# Samacheer Kalvi Books: Tamilnadu State Board Text Books Solutions

## Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 1 Relations and Functions Notes Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 1 Relations and Functions Notes

## Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Notes PDF Download

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10th Maths Exercise 1.1 Samacheer Kalvi Question 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)
Solution:
(i) A = {2,-2,3}, B = {1,-4}
A × B = {(2, 1), (2, -4), (-2, 1), (-2, -4), (3,1) , (3,-4)}
A × A = {(2, 2), (2,-2), (2, 3), (-2, 2), (-2, -2), (-2, 3), (3, 2), (3, -2), (3,3) }
B × A = {(1, 2), (1, -2), (1, 3), (-4, 2), (-4, -2), (-4,3)}

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

(iii) A = {m,n} × Φ
A × B = { }
A × A = {(m, m), (m, n), (n, m), (n, n)}
B × A = { }

10th Maths Exercise 1.1 Question 2.
Let A= {1,2,3} and B = {× | x is a prime number less than 10}. Find A × B and B × A.
A = {1,2,3}, B = {2, 3, 5, 7}
A × B = {1,2,3} × {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)}
B × A = {2, 3, 5, 7} × {1,2,3}
= {(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(3, 3) (5, 1)(5, 2)(5, 3) (7, 1) (7,2)(7, 3)}

10th Maths Exercise 1.1 In Tamil Question 3.
If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4),(3, 3), (3, 4)} find A and B.
Solution:
B × A ={(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)}
A = {3, 4), B = { -2, 0, 3}

Samacheer Kalvi 10th Maths Exercise 1.1 Question 4.
If A= {5, 6}, B = {4, 5 ,6}, C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C)
A ={5,6}, B = {4,5,6}, C = {5, 6,7}
A × A = {5, 6} × {5,6}
= {(5, 5) (5, 6) (6, 5) (6, 6)} ….(1)
B × B = {4, 5, 6} × {4, 5, 6}
= {(4, 4)(4, 5)(4, 6)(5, 4)(5, 5) (5, 6) (6, 4)(6, 5) (6, 6)}
C × C = {5,6,7} × {5,6,7}
= {(5, 5)(5, 6)(5, 7)(6, 5)(6, 6) (6, 7)(7, 5)(7, 6) (7, 7)}
(B × B) ∩ (C × C) = {(5, 5)(5, 6)(6, 5)(6, 6)} ….(2)
From (1) and (2) we get
A × A = (B × B) ∩ (C × C)

Ex 1.1 Class 10 Samacheer Question 5.
Given A ={1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}, check if (A ∩ C) x (B ∩ D) = (A × B) ∩ (C × D) is true?
Solution:
LHS = {(A∩C) × (B∩D)
A ∩C = {3}
B ∩D = {3, 5}
(A ∩ C) × (B ∩ D) = {(3, 3) (3, 5)} ………….. (1)
RHS = (A × B) ∩ (C × D)
A × B = {(1, 2), (1, 3), (1, 5), (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5)}
C × D = {(3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5)}
(A × B) ∩ (C × D) = {(3, 3), (3, 5)} …(2)
∴ (1) = (2) ∴ It is true.

10th Maths Book Exercise 1.1 Question 6.
Let A = {x ∈ W | x < 2},
B = {x ∈ N | 1 < 1 < × < 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)
(i) A = {0, 1}
B = {2,3,4}
C = {3,5}
(i) A × (B ∪ C) = (A × B) ∪ (A × c)
B ∪ C = {2, 3,4} ∪ {3,5}
= {2, 3, 4, 5}
A × (B ∪ C) = {0, 1} × {2, 3, 4, 5}
= {(0, 2) (0, 3) (0, 4) (0, 5) (1, 2) (1, 3)(1, 4)(1, 5)} ….(1)
A × B = {0, 1} × {2,3,4}
= {(0,2) (0,3) (0,4) (1,2) (1,3) (1,4) }
A × C = {0, 1} × {3, 5}
{(0, 3) (0, 5) (1,3) (1,5)}
(A × B) ∪ (A × C) = {(0, 2) (0, 3) (0, 4) (0, 5) (1, 2)(1, 3)(1, 4)(1, 5)} ….(2)
From (1) and (2) we get
A × (B ∪ C) = (A × B) ∪ (A × C)

(ii) A × (B n C) = (A × B) n (A × C)
B ∩ C = {2,3,4} ∩ {3,5}
= {3}
A × (B ∩ C) = {0, 1} × {3}
= {(0,3) (1,3)} ….(1)
A × B = {0,1} × {2,3,4}
= {(0, 2) (0, 3) (0, 4) (1,2) (1,3) (1,4)}
A × C = {0,1} × {3,5}
{(0, 3) (0, 5) (1,3) (1,5)}
(A × B) n (A × C) = {(0, 3) (1, 3)} ….(2)
From (1) and (2) we get
A × ( B n C) = (A × B) n (A × C)

