Class 9 Advanced Maths Chapter 3 Relations and Functions Ex 3.4 Solutions - #NCSOLVE 📚

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Ex 3.4 Class 9 Advanced Maths Solutions

Advanced Maths Class 9 Exercise 3.4 Solutions

Exercise 3.4 Class 9 Advanced Maths 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) We have, R = {(5, 1), (4, 1), (3, 1), (2, 0)}
We know that the domain is the set of all first elements of the ordered pairs.
∴ Domain = {5, 4, 3, 2}
and the range is the set of all second elements of the ordered pairs.
∴ Range = {1, 0}
Here, each element of the domain has exactly one image in the co-domain.
Hence, the given relation is a function.

(b) We have.R = {(1, -1), (2, -2), (3, -3), (4, -4), (5, -5)}
Here, Domain = {1, 2, 3, 4, 5}
and Range = {-1, -2, -3, -4, -5}
Here, each element of the domain has exactly one image in the co-domain.
Hence, the given relation is a function.

(c) We have, R = {(3, -1), (3, 0), (3, 1), (3, 2)}
Here, Domain = {3}
and Range = {-1, 0, 1, 2}
Here, the element 3 has more than one image in the co-domain.
Hence, the given relation is not a function.

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 = -√x , where x ∈ (0, ∞)}
Solution:
(a) We have, R = {(x, y) : xy = 8, where x, y, ∈ Z}

Now, xy = 8 gives integer factor pairs of 8.
So, R = {(1, 8), (2, 4), (4, 2), (8, 1), (-1 -8), (-2, -4), (-4, -2), (-8, -1)}
Hence, Domain = {-8, – 4, -2, -1, 1, 2, 4, 8}
and Range = {-8, – 4, -2, -1, 1, 2, 4, 8}

(b) We have, R = {(x, y) : x = |y|, x ∈ Z, 0 ≤ x ≤ 5}
Now, x = |y| ⇒ y = ±x
Given, x = 0, 1, 2, 3, 4, 5
So, the ordered pairs are
R = {(0, 0), (1, 1), (1, -1), (2, 2), (2, -2), (3, 3), (3, -3), (4, 4), (4, -4), (5, 5), (5, -5)}.
Hence, Domain = {0, 1, 2, 3, 4, 5}
and Range = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

(c) R = {(x, y) : y = -√x, x ∈ (0, ∞)}
Here, x > 0
Since, √x > 0
∴ y = -√x < 0
Hence, Domain = (0, ∞) and Range = (-∞, 0)

Question 3.
Draw the graph of the functions f, 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:
Given, f(x) = |x|, g(x) – |x| -1, and h(x) = |x| + 1
For f(x) = |x|,

x	y = |x|
-2	2
-1	1
0	0
1	1
2	2

For g(x) = |x| – 1

x	y = |x| – 1
-2	1
-1	0
0	-1
1	0
2	1

For h(x) = |x| + 1,

x	y = |x| +  1
-2	3
-1	2
0	1
1	2
2	3

Hence, all three graphs are V-shaped and are symmetric about the Y-axis.
The graph of g(x) = |x| – 1 is obtained by shifting the graph of f(x) = |x| one unit downward.
Also, the graph of h (x) = |x| + 1 is obtained by shifting the graph of f(x) = |x| one unit upward.

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

Describe the relationship among the graphs of f, g and h Are there domain and range equal?
Solution:
Given, f(x) = x², g(x) = (x – 1)² and h(x) = (x + 2)²
For f (x) = x²,

x	y = x2
-2	4
-1	1
0	0
1	1
2	4

For g(x) = (x – 1)²,

x	y = (x- 1) 2
-2	9
-1	4
0	1
1	0
2	0

For h(x) = {x + 2)²,

x	y = (x + 2)2
-2	0
-1	1
0	4
1	9
2	16

Now, the graph of f(x) = x² has vertex at (0, 0).
Also, the graph of g(x) = (x – 1))², is a horizontal shift of f(x) to the right by 1 unit and h(x) = (x + 2))², is a horizontal shift of f(x) to the left by 2 units.
Hence, all three graphs have the same shape, but their positions are different.
The domain of all three functions is the set of all real numbers, that isR and the range of all three functions is (0, ∞).
Hence, their domains are equal and their ranges are also equal.

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 = √-x
Solution:
(a) We have, y = \(\frac{1}{x^2}\)
For the given function, y = \(\frac{1}{x^2}\)
∵ The denominator cannot be zero.
So, x² ≠ 0 which gives x ≠ 0.
Hence, the domain of the function is R – {0}.
Also, since x² > 0 for every x ≠ 0,
Therefore, \(\frac{1}{x^2}\) > 0
Thus, the value of y is always positive and can never be zero.
Hence, the range of the function is (0, ∞).

(b) We have, y = 2 – 1 |x|
For the given function, y = 2 – |x|
∵ The modulus function is defined for every real number.
Hence, the domain of the function is R.
Now, |x| ≥ 0
Therefore, 2 – |x|≤ 2
Also, as the value of |x| increases, the value of 2 – |x|
decreases without bound.
Hence, the greatest value of y is 2.
Therefore, the range of the function is (-∞, 2].

(c) We have,y = (x – 1)³
For the given function,
y = (x – 1)³
∴ The cubic expression is defined for every real number. Hence, the domain of the function is R.
Also, a cubic function can take every real value from
-∞ to + ∞.
Hence, the range of the function is R.

(d) We have, y = √-x
For the square root to exist,
-x ≥ 0
On multiplying by – 1, we get x ≤ 0
Hence, the domain of the function is (-∞,0].
Also, the square root of a number is always non-negative.
Therefore, y ≥ 0
Hence, the range of the function is [0, ∞).

The post Class 9 Advanced Maths Chapter 3 Relations and Functions Ex 3.4 Solutions appeared first on Learn CBSE.

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