You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Question 1.
Let A = {1, 2, 3, 4} and B = {-1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Let R = {(1, 3), (2, 6), (3,10), (4, 9)} ⊂ A × B be a relation. Show that R is a function and find its domain, co-domain and the range of R.
Answer:
Domain of R = {1, 2, 3, 4}
Co-domain of R = B = {-1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12}
Range of R= {3, 6, 10, 9}

Question 2.
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 and (iv) a graph.
Solution:
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) A set of ordered pairs.
f = {(0, 1), (1, 3), (2, 5), (3, 7)}
(ii) A table
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Additional Questions 1
(iii) An arrow diagram
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Additional Questions 2

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

Question 3.
State whether the graph represent a function. Use vertical line test.
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Additional Questions 3
Solution:
It is not a function as the vertical line PQ cuts the graph at two points.

Question 4.
Let f = {(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5, 3)} be a function from A = {-1, 0, 2, 3, 5, 7} to B = {2, 3, 4, 6, 7, 9}. Is this (i) an one-one function (ii) an onto function, (iii) both one- one and onto function?
Solution:
It is both one-one and onto function.
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Additional Questions 4
All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

Question 5.
A function f: (-7,6) → R is defined as follows.
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Additional Questions 5
Find (i) 2f(-4) + 3 f(2)
(ii) f(-7) – f(-3)
Solution:
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Additional Questions 6
(i) 2f(-4) + 3f(2)
f(-4) = x + 5 = -4 + 5 = 1
2f(-4) = 2 × 1 = 2
f(2) = x + 5 = 2 + 5 = 7
3f(2) = 3(7) = 21
∴ 2f(-4) + 3f(2) = 2 + 21 = 23

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

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

Question 6.
If A = {2,3, 5} and B = {1, 4} then find
(i) A × B
(ii) B × A
Answer:
A = {2, 3, 5}
B = {1, 4}

(i) A × B = {2,3,5} × {1,4}
= {(2, 1) (2, 4) (3, 1) (3, 4) (5,1) (5, 4)}.

(ii) B × A = {1,4} × {2,3,5}
= {(1,2) (1,3) (1,5) (4, 2) (4, 3) (4, 5)}

Question 7.
Let A = {5, 6, 7, 8};
B = {- 11, 4, 7, -10, -7, – 9, -13} and
f = {(x,y): y = 3 – 2x, x ∈ A, y ∈ B}.
(i) Write down the elements of f.
(ii) What is the co-domain?
(iii) What is the range?
(iv) Identify the type of function.
Answer:
Given, A = {5, 6, 7, 8},
B = {- 11,4, 7,-10,-7,-9,-13}
y = 3 – 2x
ie; f(x) = 3 – 2x
f(5) = 3 – 2 (5) = 3 – 10 = – 7
f(6) = 3 – 2 (6) = 3 – 12 = – 9
f(7) = 3 – 2(7) = 3 – 14 = – 11
f(8) = 3 – 2 (8) = 3 – 16 = – 13
(i) f = {(5, – 7), (6, – 9), (7, – 11), (8, – 13)}
(ii) Co-domain (B)
= {-11,4, 7,-10,-7,-9,-13} i
(iii) Range = {-7, – 9, -11,-13}
(iv) It is one-one function.

Question 8.
A function f: [1, 6] → R is defined as follows:
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Additional Questions 7
Find the value of (i) f(5)
(ii) f(3)
(iii) f(2) – f(4).
Solution:
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Additional Questions 8
(i) f(5) = 3x2 – 10
= 3 (52) – 10 = 75 – 10 = 65
(ii) f(3) = 2x – 1
= 2(3) – 1 = 6 – 1 = 5
(ii) f(2) – f(4)
f(2) = 2x – 1
= 2(2) – 1 = 3
f(4) = 3x2 – 10
= 3(42) – 10 = 38
∴ f(2) – f(4) = 3 – 38 = 35

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

Question 9.
The following table represents a function from A = {5, 6, 8, 10} to B = {19, 15, 9, 11}, where f(x) = 2x – 1. Find the values of a and b.
Solution:
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Additional Questions 9
A = {5, 6, 8, 10}, B = {19, 15, 9, 11}
f(x) = 2x – 1
f(5) = 2(5) – 1 = 9
f(8) = 2(8) – 1 = 15
∴ a = 9, b = 15

Question 10.
If R = {(a, -2), (-5, 6), (8, c), (d, -1)} represents the identity function, find the values of a,b,c and d.
Solution:
R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.