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## Tamilnadu Samacheer Kalvi 10th 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.

Answer:

The vertical line cuts the graph at A and B. The given graph does not represent a function.

The vertical line cuts the graph at most one point P. The given graph represent a function.

The vertical line cuts the graph at three points S,T and U. The given graph does not represent a function.

The vertical line cuts the graph at most one point D. The given graph represents a function.

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 f by

(i) set of ordered pairs

(ii) a table

(iii) an arrow diagram

(iv) a graph

Answer:

A = {2,4,6, 10, 12}

B = {0,1, 2, 4, 5, 9}

f(x) = \(\frac { x }{ 2 } \) – 1

f(2) = \(\frac { 2 }{ 2 } \) – 1 = 1 – 1 = 0

f(4) = \(\frac { 4 }{ 2 } \) – 1 = 2 – 1 = 1

f(6) = \(\frac { 6 }{ 2 } \) – 1 = 3 – 1 = 2

f(10) = \(\frac { 10 }{ 2 } \) – 1 = 5 – 1 = 4

f(12) = \(\frac { 12 }{ 2 } \) – 1 = 6 – 1 = 5

(i) Set of ordered pairs

f = {(2, 0) (4, 1) (6, 2) (10, 4) (12, 5}

(ii) Table

X | 2 | 4 | 6 | 10 | 12 |

f(x) | 0 | 1 | 2 | 4 | 5 |

(iii) Arrow diagram

(iv) Graph

Question 3.

Represent the function f = {(1,2), (2,2), (3,2), (4,3),(5,4)} through (i) an arrow diagram (it) a table form (iii) a graph.

Answer:

f = {(1, 2) (2, 2) (3, 2) (4, 3) (5,4)}

Let A = {1,2, 3, 4, 5}

B = {2, 3, 4}

(i) Arrow diagram

(ii) Table form

X | 1 | 2 | 3 | 4 | 5 |

f(x) | 2 | 2 | 2 | 3 | 4 |

(iii) Graph

Question 4.

Show that the function f : N → N defined by f(x) = 2x – 1 is one-one but not onto.

Answer:

f: N → N

N = {1,2,3,4,5,… }

f(x) = 2x – 1

f(1) = 2(1) – 1 = 2 – 1 = 1

f(2) = 2(2) – 1 = 4 – 1 = 3

f(3) = 2(3) – 1 = 6 – 1 = 5

f(4) = 2(4) – 1 = 8 – 1 = 7

f(5) = 2(5) – 1 = 10 – 1 = 9

f = {(1,1) (2, 3) (3, 5) (4, 7) (5,9) …..}

(i) Different elements has different images. This function is one to one function.

(ii) Here Range is not equal to co-domain. This function not an onto function.

∴ The given function is one-one but not an onto.

Question 5.

Show that the function f: N ⇒ N defined by f(m) = m^{2} + m + 3 is one-one function.

Answer:

N = {1,2,3, 4,5, ….. }

f(m) = m^{2} + m + 3

f(1) = 1^{2} + 1 + 3 = 5

f(2) = 2^{2} + 2 + 3 = 9

f(3) = 3^{2 }+ 3 + 3 = 15

f(4) = 4^{2} + 4 + 3 = 23

f = {(1,5) (2, 9) (3, 15) (4, 23)}

From the diagram we can understand different elements in (N) in the domain, there are different images in (N) co-domain.

∴ The function is a one-one function.

Question 6.

Let A = {1, 2, 3, 4) and B = N. Letf: A → B be

defined by f(x) = x^{3} 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) = x^{3}

(i) f(1) = 1^{3} = 1

f(2) = 2^{3} = 8

f(3) = 3^{3} = 27

f(4) = 4^{3} = 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 – 4x^{2}

Answer:

(i) f(x) = 2x + 1

f(0) = 2(0) + 1 = 0 + 1 = 1

f(1) = 2(1) + 1 = 2 + 1 = 3

f(2) = 2(2) + 1 = 4 + 1 = 5

f(3) = 2(3) + 1 = 6 + 1 = 7

Different elements has different images

∴ It is an one-one function.

It is also an onto function. The function is one-one and onto function.

∴ It is a bijective function.

(ii) f(x) = 3 – 4x^{2}

f(1) = 3 – 4(1)^{2}

= 3 – 4 = -1

f(2) = 3 – 4(2)^{2} = 3 – 16 = – 13

f(3) = 3 – 4(3)^{2} = 3 – 36 = – 33

f(4) = 3 – 4(4^{2}) = 3 – 64 = – 61

It is not a bijective function. The positive numbers “R” do not have negative pre – image in X in R.

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.

