Relations-and-Functions
Case Study Questions-
1. Consider the mapping f: A → B is defined by f(x) = x – 1 such that f is a bijection.
Based on the above information, answer the following questions.
2. A relation R on a set A is said to be an equivalence relation on A iff it is:
Based on the above information, answer the following questions.
Answer Key-
Multiple Choice questions-
(b) R is reflexive and transitive but not symmetric
(c) (6, 8) ∈ R
(a) 1
(b) 2
(d) f is neither one-one nor onto.
(a) f is one-one onto
(c) x
(b) g(y) = 4y4 − 3y
(b) Transitive and symmetric
(c) 24
Very Short Answer:
[∵ When x = 2, then y = 3, when x = 4, then y = 2, when x = 6, then y = 1]
∴ under ‘f’, all the three elements of {1, 2, 3} should correspond to three different elements of the co-domain {1, 2, 3}.
Hence, ‘f’ is onto.
than f(x) = x − 1x − 1 = 1.
When x < 1,
than f(x) = −(x − 1)x − 1 = -1
Hence, Rf = {-1, 1}.
Let x1, x2∈ N.
Now, f(x1) = f(x2)
⇒ 2x1 = 2x2
⇒ x1 = x2
⇒ f is one-one.
Now, f is not onto.
∵ For 1 ∈ N, there does not exist any x ∈ N such that f(x) = 2x = 1.
Hence, f is ono-one but not onto.
f(f(x)) = 3 f(x) + 2
= 3(3x + 2) + 2 = 9x + 8.
fof(x) = f(f(x)) = (3-(f(x))3)1/3
= (3 – ((3 – x3)1/3)3)1/3
= (3 – (3 – x3))1/3 = (x3)1/3 = x.
Consider f: {1, 2, 3, 4} → {1, 2, 3, 4}
and g: {1, 2, 3,4} → {1,2.3} defined by:
f(1) = 1, f(2) = 2, f(3) = f(4) = 3
g (1) = 1, g (2) = 2, g (3) = g (4) = 3.
∴gof = g (f(x)) {1, 2,3}, which is onto
But f is not onto.
[∵ 4 is not the image of any element]
Short Answer:
(i) Here R = {(a, b): a is sister of b}.
Since the school is a Boys’ school,
∴ no student of the school can be the sister of any student of the school.
Thus R = Φ Hence, R is an empty relation.
(ii) Here R’ = {(a,b): the difference between heights of a and b is less than 3 metres}.
Since the difference between heights of any two students of the school is to be less than 3 metres,
∴ R’ = A x A. Hence, R’ is a universal relation.
For each a ∈ X, (a, a) ∈ R.
Thus R is reflexive. [∵ f (a) = f(a)]
Now (a, b) ∈ R
⇒ f(a) = f(b)
⇒ f(b) = f (a)
⇒ (b, a) ∈ R.
Thus R is symmetric.
And (a, b) ∈ R
and (b, c) ∈ R
⇒ f(a) = f(b)
and f(b) = f(c)
⇒ f(a)= f(c)
⇒ (a, c) ∈ R.
Thus R is transitive.
Hence, R is an equivalence relation.
Let 2 divide (a – b) and 2 divide (b – c), where a,b,c∈ Z
⇒ 2 divides [(a – b) + (b – c)]
⇒ 2 divides (a – c).
Hence, R is transitive.
And [0] = {0, ± 2, ± 4, ± 6,…].
Since f(1) = f(2) = 1,
∴f(1) = f(2), where 1 ≠ 2.
∴‘f’ is not one-one.
Let y ∈ N, y ≠ 1,
we can choose x as y + 1 such that f(x) = x – 1
= y + 1 – 1 = y.
Also 1 ∈ N, f(1) = 1.
Thus ‘f ’ is onto.
Hence, ‘f ’ is onto but not one-one.
We have:
f(x) = cos x and g(x) = 3x2.
∴gof (x) = g (f(x)) = g (cos x)
= 3 (cos x)2 = 3 cos2 x
and fog (x) = f(g (x)) = f(3x2) = cos 3x2.
Hence, gof ≠ fog.
We have: 4x + 36x − 4…(1)
∴fof(x) – f (f (x))
Given: (a, b) R (c, d) if and only if ad = bc.
(I) (a, b) R (a, b) iff ab – ba, which is true.
[∵ ab = ba∀ a, b ∈ N]
Thus, R is reflexive.
(II) (a, b) R (c,d) ⇒ ad = bc
(c, d) R (a, b) ⇒cb = da.
But cb = be and da = ad in N.
∴ (a, b) R (c, d) ⇒ (c, d) R (a, b).
Thus, R is symmetric.
(III) (a,b) R (c, d)
⇒ ad = bc…(1)
(c, d) R (e,f)
⇒cf = de … (2)
Multiplying (1) and (2), (ad). (cf) – (be), (de)
⇒af = be
⇒ (a,b) = R(e,f).
Thus, R is transitive.
Thus, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.
For x ∈ (0,1].
And (gof) (x) = g(f(x)) = g(1)
[∵ f(x) = 1 ∀ x > 0]
= [1] = 1
⇒ (gof) (x) = 1 ∀ x ∈ (0, 1] …(2)
From (1) and (2), (fog) and (gof) do not coincide in (0, 1].