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
A ∪ B = {0, 1} ∪ {2,3,4}
= {0,1, 2, 3, 4}
(A ∪ B) × C = {0, 1,2, 3,4} × {3,5}
= {(0, 3) (0, 5) (1, 3) (1, 5)(2, 3) (2, 5) (3, 3)(3, 5) (4, 3)(4, 5)} ….(1)
A × C = {0, 1} × {3,5}
= {(0,3) (0,5) (1,3) (1,5)}
B × C = {2,3,4} × {3,5}
= {(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)
From (1) and (2) we get
(A ∪ B) × C = (A × C) ∪ (B × C)

Maths Exercise 1.1 Class 10 Samacheer 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)
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2}
Solution:
(i)(A ∩ B) × C = (A × c) ∩ (B × C)
LHS = (A ∩ B) × C
A ∩ B = {2, 3, 5, 7}
(A ∩ B) × C = {(2, 2), (3, 2), (5, 2), (7, 2)} ………… (1)
RHS = (A × C) ∩ (B × C)
(A × C) = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}
(B × C) = {2, 2), (3, 2), (5, 2), (7, 2)}
(A × C) ∩ (B × C) = {(2, 2), (3, 2), (5, 2), (7, 2)} ……….. (2)
(1) = (2)
∴ LHS = RHS. Hence it is verified.

(ii) A × (B – C) = (A × B) – (A × C)
LHS = A × (B – C)
(B – C) = {3,5,7}
A × (B – 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)} …………. (1)
RHS = (A × B) – (A × C)
(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),
(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), (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)
(1) = (2) ⇒ LHS = RHS.
Hence it is verified.

10th Maths Exercise 1.2 Samacheer Kalvi Question 1.
Let A = {1, 2, 3, 7} and B = {3, 0, -1, 7}, which of the following are relation from A to B ?
(i) R1 = {(2, 1), (7, 1)}
(ii) R2 = {(-1, 1)}
(iii) R3 = {(2, -1), (7, 7), (1, 3)}
(iv) R4 = {(7,-1), (0, 3), (3, 3), (0, 7)}
(i) A = {1, 2, 3, 7}, B = {3, 0,-1, 7}
Solution:
R1 = {(2,1), (7,1)} It is not a relation there is no element as 1 in B.
(ii) R2 = {(-1, 1)}
It is not [∵ -1 ∉ A, 1 ∉ B]
(iii) R3 = {(2, -1), (7, 7), (1, 3)}
It is a relation.
R4 = {(7,-1), (0, 3), (3, 3), (0, 7)}
It is also not a relation. [∵ 0 ∉ A]

Ex 1.2 Class 10 Samacheer Question 2.
Let A = {1, 2, 3, 4, ….., 45} and R be the relation defined as 'is square of ' on A. Write R as a subset of A × A. Also, find the domain and range of R.
A = {1,2, 3, 4 . . . . 45}
The relation is defined as 'is square of’
R = {(1,1) (2, 4) (3, 9)
(4, 16) (5,25) (6, 36)}
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {1, 4, 9, 16, 25, 36}

Exercise 1.2 Class 10 Maths Samacheer Question 3.
A Relation R is given by the set {(x, y) /y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain and range.
Solution:
x = {0, 1, 2, 3, 4, 5}
y = x + 3 ⇒ y = {3, 4, 5, 6, 7, 8}
R = {(x, y)}
= {(0, 3),(1, 4),(2, 5),(3, 6), (4, 7), (5, 8)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {3, 4, 5, 6, 7, 8}

10th Maths Exercise 1.2 Question 4.
Represent each of the given relation by (a) an arrow diagram, (b) a graph and (c) a set in roster form, wherever possible.
(i) {(x, y)|x = 2y,x ∈ {2, 3, 4, 5},y ∈ {1, 2, 3, 4)
(ii) {(x, y)|y = x + 3, x, y are natural numbers <10}
Solution:
(i){(x, y)|x = 2y, x ∈ {2, 3, 4, 5},y ∈ {1, 2, 3, 4}} R = (x = 2y)
2 = 2 × 1 = 2
4 = 2 × 2 = 4 (c) {(2, 1), (4, 2)}
(ii) {(x, y)|y = x + 3, x,+ are natural numbers <10}
x = {1, 2, 3, 4, 5, 6, 7, 8, 9} R = (y = x + 3)
y = {1, 2, 3, 4, 5, 6, 7, 8, 9}
R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}  (c) R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

