Class 9 Advanced Maths Chapter 3 Relations and Functions Ex 3.4 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 3 Relations and Functions Ex 3.4 to verify their answers.

Advanced Maths Class 9 Exercise 3.4 Solutions

Class 9 Advanced Maths Ex 3.4 Solutions

Question 1.
What is the domain and range of each of the relations given below? Which of these relations are functions?
(a) R = {(5,1), (4,1), (3,1), (2,0)}
(b) R = {(1, -1), (2, -2), (3, -3), (4, – 4), (5, -5)}
(c) R= {(3,-1), (3,0), (3,1), (3,2)}
Solution:
(a) R = {(5,1), (4,1), (3,1), (2,0)}

1. Domain: {2, 3, 4, 5}
2. Range: {0, 1}
3. Function: Yes. Each input maps to exactly one output.

(b) R = {(1, -1), (2, -2), (3, -3), (4, -4), (5, -5)}

1. Domain: {1, 2, 3, 4, 5}
2. Range: {-5, -4, -3, -2, -1}
3. Function: Yes. Each input maps to exactly one output.

(c) R = {(3, -1), (3, 0), (3, 1), (3, 2)}

• Domain: {3}
• Range: {-1, 0,1, 2}
• Function: No. The input 3 maps to multiple outputs.

Question 2.
Draw a rough sketch of each of the following relations. Also write their domain and range.
(a) R = {(x, y): xy = 8 where x, y ∈ Z }
(b) R = {(x, y): x = |y| where x ∈ Z and 0 ≤ x ≤ 5}
(c) R = {{x, y): y = \(-\sqrt{x}\) [x where x ∈ (0, ∞)}
Solution:
(a) Relation: R = {(x, y): xy = 8 where x, y ∈ Z } Because this relation is strictly defined over integers (Z), the actual graph consists only of isolated dots. However, to help visualize the mathematical trend, the discrete points are plotted alongside the underlying smooth hyperbolic curves y = \(\frac{8}{x}\).

x Coordinate	y Coordinate (y = \(\frac{8}{x}\))	Ordered Pair (x, y)
-8	-1	(-8,-1)
-4	-2	(-4, -2)
-2	-4	(-2, -4)
-1	-8	(-1, -8)
1	8	(1, 8)
2	4	(2, 4)
4	2	(4, 2)
8	1	(8, 1)

• Domain: {±1, ±2, ±4, ±8}
• Range: {+1, ±2, ±4, ±8}

(b) Relation: R = {(x, y): x = | y | where r ∈ Z and 0 ≤ x ≤ 5} This relation contains discrete coordinate points due to the integer constraint on x. The points are superimposed on top of the continuous absolute value V-shaped side curve to highlight the structure.

x Coordinate	y Coordinate (y = ±x)	Ordered Pair (x, y)
0	0	(0, 0)
1	-1, 1	(1, -1) (1, 1)
2	-2, 2	(2, -2) (2, 2)
3	-3, 3	(3, -3) (3, 3)
4	-4, 4	(4, -4) (4, 4)
5	-5, 5	(5, -5) (5, 5)

• Domain: {0, 1, 2, 3, 4, 5}
• Range: {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

(c) Relation: R = {(x, y): y = \(-\sqrt{x}\) [x where x ∈ (0, ∞)}
This graph shows the real-valued, continuous smooth square root curve paired with prominent data marker dots at integer perfect square values along its trajectory.

x Coordinate	y Coordinate (y = –\(-\sqrt{x}\))	Ordered Pair (x, y)
1	-1	(1, -1)
2	\(-\sqrt{2}\) ≈ – 1.41	(2, -1.41)
3	\(-\sqrt{3}\) ≈ -1.73	(3, -1.73)
4	-2	(4, -2)
9	-3	(9, -3)

Note: The point (0, 0) is excluded from the graph since x > 0.

Domain: (0, ∞)
Range: (-∞, 0)

Question 3.
Draw the graph of the functions g and h on the same coordinate axes. You may fill the tables given below to draw the graphs.

Describe the relationship among the graphs of f, g and h.
Solution:

x	y = |x|	y = |x| – 1	y = |x| + 1
-2	2	1	3
-1	1	0	2
o	0	-1	1
1	1	0	2
2	2	1	3

The functions are plotted below using the calculated points:

Relationship Among the Graphs

1. Base Graph: The graph of f(x) = | x | forms a V-shape with its vertex at the origin (0, 0).
2. Vertical Shift Down: The graph of g(x) =|x| – 1 is a vertical translation of f(x) downward by 1 unit. Its vertex is at (0, -1).
3. Vertical Shift Up: The graph of h(x) = |x| + 1 is a vertical translation of f(x) upward by 1 unit. Its vertex is at (0, 1).

Question 4.
Draw the graphs of the functions f, g and h on the same coordinate axes. You may fill in the tables given below to draw the graph.

Describe the relationship among the graphs of f, g and h. Are the domain and range equal?
Solution:

x	f(x)=x2	g(x) = (x – 1)2	h(x) = (x + 2)2
-2	4	9	0
-1	1	4	1
0	0	1	4
1	1	0	9
2	4	1	16

The functions are plotted below using the calculated points:
Comparison of f(x), g(x), and h(x)

Graph Relationship

1. Base Graph: f(x) = x² represents the standard parabola centred at the origin (0, 0).
2. Horizontal Shift Right: g(x) = (x – 1)² is shifted 1 unit to the right relative to f(x).
3. Horizontal Shift Left: h(x) = (x + 2)² is shifted 2 units to the left relative to f(x)

Domain and Range Comparison

1. Domain: Equal for all three functions. Every real number can be evaluated: (-∞, ∞).
2. Range: Equal for all three functions. All yield non-negative values: [0, ∞).

Question 5.
Determine the domain and range of the following functions:
(a) y = \(\frac{1}{x^2}\)
(b) y = 2 – |x|
(c) y = (x – 1)³
(d) y = \(\sqrt{-x}\)
Solution:
(a) y = \(\frac{1}{x^2}\)
For the function to be defined, denominator cannot be zero.
x² ≠ 0
x ≠ 0
So, Domain = R – {0}
Also, \(\frac{1}{x^2}\) > 0 for every allowed value of x.
It never becomes 0.
Hence, Range = (0, ∞)

(b) y = 2 – |x|
|x| is defined for all real numbers.
So, Domain = R
Since | x | > 0,
y = 2- |x| ≤ 2
Maximum value is 2 when x = 0.
As | x | increases, y decreases without bound.
Hence, Range = (-∞, 2]

(c) y = (x – 1)³
A cubic expression is defined for all real numbers.
So, Domain = R
A cubic function can take every real value.
Thus, Range = R

(d) y = \(\sqrt{-x}\)
For square root to exist,
-x ≥ 0
x ≤ 0
Therefore, Domain = (-∞, 0]
Also, square root values are always non-negative,
y ≥ 0
Hence, Range = [0, ∞)