Chapter 2- Relations and Functions Miscellaneous Exercise Page No: 46
We can see that for 0 ≤ x < 3,
f(x) = x2
and for 3 < x ≤ 10,
f(x) = 3x
Also, at x = 3
f(x) = 32 = 9 or f(x) = 3 × 3 = 9
i.e., at x = 3, f(x) = 9 [single image for f(x) = x2 as well as f(x) = 3x ]
Hence, for 0 ≤ x ≤ 10, the images of f(x) are unique.
Therefore, the given relation f is a function.
Now,
The relation g is defined as
We can see that, for x = 2
g(x) = 22 = 4 and g(x) = 3 × 2 = 6
Thus, element 2 of the domain of the relation g corresponds to two different images, i.e., 4 and 6.
Hence, the given relation g is not a function.
We can see that the function f is defined for all real numbers except at x = 6 and x = 2,
Hence, the domain of f is R – {2, 6}.
4. Find the domain and the range of the real function f defined by
f(x) = √(x – 1).
Solution:
Given real function is,
f(x) = √(x – 1)
Here, √(x – 1) is defined for (x – 1) ≥ 0
So, the function f(x) = √(x – 1) is defined for x ≥ 1
Thus, the domain of f is the set of all real numbers greater than or equal to 1.
Hence, Domain of f = [1, ∞)
Now,
As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √(x – 1) ≥ 0
Thus, the range of f is the set of all real numbers greater than or equal to 0.
Hence, Range of f = [0, ∞)
It is given, the functions f, g: R → R is defined as
f(x) = x + 1, g(x) = 2x – 3
So,
(f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2
Therefore, (f + g) (x) = 3x – 2
(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4
therefore, (f – g) (x) = –x + 4
8. Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.
Solution:
It is given,
f = {(1, 1), (2, 3), (0, –1), (–1, –3)}
And, f(x) = ax + b
Therefore for (1, 1) ∈ f
We have, f(1) = 1
So, a × 1 + b = 1
a + b = 1 …. (i)
And for (0, –1) ∈ f
f(0) = –1
a × 0 + b = –1
b = –1…….(ii)
On substituting b = –1 in (i), we get
a + (–1) = 1 ⇒ a = 1 + 1 = 2.
Therefore,
a= 2 and b = -1
9. Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b2}. Are the following true?
(i) (a, a) ∈ R, for all a ∈ N
(ii) (a, b) ∈ R, implies (b, a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R
Justify your answer in each case.
Solution:
It is given relation R = {(a, b): a, b ∈ N and a = b2}
(i)We can see that 2 ∈ N; however, 2 ≠ 22 = 4.
Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.
(ii) We can see that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.
Now, 3 ≠ 92 = 81;
Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.
(iii) It’s clearly seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 42 and 4 = 22.
Now, 16 ≠ 22 = 4;
So,
(16, 4) ∈ R, (4, 2) ∈ R does not implies (16,2) ∈ R for relation from N to N
R = {(a, b): a, b ∈ N and a = b2}.
Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R ” is not true.
10. Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?
(i) f is a relation from A to B (ii) f is a function from A to B
Justify your answer in each case.
Solution:
It is given,
A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}
So,
A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}
Also, given that,
f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.
We can see that f is a subset of A × B.
Hence, f is a relation from A to B.
(ii) As the element, 2 corresponds to two different images (9 and 11), relation f is not a function.
11. Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.
Solution:
It is given relation f is defined as,
f = {(ab, a + b): a, b ∈ Z}
We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B.
3, 4, –3, –4 ∈ Z,
(3× 4, 3 + 4), (-3 × –4, –3 + (–4)) ∈ f
i.e., (12, 7), (12, -7) ∈ f
We can see that the same first element, 12, corresponds to two different images (7 and –7).
Hence, the relation f is not a function.
12. Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.
Solution:
It is given,
A = {9, 10, 11, 12, 13}
Now, f: A → N is defined as
f(n) = The highest prime factor of n
Prime factor of 9 = 3
Prime factors of 10 = 2, 5
Prime factor of 11 = 11
Prime factors of 12 = 2, 3
Prime factor of 13 = 13
Thus,
f(9) = The highest prime factor of 9 = 3
f(10) = The highest prime factor of 10 = 5
f(11) = The highest prime factor of 11 = 11
f(12) = The highest prime factor of 12 = 3
f(13) = The highest prime factor of 13 = 13
The range of f is the set of all f(n), where n ∈ A.
Hence,
Range of f = {3, 5, 11, 13}