10th Maths Exercise 1.2 Answers Question 5.
A company has four categories of employees given by Assistants (A), Clerks (C), Managers (M) and an Executive Officer (E). The company provide ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to the people who work in the categories A, C, M and E respectively. If A1, A2, A3, A4 and As were Assistants; C1, C2, C3, C4 were Clerks; M1, M2, M3 were managers and E1,E2 were Executive officers and if the relation R is defined by xRy, where x is the salary given to person y, express the relation R through an ordered pair and an arrow diagram.
Solution:
A – Assistants → A1, A2, A3, A4, A5
C – Clerks → C1, C2, C3, C4
D – Managers → M1, M2, M3
E – Executive officer → E1, E2
(a) R = {(10,000, A1), (10,000, A2), (10,000, A3),
(10,000, A4), (10,000, A5), (25,000, C1),
(25,000, C2), (25,000, C3), (25,000, C4),
(50,000, M1), (50,000, M2), (50,000, M3),
(1,00,000, E1), (1,00,000, E2)}

(b) ## TN Class 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 1.
Let f = {(x, y)|x, y ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?
Solution:
F = {(x, y)|x, y ∈ N and y = 2x}
x = {1, 2, 3,…}
y = {1 × 2, 2 × 2, 3 × 2, 4 × 2, 5 × 2 …}
R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10),…}
Domain of R = {1, 2, 3, 4,…},
Co-domain = {1, 2, 3…..}
Range of R = {2, 4, 6, 8, 10,…}
Yes, this relation is a function.

Question 2.
Let X = {3, 4, 6, 8}. Determine whether the relation R = {(x, f(x))|x ∈ X, f(x) = x2 + 1} is a function from X to N ?
Solution:
x = {3,4, 6, 8}
R = ((x, f(x))|x ∈ X, f(x) = X2 + 1}
f(x) = x2 + 1
f(3) = 32 + 1 = 10
f(4) = 42 + 1 = 17
f(6) = 62 + 1 = 37
f(8) = 82 + 1 = 65 R = {(3, 10), (4, 17), (6, 37), (8, 65)}
R = {(3, 10), (4, 17), (6, 37), (8, 65)}
Yes, R is a function from X to N. Question 3.
Given the function
f : x → x2 – 5x + 6, evaluate
(i) f(-1)
(ii) f(2 a)
(iii) f(2)
(iv) f(x – 1)
f(x) = x2 – 5x + 6
(i) f (-1) = (-1)2 – 5 (-1) + 6 = 1 + 5 + 6 = 12
(ii) f (2a) = (2a)2 – 5 (2a) + 6 = 4a2 – 10a + 6
(iii) f(2) = 22 – 5(2) + 6 = 4 – 10 + 6 = 0
(iv) f(x – 1) = (x – 1)2 – 5 (x – 1) + 6
= x2 – 2x + 1 – 5x + 5 + 6
= x2 – 7x + 12

Question 4.
A graph representing the function f(x) is given in figure it is clear that f(9) = 2. (i) Find the following values of the function
(a) f(0)
(b) f(7)
(c) f(2)
(d) f(10)
(ii) For what value of x is f (x) = 1?
(iii) Describe the following
(i) Domain
(ii) Range.
(iv) What is the image of 6 under f?
Solution:
From the graph
(a) f(0) = 9
(b) f(7) = 6
(c) f(2) = 6
(d) f(10) = 0
(ii) At x = 9.5, f(x) = 1
(iii) Domain = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
= {x |0 < x < 10, x ∈ R}
Range = {x|0 < x < 9, x ∈ R}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(iv) The image of 6 under f is 5. Question 5.
Let f(x) = 2x + 5. If x ≠ 0 then find f(x+2)−f(2)x
Solution:
Given f(x) = 2x + 5, x ≠ 0. Question 6.
A function fis defined by f(x) = 2x – 3
(i) find f(0)+f(1)2
(ii) find x such that f(x) = 0.
(iii) find x such that f(x) = x.
(iv) find x such that f(x) = f(1 – x).
Solution:
Given f(x) = 2x – 3
(i) find f(0)+f(1)2
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = -1
∴ f(0)+f(1)2=−3−12=−42 = -2

(ii) f(x) = 0
⇒ 2x – 3 = 0
2x = 3
x = 32

(iii) f(x) = x
⇒ 2x – 3 = x ⇒ 2x – x = 3
x = 3

(iv) f(x) = f(1 – x)
2x – 3 = 2(1 – x) – 3
2x – 3 = 2x – 2x – 3
2x + 2x = 2 – 3 + 3
4x = 2
x = 24
x = 12

