4}
f(x) = 2x – 3; x = {5,6}
(i) f(5) + f(6)
f(x) = 2x – 3
f(5) = 2(5) – 3 = 10 – 3 = 7
f(6) = 2(6) – 3 = 12 – 3 = 9
∴ f(5) + f(6) = 7 + 9 = 16

(ii) f(1) – f(-3)
f(x) = 4x2 – 1
f(1) = 4(1)2 – 1 = 4 – 1 = 3
f(-3) = [4(-3)2 – 1]
= 4 (9) – 1
= 36 – 1 = 35
∴ f(1) – f(-3) = 3 – (35) = -32

(iii) f(-2) – f(4)
f(x) = 4x2 – 1
f(-2) = 4(-2)2 – 1 = 4(4) – 1 = 16 – 1 = 15
f(x) = 3x – 2
f(4) = [3(4) – 2] = 12 – 2 = 10
∴ f(-2) – f(4) = 15 – 10 = 5

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 30

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 7.
A function f : [- 7, 6) → R is defined as follows
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 31
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 32
Answer:
Given, f(x) = x2 + 2x + 1 ; x = {-7, -6}
f(x) = x + 5 ; x = {-5, -4, -3, -2, -1, 0, 1, 2}
f(x) = x – 1; x{3, 4, 5}

(i) 2f(- 4) + 3f(2)
f(x) = x + 5
f(-4) = -4 + 5 = 1
f(2) = 2 + 5 = 7
∴ 2f(-4) + 3 f(2) = 2(1) + 3(7) = 2 + 21 = 23

(ii) f(-7) – f(-3)
f(x) = x2 + 2x + 1
f(-7) = (-7)2 + 2(-7) + 1 = 49 – 14 + 1 = 36
f(x) = x + 5
f(-3) = -3 + 5 = 2
∴ f(-7) – f(-3) = 36 – 2 = 34

(iii) \(\frac{4 f(-3)+2 f(4)}{f(-6)-3 f(1)}\)
f(x) = x + 5
f(-3) = -3 + 5 = 2
f(x) = x – 1
f(4) = 4 – 1 = 3
f(x) = x2 + 2x + 1
f(-6) = (-6)2 + 2(-6) + 1 = 36 – 12 + 1 = 25
f(x) = x + 5
f(1) = 1 + 5 = 6
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 33

Question 8.
Let A= { 0,1, 2, 3 } and B = {1, 3, 5, 7, 9 } be two sets. Let f: A → B be a function given by f (x) = 2x + 1. Represent this function as
(i) a set of ordered pairs
(ii) a table
(iii) an arrow diagram and
(iv) a graph.
Answer:
A = {0, 1, 2, 3}, B = { 1, 3, 5, 7, 9 },f(x) = 2x + 1
f(0) = 2(0) + 1 = 1, f(1) = 2(1) + 1 = 3 ,f(2) = 2(2) + 1 = 5, f(3) = 2(3) + 1 = 7

(i) Set of ordered pairs
The given function/can be represented as a set of ordered pairs as
f = {(0, 1), (1, 3), (2, 5), (3,7)}

(ii) Table form
Let us represent f using a table as shown below.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 34

(iii) Arrow Diagram
Let us represent f by an arrow diagram.
We draw two closed curves to represent the sets A and B. Here each element of A and its unique image element in B are related with an arrow.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 35

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

(iv) Graph
We are given that
f = {(x,f(x)) | x ∈ A} = {(0,1), (1, 3), (2, 5), (3, 7)}. Now, the points (0, 1), (1, 3), (2, 5) and (3, 7) are plotted on the plane as shown below.
The totality of all points represent the graph of the function.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 36

Question 9.
A. function f: [1, 6) → R is defined as follows
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions 37
(Here, [1, 6) = {x ∈ R : 1 ≤ x < 6})
Find the value of
(i) f(5)
(ii) f(3)
(iii) f(1)
(iv) f(2) – f(4)
(v) 2f(5) – 3f(1).
Answer:
(i) Let us find f(5). Since 5 lies between 4 and 6, we have to use f(x) = 3x2 – 10.
Thus, f(5) = 3(52) – 10 = 65.

(ii) To find f(3), note that 3 lies between 2 and 4.
So, we use f(x) = 2x – 1 to calculate f(3).
Thus, f(3) = 2(3) – 1 = 5.