Long Answer:
We have: R = {(a, b)} = a ≤ b}.
Since, a ≤ a ∀ a ∈ R,
∴ (a, a) ∈ R,
Thus, R reflexive.
Now, (a, b) ∈ R and (b, c) ∈ R
⇒ a ≤ b and b ≤ c
⇒ a ≤ c
⇒ (a, c) ∈ R.
Thus, R is transitive.
But R is not symmetric
[∵ (3, 5) ∈ R but (5, 3) ∉ R as 3 ≤ 5 but 5 > 3]
Solution:
Let x1, x2∈ N.
Now, f(x1) = f(x2)
Thus, f is one-one.
Let y ∈ N, then for any x,
f(x) = y if y = x2 + x + 1
We have:
R = {(a, b): a, b ∈ A; |a – b| is divisible by 4}.
(1) Reflexive: For any a ∈ A,
∴ (a, b) ∈ R.
|a – a| = 0, which is divisible by 4.
Thus, R is reflexive.
Symmetric:
Let (a, b) ∈ R
⇒ |a – b| is divisible by 4
⇒ |b – a| is divisible by 4
Thus, R is symmetric.
Transitive: Let (a, b) ∈ R and (b, c) ∈ R
⇒ |a – b| is divisible by 4 and |b – c| is divisible by 4
⇒ |a – b| = 4λ
⇒ a – b = ±4λ………….(1)
and |b – c| = 4µ
⇒ b – c = ± 4µ………….(2)
Adding (1) and (2),
(a-b) + (b-c) = ±4(λ + µ)
⇒ a – c = ± 4 (λ + µ)
⇒ (a, c) ∈ R.
Thus, R is transitive.
Now, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.
(ii) Let ‘x’ be an element of A such that (x, 1) ∈ R
⇒ |x – 1| is divisible by 4
⇒ x – 1 = 0,4, 8, 12,…
⇒ x = 1, 5, 9, 13, …
Hence, the set of all elements of A which are related to 1 is {1, 5, 9}.
(iii) Let (x, 2) ∈ R.
Thus |x – 2| = 4k, where k ≤ 3.
∴ x = 2, 6, 10.
Hence, equivalence class [2] = {2, 6, 10}.
Let y∈ R.
For any x, f(x) = y if y = 9x2 + 6x – 5
⇒ y = (9x2 + 6x + 1) – 6
= (3x + 1)2 – 6
Assertion and Reason Answers-
1. (a)Both A and R are true and R is the correct explanation of A.
2. (a) Both A and R are true and R is the correct explanation of A.
Case Study Answers-
1. Answer :
Solution:
For f(x) to be defined x – 2; ≠ 0 i.e., x; ≠ 2.
∴ Domain of f = R – {2}
Solution:
Let y = f(x), then 
⇒xy – 2y = x – 1 ⇒xy – x = 2y –
Since, x ∈∈ R – {2}, therefore y ≠ 1
Hence, range of f = R – {1}
Solution:
We have, g(x) = 2f(x) – 1
Solution:
⇒ x1x2 – 2x1 = x1x2 – 2x2 ⇒ 2x1 = 2x2 ⇒ x1 = x2
Thus, g(x1) = g(x2) ⇒ x1 = x2
Hence, g(x) is one-one.
2. Answer :
Solution:
Clearly, (1, 1), (2, 2), (3, 3), ∈ R. So, R is reflexive on A.
Since, (1, 2) ∈ R but (2, 1) ∉ R. So, R is not symmetric on A.
Since, (2, 3), ∈ R and (3, 1) ∈ R but (2, 1) ∉ R. So, R is not transitive on A.
Solution:
Since, (1, 1), (2, 2) and (3, 3) are not in R. So, R is not reflexive on A.
Now, (1, 2) ∈ R ⇒ (2, 1) ∈ R and (1, 3) ∈ R ⇒ (3, 1) ∈ R. So, R is symmetric,
Clearly, (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∉ R. So, R is not transitive on A.
Solution:
We have, R = {(x, y): y = x + 5 and x < 4}, where x, y ∈ N.
∴ R = {(1, 6), (2, 7), (3, 8)}
Clearly, (1, 1), (2, 2) etc. are not in R. So, R is not reflexive.
Since, (1, 6) ∈ R but (6, 1) ∉ R. So, R is not symmetric.
Since, (1, 6) ∈ R and there is no order pair in R which has 6 as the first element.
Same is the case for (2, 7) and (3, 8). So, R is transitive.
Solution:
We have, R = {(x, y): 3x – y = 0}, where x, y ∈ A = {1,2, ……, 14}.
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Clearly, (1, 1) ∉ R. So, R is not reflexive on A.
Since, (1, 3) ∈ R but (3, 1) ∉ R. So, R is not symmetric on A.
Since, (1, 3) ∈ Rand (3, 9) ∈ R but (1, 9) ∉ R. So, R is not transitive on A.
Solution:
Clearly, (1, 1), (2, 2), (3, 3) ∈ R. So, R is reflexive on A.
We find that the ordered pairs obtained by interchanging the components of ordered pairs in R are also in R. So, R is symmetric on A. For 1, 2, 3 ∈ A such that (1, 2) and (2, 3) are in R implies that (1, 3) is also, in R. So, R is transitive on A. Thus, R is an equivalence relation.