Question 7.
An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown in figure. Express the volume V of the box as a function of x. Solution:
Volume of the box = Volume of the cuboid
= l × b × h cu. units
Here l = 24 – 2x
b = 24 – 2x
h = x
∴ V = (24 – 2x) (24 – 2x) × x
= (576 – 48x – 48x + 4x2)x
V = 4x3 – 96x2 + 576x

Question 8.
A function f is defined bv f(x) = 3 – 2x . Find x such that f(x2) = (f(x))2.
Solution:
f(x) = 3 – 2x
f(x2) = 3 – 2x2  Question 9.
A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time r in hours.
Speed of the plane = 500 km/hr
Distance travelled in 't' hours
= 500 × t (distance = speed × time)
= 500 t

Question 10.
The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length(x) as y = ax + b, where a, b are constants. (i) Check if this relation is a function.
(ii) Find a and b.
(iii) Find the height of a woman whose forehand length is 40 cm.
(iv) Find the length of forehand of a woman if her height is 53.3 inches.
Solution:
(i) Given y = ax + b …………. (1)
The ordered pairs are R = {(35, 56) (45, 65) (50, 69.5) (55, 74)}
∴ Hence this relation is a function. Substituting a = 0.9 in (2) we get
⇒ 65 = 45(.9) + b
⇒ 65 = 40.5 + b
⇒ b = 65 – 40.5
⇒ b = 24.5
∴ a = 0.9, b = 24.5
∴ y = 0.9x + 24.5
(iii) Given x = 40 , y = ?
∴ (4) → y = 0.9 (40) + 24.5
⇒ y = 36 + 24.5
⇒ y = 60.5 inches
(iv) Given y = 53.3 inches, x = ?
(4) → 53.3 = 0.9x + 24.5
⇒ 53.3 – 24.5 = 0.9x
⇒ 28.8 = 0.9x
⇒ x = 28.80.9 = 32 cm
∴ When y = 53.3 inches, x = 32 cm

Exercise 1.4 Class 10 Maths Samacheer Question 1.
Determine whether the graph given below represent functions. Give reason for your answers concerning each graph. Solution: (i) It is not a function. The graph meets the vertical line at more than one points.
(ii) It is a function as the curve meets the vertical line at only one point.
(iii) It is not a function as it meets the vertical line at more than one points.
(iv) It is a function as it meets the vertical line at only one point.

10th Maths Exercise 1.4 Samacheer Kalvi Question 2.
Let f :A → B be a function defined by f(x) = x2 – 1, Where A = {2, 4, 6, 10, 12},
B = {0, 1, 2, 4, 5, 9}. Represent f by
(i) set of ordered pairs;
(ii) a table;
(iii) an arrow diagram;
(iv) a graph
Solution:
f: A → B
A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9} (i) Set of ordered pairs
= {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
(ii) a table (iii) an arrow diagram; Ex 1.4 Class 10 Samacheer Question 3.
Represent the function f = {(1, 2),(2, 2),(3, 2), (4,3), (5,4)} through
(i) an arrow diagram
(ii) a table form
(iii) a graph
Solution:
f = {(1, 2), (2, 2), (3, 2), (4, 3), (5, 4)}
(i) An arrow diagram. 10th Maths Exercise 1.4 Question 4.
Show that the function f : N → N defined by f{x) = 2x – 1 is one – one but not onto.
Solution:
f: N → N
f(x) = 2x – 1
N = {1, 2, 3, 4, 5,…}
f(1) = 2(1) – 1 = 1
f(2) = 2(2) – 1 = 3
f(3) = 2(3) – 1 = 5
f(4) = 2(4) – 1 = 7
f(5) = 2(5) – 1 = 9 In the figure, for different elements in x, there are different images in f(x).
Hence f : N → N is a one-one function.
A function f: N → N is said to be onto function if the range of f is equal to the co-domain of f
Range = {1, 3, 5, 7, 9,…}
Co-domain = {1, 2, 3,..}
But here the range is not equal to co-domain. Therefore it is one-one but not onto function.

10th Maths 1.4 Exercise Question 5.
Show that the function f: N → N defined by f (m) = m2 + m + 3 is one – one function.
Solution:
f: N → N
f(m) = m2 + m + 3
N = {1, 2, 3, 4, 5…..}m ∈ N
f{m) = m2 + m + 3
f(1) = 12 + 1 + 3 = 5
f(2) = 22 + 2 + 3 = 9
f(3) = 32 + 3 + 3 = 15
f(4) = 42 + 4 + 3 = 23 In the figure, for different elements in the (X) domain, there are different images in f(x). Hence f: N → N is a one to one but not onto function as the range of f is not equal to co-domain.
Hence it is proved.

