10th Maths - Book Back Answers - Chapter 1 Excercise 1.4 - English Medium Guides

  

 

SSLC / 10th - Maths - Book Back Answers - Chapter 1 Excercise 1.4 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 1 Excercise 1.4: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 1 Excercise 1.4 – Maths from the Tamil Nadu State Board 10th Standard Maths textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 1 Excercise 1.4 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 1 Relations and functions Ex 1.4

Question 1.
Determine whether the graph given below represent functions. Give reason for your Answer concerning each graph.

 
 

 
 

 

 

 

 

 

 

 

 

Answer Key:

 

 

 

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) = 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 Key:

A = {2,4,6, 10, 12}
B = {0,1, 2, 4, 5, 9}
f(x) = x/2 – 1
f(2) = 2/2 – 1 = 1 – 1 = 0
f(4) = 4/2 – 1 = 2 – 1 = 1
f(6) = 6/2 – 1 = 3 – 1 = 2
f(10) = 10/2 – 1 = 5 – 1 = 4
f(12) = 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)	2	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 Key:

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

x	1	2	3	4	4
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 Key:

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) = m2 + m + 3 is one-one function.

Answer Key:

N = {1,2,3, 4,5, ….. }
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
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) = x3 then,
(i) find the range off
(ii) identify the type of function

Answer Key:

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 Key.
(i) f: R → R defined by f (x) = 2x + 1
(ii) f: R → R defined by f(x) = 3 – 4x2

Answer Key:

(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 – 4x2
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(42) = 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 Key:

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
a-b = 0             …..(1)

a+b = 2            …..(2)

________________

(1)+ (2) ⟹2a = 2

              a =2/3 = 1
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 Key:

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:

 

 

Find the values of 

(i) f (−3)+ f (2) 

(ii) f (7)- f (1) 

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

 

Answer Key: 

 

f(x) = 6x + 1 ; x = {-5,-4,-3,-2,-1,0,1}
f(x) = 5x2 – 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) = 5x2 – 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) = 5x2 – 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) = 5x2 – 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) = 1/2 gt2 + at + b where, (g is the acceleration due to gravity), a, b are constants. Check if the function S (t)is one-one.

Answer Key:

S(t) = 1/2 gt2 + at + b
Let t be 1, 2, 3, ………, seconds
S(1) = 1/2 g(12) + a(1) + b = 12 g + a + b
S(2) = 1/2 g(22) + 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 pre images 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 = 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 Key:

Given t(C) = 9C/5 + 32
(i) t(0) = 9(0)/5 + 32
= 32° F

(ii) t(28) = 9(28)/5 + 32
= 252/5 + 32
= 50.4 + 32
= 82.4° F

(iii) t(-10) = 9(−10)/5 + 32
= -18 + 32
= 14° F

(iv) t(C) = 212
9C/5 + 32 = 212
9C/5 = 212 – 32
= 180
9C = 180 × 5
C = 180×5/9
= 100° C

(v) consider the value of C be “x”
t(C) = 9C/5 + 32
x = 9x/5 + 32
5x = 9x + 160
-160 = 9x – 5x
-160 = 4x
x = −160/4 = -40
The temperature when the Celsius value is equal to the fahrenheit value is -40°