Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.4

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.

(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) = x/2 – 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
Answer:
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 Standard Maths Exercise 1.4 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

10th Maths 1.4 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 10th Sum 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 Relation And Function Question 11
The distance S an object travels under the influence of gravity in time t seconds is 1 2 given by S(t) = 1/3gt² + at + b, where, (g is the acceleration due to gravity), a, b are constants. Check if the function S(t) is one-one.
Answer:
S(t) = 1/2gt² + at + b
Let the time be 1, 2, 3 …. n seconds
S(1) = 1/2g(1)² + a(1) + b
= g/2 + a + b
S(2) = 1/2 g(2)² + a(2) + b
= 4g/2 + 2a + b
= 2g + 2a + b
S(3) = 1/2 g(3)² + a(3) + 6
= 9/2 g + 3a + b
For every different value of t, there will be different distance.
∴ It is a one-one function.

Samacheer Kalvi Guru 10th Maths Question 12.
The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by t(C)= F where F = 9/5 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: