Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 1.
Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 1
Solution:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 2
(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.

Question 2.
Let f :A → B be a function defined by f(x) = \(\frac { x }{ 2 } \) – 1, Where A = {2,4,6,10,12},
B = {0,1, 2, 4, 5, 9}. Represent/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}
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 3Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 3
(i) Set of ordered pairs
= {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
(ii) a table
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 4
(iii) an arrow diagram;
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 5
(iv) a graph
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 6

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.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 7
(ii) a table form
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 8
(iii) A graph representation.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 9

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
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 10
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.

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
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 11
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.

Question 6.
Let A = {1, 2,3,4) and B = N. Letf: A → B be
defined by f(x) = x3 then,
(i) find the range off
(ii) identify the tpe of function
Solution:
A = {1,2,3,4}
B = N
f: A → B,f(x) = x3
(i) f(1) = 13 = 1
f(2) = 23 = 8
f(3) = 33 = 27
f(4) = 43 = 64
(ii) Therange of f = {1,8,27,64 )
(iii) It is one-one and into function.

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

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
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 13

Question 9.
If the function f is defined by
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 14
(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
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 154

Question 10.
A function f: [-5,9] → R is defined as follows:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 15
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
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 16

Question 11
The distance S an object travels under the influence of gravity in time t seconds is given by S(t) = \(\frac { 1 }{ 2 } \) gt2 + at + b where, (g is the 2 acceleration due to gravity), a, b are constants. Check if the function S (t)is one-one.
Solution:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 17
Yes, for every different values of t, there will be different values as images. And there will be different preimages for the different values of the range. Therefore it is one-one function.

Question 12.
The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by ty(C)= F where F = \(\frac { 9 }{ 5 } \) 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:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 18
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 19

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