10th Maths Ex 1.4 Question 6.
Let A = {1,2, 3, 4} and B = N .
Let f: A → B be defined by f(x) = x3 then,
(i) find the range of f
(ii) identify the type of function
A = {1,2, 3,4}
B = {1,2, 3, 4, 5,….}
f(x) = x3
f(1) = 13 = 1
f(2) = 23 = 8
f(3) = 33 = 27
f(4) = 43 = 64
(i) Range = {1,8, 27, 64}
(ii) one -one and into function.

10th Maths Exercise 1.4 Answers Question 7.
In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) f: R → R defined by f(x) = 2x + 1
(ii) f: R → R defined by f(x) = 3 – 4x2
Solution:
(i) f : R → R
f(x) = 2x + 1
f(1) = 2(1) + 1 = 3
f(2) = 2(2) + 1 = 5
f(-1) = 2(-1) + 1 = -1
f(0) = 2(0) + 1 = 1
It is a bijective function. Distinct elements of A have distinct images in B and every element in B has a pre-image in A.
(ii) f: R → R; f(x) = 3 – 4x2
f(1) = 3 – 4(12) = 3 – 4 = -1
f(2) = 3 – 4(22) = 3 – 16 = -13
f(-1) = 3 – 4(-1)2 = 3 – 4 = -1
It is not bijective function since it is not one-one

Samacheer Kalvi 10th Maths Book Graph Solution Question 8.
Let A = {-1, 1} and B = {0, 2}. If the function f: A → B defined by f(x) = ax + b is an onto function? Find a and b.
Solution:
A= {-1, 1},B = {0, 2}
f: A → B, f(x) = ax + b
f(-1) = a(-1) + b = -a + b
f(1) = a(1) + b = a + b
Since f(x) is onto, f(-1) = 0
⇒ -a + b = 0 …(1)
& f(1) = 2
⇒ a + b = 2 …(2)
-a + b = 0 10th Maths Exercise 1.4 In Tamil Question 9.
If the function f is defined by (i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(-2)
Solution:
(i) f(3) ⇒ f(x) = x + 2 ⇒ 3 + 2 = 5
(ii) f(0) ⇒ 2
(iii) f (- 1.5) = x – 1
= -1.5 – 1 = -2.5
(iv) f(2) + f(-2)
f(2) = 2 + 2 = 4 [∵ f(x) = x + 2]
f(-2) = -2 – 1 = -3 [∵ f(x) = x – 1]
f(2) + f(-2) = 4 – 3 = 1

Samacheer Kalvi 10th Maths Exercise 1.4 Question 10.
A function f: [-5,9] → R is defined as follows: Solution:
f : [-5, 9] → R
(i) f(-3) + f(2)
f(-3) = 6x + 1 = 6(-3) + 1 = -17
f(2) = 5 × 2 – 1 = 5(22) – 1 = 19
∴ f(-3) + f(2) = -17 + 19 = 2

(ii) f(7) – f(1)
f(7) = 3x – 4 = 3(7) – 4 = 17
f(1) = 6x + 1 = 6(1) + 1 = 7
f(7) – f(1) = 17 – 7 = 10

(iii) 2f(4) + f(8)
f(4) = 5x2 – 1 = 5 × 42 – 1 = 79
f(8) = 3x – 4 = 3 × 8 – 4 = 20
∴ 2f(4) + f(8) = 2 × 79 + 20 = 178 10th Maths 1.4 Question 11
The distance S an object travels under the influence of gravity in time t seconds is 1 2 given by S(t) = 13gt2 + at + b, where, (g is the acceleration due to gravity), a, b are constants. Check if the function S(t) is one-one.
S(t) = 12gt2 + at + b
Let the time be 1, 2, 3 …. n seconds
S(1) = 12g(1)2 + a(1) + b
= g2 + a + b
S(2) = 12 g(2)2 + a(2) + b
= 4g2 + 2a + b
= 2g + 2a + b
S(3) = 12 g(3)2 + a(3) + 6
= 92 g + 3a + b
For every different value of t, there will be different distance.
∴ It is a one-one function.

Samacheer Kalvi 10th Maths Book Graph Solutions Question 12.
The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by t(C)= F where F = 95 C + 32 . Find,
(i) t(0)
(ii) t(28)
(iii) t(-10)
(iv) the value of C when t(C) = 212
(v) the temperature when the Celsius value is equal to the Farenheit value.
Solution:  f(x) = 2x – 1, g(x) = x+12, fog = gof = x Question 4.
(i) If f (x) = x2 – 1, g(x) = x – 2 find a, if g o f(a) = 1.
(a) Find k, if f(k) = 2k -1 and
fof (k) = 5.
(i) f(x) = x2 – 1 ; g(x) = x – 2 .
gof = g [f(x)]
= g(x2 – 1)
= x2 – 1 – 2
= x2 – 3
given gof (a) = 1
a2 – 3 = 1 [But go f(x) = x2 – 3]
a2 = 4
a = 4 = ± 2
The value of a = ± 2