Answer:

A = {-1, 1}; B = {0,2}

f(x) = ax + b

f(-1) = a(-1) + b

0 = -a + b

a – b = 0 ….(1)

f(1) = a(1) + b

2 = a + b

a + b = 2 ….(2)

Solving (1) and (2) we get

Substitute a = 1 in (1)

The value of a = 1 and b = 1

Question 9.

If the function f is defined by

find the value of

(i) f(3)

(ii) f(0)

(iii) f(1. 5)

(iv) f(2) + f(-2)

Answer:

f(x) = x + 2 when x = {2,3,4,……}

f(x) = 2

f(x) = x – 1 when x = {-2}

(i) f(x) = x + 2

f(3) = 3 + 2 = 5

(ii) f(x) = 2

f(0) = 2

(iii) f(x) = x – 1

f(-1.5) = -1.5 – 1 = -2.5

(iv) f(x) = x + 2

f(2) = 2 + 2 = 4

f(x) = x – 1

f(-2) = – 2 – 1 = – 3

f(2) + f(-2) = 4 – 3

= 1

Question 10.

A function f: [-5, 9] → R is defined as follows:

Answer:

f(x) = 6x + 1 ; x = {-5,-4,-3,-2,-1,0,1}

f(x) = 5x^{2} – 1 ; x = {2, 3, 4, 5}

f(x) = 3x – 4 ; x = {6, 7, 8, 9}

(i) f(-3) + f(2)

f(x) = 6x + 1

f(-3) = 6(-3) + 1 = -18 + 1 = -17

f(x) = 5x^{2} – 1

f(2) = 5(2)^{2} – 1 = 20 – 1 = + 19

f(-3) + f(2) = – 17 + 19

= 2

(ii) f(7) – f(1)

f(x) = 3x – 4

f(7) = 3(7) – 4 = 21 – 4 = 17

f(x) = 6x + 1

f(1) = 6(1) + 1 = 6 + 1 = 7

f(7) – f(1) = 17 – 7

= 10

(iii) 2f(4) + f(8)

f(x) = 5x^{2} – 1

f(4) = 5(4)2 – 1 = 5(16) – 1

= 80 – 1 = 79

f(x) = 3x – 4

f(8) = 3(8) – 4 = 24 – 4 = 20

2f(4) + f(8) = 2(79) + 20

= 158 + 20

= 178

f(x) = 6x + 1

f(-2) = 6(-2) + 1 = -12 + 1 = -11

f(x) = 3x – 4

f(6) = 3(6) – 4 = 18 – 4 = 14

f(x) = 5x^{2} – 1

f(4) = 5(4)^{2} – 1 = 5(16) – 1

= 80 – 1 = 79

f(x) = 6x + 1

f(-2) = 6(-2) + 1 = -12 + 1 = -11

Question 11.

The distance S an object travels under the influence of gravity in time t seconds is given by S(t) = \(\frac{1}{2}\) gt^{2} + at + b where, (g is the acceleration due to gravity), a, b are constants. Check if the function S (t)is one-one.

Solution:

S(t) = \(\frac{1}{2}\) gt^{2} + at + b

Let t be 1, 2, 3, ………, seconds

S(1) = \(\frac{1}{2}\) g(1^{2}) + a(1) + b = \(\frac{1}{2}\) g + a + b

S(2) = \(\frac{1}{2}\) g(2^{2}) + a(2) + b = 2g + 2a + b

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

t(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 Fahrenheit value.

Answer:

Given t(C) = \(\frac { 9C }{ 5 } \) + 32

(i) t(0) = \(\frac { 9(0) }{ 5 } \) + 32

= 32° F

(ii) t(28) = \(\frac { 9(28) }{ 5 } \) + 32

= \(\frac { 252 }{ 5 } \) + 32

= 50.4 + 32

= 82.4° F

(iii) t(-10) = \(\frac { 9(-10) }{ 5 } \) + 32

= -18 + 32

= 14° F

(iv) t(C) = 212

\(\frac { 9C }{ 5 } \) + 32 = 212

\(\frac { 9C }{ 5 } \) = 212 – 32

= 180

9C = 180 × 5

C = \(\frac{180 \times 5}{9}\)

= 100° C

(v) consider the value of C be “x”

t(C) = \(\frac { 9C }{ 5 } \) + 32

x = \(\frac { 9x }{ 5 } \) + 32

5x = 9x + 160

-160 = 9x – 5x

-160 = 4x

x = \(\frac { -160 }{ 4 } \) = -40

The temperature when the Celsius value is equal to the fahrenheit value is -40°

Composition of two Functions

Let f: A → B and g: B → C be two functions. Then the composition of f and g denoted by gof is defined as the function gof (x) = g[f(x)] for all x ∈ A.

Composition of three Functions

Let A, B, C, D be four sets and let f: A → B; g : B → C and h : C → D be three functions, using composite functions fog and goh, we get two new functions like (fog) oh and fo (goh).

Note: Composition of three function is always associative.