Tamil Nadu Board 10th Standard Maths - Chapter 1 Excercise 1.3: Book Back Answers and Solutions
This post covers the book back answers and solutions for Chapter 1 Excercise 1.3 – 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 1.3 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 1.3
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?
Answer Key:
X = {1,2,3,….}
Y = {1,2,3,….}
f = {(1,2) (2, 4) (3, 6) (4, 8) ….}
Domain = {1, 2, 3, 4 ….}
Co – Domain = {1, 2, 3, 4 ….}
Range = {2, 4, 6, 8 }
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?
Answer Key:
f(x) = x2 + 1
f(3) = 32 + 1 = 9 + 1 = 10
f(4) = 42 + 1 = 16 + 1 = 17
f(6) = 62 + 1 = 36 + 1 = 37
f(8) = 82 + 1 = 64 + 1 = 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(2a)
(iii) f(2)
(iv) f(x – 1)
Answer Key:
Give the function 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 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)
Answer Key:
(a) f (0) = 9
(b) f (7) = 6
(c) f (2) = 6
(d) f(10) = 0
(ii) For what value of x is f(x) = 1 ?
Answer Key:
When f(x) = 1 the value of x is 9.5
(iii) Describe the following
(i) Domain
(ii) Range.
Answer Key:
Domain = {0, 1, 2, 3,… .10}
= {x / 0 < x < 10, x ∈ R}
Range = {0,1,2,3,4,5,6,7,8,9}
= {x / 0 < x < 9, x ∈ R}
(iv) What is the image of 6 under f?
Answer Key:
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
Answer Key:
f(x) = 2x + 5
f(x + 2) = 2(x + 2) + 5
= 2x + 4 + 5
= 2x + 9
A function/is 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/ (A:) = x.
(iv) find x such that fix) =/(l – x).
Answer Key:
(i) f(x) = 2x – 3
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = 2 – 3 = -1
(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(1 – x) = 2(1 – x) – 3
= 2 – 2x – 3
= – 2x – 1
f(x) = f(1 – x)
2x – 3 = – 2x – 1
2x + 2x = 3 – 1
4x = 2
x = 24 = 12
Question 7.
square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume V of the box as a function of x.
Answer Key:
length of the cuboid (l) = 24 – 2x
breadth of the cuboid (b) = 24 – 2x
height of the cuboid (h) = 2x
Volume of the box = Volume of the cuboid
V = (24 – 2x)(24 – 2x) (x)
= (24 – 2x)2 (x)
= (576 + 4x2 – 96x) x
= 576x + 4x3 – 96x2
V = 4x3 – 96x2 + 576x
V(x) = 4x3 – 96x2 + 576x
Question 8.
A function f is defined by f(x) = 3 – 2x. Find x such that f(x2) = (f (x))2.
Answer Key:
f(x) = 3 – 2x
f(x2) = 3 – 2 (x2)
= 3 – 2x2
(f (x))2 = (3 – 2x)2
= 9 + 4x2 – 12x
But f(x2) = (f(x))2
3 – 2 x2 = 9 + 4x2 – 12x
-2x2 – 4x2 + 12x + 3 – 9 = 0
-6x2 + 12x – 6 = 0
(÷ by – 6) ⇒ x2 – 2x + 1 = 0
(x – 1) (x – 1) = 0
x – 1 = 0 or x – 1 = 0
x = 1
The value of x = 1
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 t in hours.
Answer Key:
Speed = distance covered / time taken
⇒ distance = Speed × time
⇒ d = 500 × t [ ∵ time = t hrs]
⇒ d = 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.
Length ‘x’ of forehand (in cm) |
Height y (in inches) |
35 |
56 |
45 |
65 |
50 |
69.5 |
55 |
74 |
Answer Key:
The relation is y = 0.9x + 24.5
(i) Yes the relation is a function.
(ii) When compare with y = ax + b
a = 0.9, b = 24.5
(iii) When the forehand length is 40 cm, then height is 60.5 inches.
Hint: y = 0.9x + 24.5
= 0.9 × 40 + 24.5
= 36 + 24.5
= 60.5 feet
(iv) When the height is 53.3 inches, her forehand length is 32 cm
Hint: y = 0.9x + 24.5
53.3 = 0.9x + 24.5
53.3 – 24.5 = 0.9 x
28.8 = 0.9 x
x = 28.80.9
x = 32 cm
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