(ii) f(k) = 2k – 1 ; fof(k) = 5
fof = f[f(k)]
= f(2k – 1)
= 2(2k – 1) – 1
= 4k – 2 – 1
= 4k – 3
fof (k) = 5
4k – 3 = 5
4k = 5 + 3
4k = 8
k = 84 = 2
The value of k = 2

Question 5.
Let A,B,C ⊂ N and a function f: A → B be defined by f(x) = 2x + 1 and g : B → C be defined by g(x) = x2. Find the range of fog and gof
Solution:
f(x) = 2x + 1
g(x) = x2
fog(x) = fg(x)) = f(x2) = 2x2 + 1
gof(x) = g(f(x)) = g(2x + 1) = (2x + 1)2
= 4x2 + 4x + 1
Range of fog is
{y/y = 2x2 + 1, x ∈ N}
Range of gof is
{y/y = (2x + 1)2, x ∈ N}.

Question 6.
Let f(x) = x2 – 1. Find (i) fof (ii) fofof
f(x) = x2 – 1
(i) fof = f[f{x)]
= f(x2 – 1)
= (x2 – 1)2 – 1
= x4 – 2x2 + 1 – 1
= x4 – 2x2

(ii) fofof = fof[f(x)]
= fof (x2 – 1)
= f(x2 – 1)2 – 1
= f(x4 – 2x2 + 1 – 1)
= f (x4 – 2x2)
fofof = (x4 – 2x2)2 – 1

Question 7.
If f: R → R and g : R → R are defined by f(x) = x5 and g(x) = x4 then check if f,g are one-one and fog is one-one?
Solution:
f(x) = x5
g(x) = x4
fog = fog(x) = f(g(x)) = f(x4)
= (x4)5 = x20
f is one-one, g is not one-one.
∵ g(1) = 14 = 1
g(-1) = ( -1)4 = 1
Different elements have same images
fog is not one-one. [∵ fog (1) = fog (-1) = 1]

Question 8.
Consider the functions f(x), g(x), h(x) as given below. Show that
(f o g) o h = f o(g o h) in each case.
(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x2
(ii) f(x) = x2, g(x) = 2x and h(x) = x + 4
(iii) f(x) = x – 4, g(x) = x2 and h(x) = 3x – 5
(i) f(x) = x – 1, g (x) = 3x + 1, h(x) = x2
fog (x) = f[g(x)]
= f(3x + 1)
= 3x + 1 – 1
fog = 3x
(fog) o h(x) = fog [h(x)] ,
= fog (x2)
= 3(x2)
(fog) oh = 3x2 …..(1)
goh (x) = g[h(x)]
= g(x2)
= 3(x2) + 1
= 3x2 +1
fo(goh) x = f [goh(x)]
= f[3x2 + 1]
= 3x2 + 1 – 1
= 3x2 ….(2)
From (1) and (2) we get
(fog) oh = fo (goh)
Hence it is verified

(ii) f(x) = x2 ; g (x) = 2x and h(x) = x + 4
(fog) x = f[g(x)]
= f (2x)
= (2x)2
= 4x2
(fog) oh (x) = fog [h(x)]
= fog (x + 4)
= 4(x + 4)2
= 4[x2 + 8x + 16]
= 4x2 + 32x + 64 …. (1)
goh (x) = g[h(x)]
= g(x + 4)
= 2(x + 4)
= 2x + 8
fo(goh) x = fo [goh(x)]
= f[2x + 8]
= (2x + 8)2
= 4×2 + 32x + 64 …. (2)
From (1) and (2) we get
(fog) oh = fo(goh)

(iii) f(x) = x – 4 ; g (x) = x2; h(x) = 3x – 5
fog (x) = f[g(x)]
= f(x2)
= x2 – 4
(fog) oh (x) = fog [h(x)]
= fog (3x – 5)
= (3x – 5)2 – 4
= 9x2 – 30x + 25 – 4
= 9x2 – 30x + 21 ….(1)
goh (x) = g[h(x)]
= g(3x – 5)
= (3x – 5)2
= 9x2 + 25 – 30x
fo(goh)x = f[goh(x)]
= f[9x2 – 30x + 25]
= 9x2 – 30x + 25 – 4
= 9x2 – 30x + 21 ….(2)
From (1) and (2) we get
(fog) oh = fo(goh)