(iii) Let us find f(1).
Now, 1 is in the interval 1 < x < 2
Thus, we have to use f(x) = 1 + x to obtain f(1) = 1 + 1 = 2.

(iv) f (2) – f(4)
Now, 2 is in the interval 2 < x < 4 and so, we use f(x) = 2x – 1.
Thus, f(2) = 2(2) -1 = 3.
Also, 4 is in the interval 4 < x < 6. Thus, we use f(x) = 3x2 – 10
Therefore, f(4) = 3(42) – 10 = 3(16) – 10 = 48 – 10 = 38.
Hence, f(2) – f(4) = 3 – 38 = -35.

(v) To calculate 2 f (5) – 3f (1), we shall make use of the values that we have already calculated in (i) and (iii). Thus, 2f(5) – 3f(1) = 2(65) – 3(2) = 130 – 6 – 124.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 10.
Given f(x) = 5x + 2; g(x) = 2x – 3;
h(x) = 3x + 1. Verify fo (goh) = (fog) oh
Answer:
f(x) = 5x + 2 ; g(x) = 2x – 3; h(x) = 3x + 1
L.H.S. = fo (goh)
goh = g[h(x)]
= g(3x + 1)
= 2(3x + 1) – 3
= 6x – 1
fo (goh) = f[goh (x)]
= f(6x – 1)
= 5 (6x – 1) + 2
= 30 x – 5 + 2
fo (goh) = 30x – 3 ….(1)
R.H.S. = (fog) oh
fog = f[g(x)]
= f(2x – 3)
= 5(2x – 3) + 2
= 30x – 5 + 2
fo (goh) = 30x – 3 …..(1)
R.H.S. = (fog) oh
fog = f[g(x)]
= f(2x – 3)
= 5 (2x – 3)
= 5 (2x – 3) + 2
= 10x – 15 + 2
= 10x – 13
(fog) oh = fog [h(x)]
= fog (3x + 1)
= 10 (3x + 1) – 13
= 30x + 10 – 13
= 30x – 3 ….(2)
From (1) and (2) we get L.H.S. = R.H.S.
fo(goh) = (fog) oh

Question 11.
Given f(x) = x2 + 4; g(x) = 3x – 2;
h(x) = x – 5. Show that the composition of functions is associative.
Answer:
f(x) = x2 + 4 ; g(x) – 3x – 2; h(x) = x – 5
To prove fo (goh) = (fog) oh
L.H.S. fo (goh)
goh = g[h(x)]
= g(x – 5)
= 3(x – 5) – 2
= 3x – 15 – 2
goh = 3x – 17
fo (goh) = f [goh (x)]
= f(3x – 17)
= (3x – 17)2 + 4
= 9x2 + 289 – 102 x + 4
= 9x2 – 102x + 293 ….(1)
R.H.S. = (fog) oh
fog – f[g(x)]
= f(3x-2)
= (3x – 2)2 + 4
= 9×2 + 4 – 12x + 4
= 9×2 – 12x + 8
(fog) oh = fog [h(x)]
= fog (x – 5)
= 9(x – 5)2 – 12 (x – 5) + 8
= 9(x2 + 25 – 10x) – 12x + 60 + 8
= 9x2 + 225 – 90x – 12x + 60 + 8
= 9x2 – 102x + 293 ….(2)
From (1) and (2) we get fo (goh) = (fog) oh.
Composition of function is associative

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Additional Questions

Question 12.
Given f(x) = x – 2; g(x) = 3x + 5; h(x) = 2x – 3. Verify that (goh) of = go (hof)
Answer:
f(x) = x – 2 ; g(x) = 3x + 5; h(x) = 2x – 3
L.H.S. (goh) of
goh = g[h(x)]
= g(2x – 3)
= 3(2x – 3) + 5
= 6x – 9 + 5
= 6x – 4
(goh) of = goh [f(x)]
= goh (x – 2)
= 6(x – 2) – 4
= 6x – 12 – 4
= 6x – 16 ….(1)

R.H.S. go(hof)
hof = h[f(x)]
= h(x- 2)
= 2(x – 2) – 3
= 2x – 4 – 3
= 2x – 7
go(hof) = g [hof (x)]
= g (2x – 7)
= 3(2x – 7) + 5
= 6x – 21 + 5
= 6x – 16 ….(2)
From (1) and (2) we get (goh) of = go(hof)

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