Question 9.
Let f ={(-1, 3),(0, -1),(2, -9)} be a linear function from Z into Z . Find f(x).
Solution:
f ={(-1, 3), (0, -1), 2, -9)
f(x) = (ax) + b ………… (1)
is the equation of all linear functions.
∴ f(-1) = 3
f(0) = -1
f(2) = -9
f(x) = ax + b
f(-1) = -a + b = 3 …………… (2)
f(0) = b = -1
-a – 1 = 3 [∵ substituting b = – 1 in (2)]
-a = 4
a = -4
The linear function is -4x – 1. [From (1)]

Question 10.
In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at1 + bt2) = aC(t1) + bC(t2), where a, b are constants. Show that the circuit C(t) = 31 is linear.
Given C(t) = 3t
C(at1) = 3at1 …. (1)
C(bt2) = 3 bt2 …. (2)
C(at1) + C(bt2) = 3at1 + 3bt2
C(at1 + bt2) = 3at1 + 3bt2
= Cat1 + Cbt2 [from (1) and (2)]
∴ C(at1 + bt2) = C(at1 + bt2)
Superposition principle is satisfied.
∴ C(t) = 3t is a linear function.

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

Question 1.
If n(A × B) = 6 and A = {1, 3} then n(B) is
(1) 1
(2) 2
(3) 3
(4) 6
(3) 3
Hint:
If n(A × B) = 6
A = {1, 1}, n(A) = 2
n(B) = 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
(3) 12
Hint: A ∪ C = [a, b, p] ∪ [p, q, r, s]
= [a, b, p, q, r, s]
n (A ∪ C) = 6
n(B) = 2
∴ n [(A ∪ C)] × B] = 6 × 2 = 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)
(1) (A × C) ⊂ (B × D)]
Hint:
A = {1, 2}, B = {1, 2, 3, 4},
C = {5, 6}, D ={5, 6, 7, 8}
A × C ={(1,5), (1,6), (2, 5), (2, 6)}
B × D = {(1, 5),(1, 6),(1, 7),(1, 8),(2, 5),(2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8)}
∴ (A × C) ⊂ B × D it is true

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) 8
(2) 2
Hint: n(A) = 5
n(A × B) = 10
(consider 1024 as 10)
n(A) × n(B) = 10
5 × n(B) = 10
n(B) = 105 = 2
n(B) = 2

Question 5.
The range of the relation R = {(x, x2)|x is 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}
(3) {4, 9, 25, 49, 121}]
Hint:
R = {(x, x2)/x is a prime number < 13}
The squares of 2, 3, 5, 7, 11 are
{4, 9, 25, 49, 121}

Question 6.
If the ordered pairs (a + 2,4) and (5, 2a + 6) are equal then (a, b) is ………
(1) (2, -2)
(2) (5, 1)
(3) (2, 3)
(4) (3, -2)
(4) (3, -2)
Hint: The value of a = 3 and 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
(4) 2mn
Hint:
n(A) = m, n(B) = n
n(A × B) = 2mn

Question 8.
If {(a, 8),(6, b)} represents an identity function, then the value of a and 6 are respectively
(1) (8,6)
(2) (8,8)
(3) (6,8)
(4) (6,6)
(1) (8,6)
Hint: f = {{a, 8) (6, 6)}. In an identity function each one is the image of it self.
∴ 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
(3) One-to one function
Hint:
A = {1, 2, 3, 4), B = {4, 8, 9,10} Question 10.
If f(x) = 2x2 and g (x) = 13x, Then fog is (3) 29x2
Hint:
f(x) = 2x2
g(x) = 13x
fog = f(g(x)) = f(13x)=2(13x)2
= 2 × 19x2=29x2

Queston 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
(1) 7
Hint:
n(B) = 7
Since it is a bijective function, the function is one – one and also it is onto.
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 fog is
(1) {0, 2, 3, 4, 5}
(2) {-4, 1, 0, 2, 7}
(3) {1, 2, 3, 4, 5}
(4) {0, 1, 2}
(4) {0, 1, 2}
Hint:
gof = g(f(x))
fog = f(g(x))
= {(0, 2),(1, 0),(2, 4),(-4, 2),(7, 0)}
Range of fog = {0, 1, 2}

Question 13.
Let f (x) = 1+x2−−−−−√ 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
(3) 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)
(2) (2,-1)
Hint:
g(x) = αx + β
α = 2
β = -1
g(x) = 2x – 1
g(1) = 2(1) – 1 = 1
g(2) = 2(2) – 1 = 3
g(3) = 2(3) – 1 = 5
g(4) = 2(4) – 1 = 7

Question 15.
f(x) = (x + 1)3 – (x – 1)3 represents a function which is …………….
(1) linear
(2) cubic
(3) reciprocal
Hint: f(x) = (x + 1)3 – (x – 1)3
[using a3 – b3 = (a – b)3 + 3 ab (a – b)]
= (x + 1 – x + 1)3 + 3(x + 1) (x – 1)
(x + 1 – x + 1)
= 8 + 3 (x2 – 1)2
= 8 + 6 (x2 – 1)
= 8 + 6x2 – 6
= 6x2 + 2

## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Unit Exercise 1

10th Maths Unit Exercise 1 Question 1.
If the ordered pairs (x2 – 3x, y2 + 4y) and (-2, 5) are equal, then find x and y.
Solution:
(x2 – 3x, y2 + 4y) = (-2, 5)
x2 – 3x = -2
x2 – 3x + 2 = 0 10th Maths Chapter 1 Unit Exercise 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.
n(A × A) = 9
n(A) = 3
A = {-1,0,1}
A × A = {-1, 0, 1} × {-1, 0, 1}
A × A = {(-1,-1)(-1, 0) (-1, 1)
(0, -1) (0, 0) (0, 1)
(1,-1) (1, 0) (1, 1)}
The remaining elements of A × A =
{(-1, -1) (-1, 1) (0, -1) (0, 0) (1,-1) (1,0) (1,1)}

10th Maths Unit 1 Question 3.
Given that (i) f(0)
(ii) f(3)
(iii) f(a + 1) in terms of a.(Given that a > 0)
Solution:
(i) f(0) = 4
(ii) f(3) = 31−−−−√=2–√
(iii) f(a + 1) = a+11−−−−−−−−√=a−−√

Samacheer Kalvi 10th Maths Book Solutions 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.
A= {9, 10, 11, 12, 13, 14, 15, 16, 17}
f: A → N
f(x) = the highest prime factor n ∈ A
f = {(9, 3) (10, 5) (11, 11) (12, 3) (13, 13) (14, 7) (15, 5) (16, 2) (17, 17)}
Range of f = {3, 5, 11, 13, 7, 2, 17}
= {2, 3, 5, 7, 11, 13, 17}

10th Maths Exercise 1.1 Samacheer Kalvi Question 5.
Find the domain of the function f(x) = 1+11x2−−−−−√−−−−−−−−−−√−−−−−−−−−−−−−−−√
Solution:
f(x) = 1+11x2−−−−−√−−−−−−−−−−√−−−−−−−−−−−−−−−√
Domain of f(x) = {-1, 0, 1}
(x2 = 1, -1, 0, because 1x2−−−−−√ should be +ve, or 0)

10th Maths Unit Exercise Solutions Question 6.
If f (x) = x2, g(x) = 3x and h(x) = x – 2, Prove that (f o g)o h = f o(g o h).
f(x) = x2 ; g(x) = 3x and h(x) = x – 2
L.H.S. = (fog) oh
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)
R.H.S. = fo(goh)
goh = g [h(x)]
= g(x – 2)
= 3(x – 2)
= 3x – 6
fo(goh) = fo [goh (x)]
= f(3x – 6)
= (3x – 6)2
= 9x2 – 36x + 36 ….(2)
From (1) and (2) we get
L.H.S. = R.H.S.
(fog) oh = fo {goh)

Samacheer Kalvi 10th Maths Exercise 1.1 Question 7.
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?
Solution:
A = {1, 2), B = (1, 2, 3, 4)
C = {5, 6}, D = {5, 6, 7, 8)
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
B × D = {(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) It is proved.

10th Maths Unit 1 Question Paper Question 8.
If f(x) = x1x+1, x ≠ 1 show that f(f(x)) = 1x, Provided x ≠ 0.
Solution: Hence it is proved.

10th Maths Unit Exercise Question 9.
The function/and g are defined by f(x) = 6x + 8; g(x) = x23.
(i) Calculate the value of gg(12)
(ii) Write an expression for g f(x) in its simplest form.
Solution: 10th Maths Chapter 1 Samacheer Kalvi Question 10.
Write the domain of the following real functions Solution:
(i) f(x) = 2x+1x9
The denominator should not be zero as the function is a real function.
∴ The domain = R – {9}
(ii) p(x) = 54x2+1
The domain is R.
(iii) g(x) = x2−−−−−√
The domain = [2, ∝)
(iv) h(x) = x + 6
The domain is R.

## 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.
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 (iii) An arrow diagram Question 3.
State whether the graph represent a function. Use vertical line test. 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. 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. Find (i) 2f(-4) + 3 f(2)
(ii) f(-7) – f(-3)
Solution: (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

Question 6.
If A = {2,3, 5} and B = {1, 4} then find
(i) A × B
(ii) B × A
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.
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: Find the value of (i) f(5)
(ii) f(3)
(iii) f(2) – f(4).
Solution: (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

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: 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.

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