Ch 1: Relations and Functions (NCERT Solutions)

Solution:

• (i) R in A = {1, 2, ..., 14} defined as R = {(x, y) : 3x - y = 0}:
  R = {(1,3), (2,6), (3,9), (4,12)}.
  - Reflexive: (1,1) ∉ R. Not reflexive.
  - Symmetric: (1,3) ∈ R but (3,1) ∉ R. Not symmetric.
  - Transitive: (1,3) ∈ R and (3,9) ∈ R but (1,9) ∉ R. Not transitive.
• (ii) R in N defined as R = {(x, y) : y = x + 5 and x < 4}:
  R = {(1,6), (2,7), (3,8)}.
  - Not reflexive ((1,1) ∉ R). Not symmetric ((1,6) ∈ R, (6,1) ∉ R). Transitive (Vacuously true, as there are no pairs (x,y) and (y,z)).
• (iii) R in A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}:
  - Reflexive: (x,x) ∈ R since x is divisible by x. Yes.
  - Symmetric: (1,2) ∈ R but (2,1) ∉ R. No.
  - Transitive: If y is divisible by x, and z is divisible by y, then z is divisible by x. Yes.
• (iv) R in Z as R = {(x, y) : x - y is an integer}:
  Reflexive (x-x=0 is int), Symmetric (x-y is int → y-x is int), Transitive (x-y & y-z are ints → (x-y)+(y-z) = x-z is int). It is an equivalence relation.
• (v) R in set of human beings in a town:
  (a) x and y work at same place: Equivalence.
  (b) x and y live in same locality: Equivalence.
  (c) x is exactly 7 cm taller than y: Not ref, Not sym, Not trans.
  (d) x is wife of y: Not ref, Not sym, Not trans.
  (e) x is father of y: Not ref, Not sym, Not trans.

Solution:

- Reflexive: Let a = 1/2. Is 1/2 ≤ (1/2)²? No, 1/2 is not ≤ 1/4. Not reflexive.

- Symmetric: Let a = 1, b = 4. 1 ≤ 4² (1 ≤ 16) is true. But 4 ≤ 1² (4 ≤ 1) is false. Not symmetric.

- Transitive: Let a = 3, b = 2, c = 1.5.
a ≤ b² → 3 ≤ 4 (True).
b ≤ c² → 2 ≤ 2.25 (True).
But a ≤ c² → 3 ≤ 2.25 (False). Not transitive.

Solution:

R = {(1,2), (2,3), (3,4), (4,5), (5,6)}

- (1,1) ∉ R → Not Reflexive.
- (1,2) ∈ R but (2,1) ∉ R → Not Symmetric.
- (1,2) ∈ R and (2,3) ∈ R, but (1,3) ∉ R → Not Transitive.

Solution:

- Reflexive: a ≤ a is always true for any real number a. So, (a,a) ∈ R.

- Transitive: If a ≤ b and b ≤ c, then mathematically a ≤ c. So, (a,c) ∈ R.

- Symmetric: Let a = 2, b = 4. 2 ≤ 4 is true, but 4 ≤ 2 is false. So, (2,4) ∈ R but (4,2) ∉ R. Not symmetric.

Solution:

- Reflexive: Let a = 1/2. 1/2 ≤ (1/2)³ → 1/2 ≤ 1/8 (False). Not reflexive.

- Symmetric: Let a = 1, b = 2. 1 ≤ 2³ is true, but 2 ≤ 1³ is false. Not symmetric.

- Transitive: Let a = 10, b = 3, c = 2.
10 ≤ 3³ (10 ≤ 27) → True.
3 ≤ 2³ (3 ≤ 8) → True.
But 10 ≤ 2³ (10 ≤ 8) → False. Not transitive.

Solution:

Set A = {1, 2, 3}. R = {(1, 2), (2, 1)}.

- (1,1) ∉ R → Not reflexive.

- (1,2) ∈ R and (2,1) ∈ R → Symmetric.

- (1,2) ∈ R and (2,1) ∈ R, but (1,1) ∉ R → Not transitive.

Solution:

- Reflexive: Book x has the same number of pages as itself. (x,x) ∈ R.

- Symmetric: If x has the same pages as y, then y has the same pages as x. (x,y) ∈ R → (y,x) ∈ R.

- Transitive: If x and y have the same pages, and y and z have the same pages, then x and z have the same pages. (x,y) ∈ R and (y,z) ∈ R → (x,z) ∈ R.

Since it is Reflexive, Symmetric, and Transitive, it is an Equivalence Relation.

Solution:

Equivalence Proof:
- Reflexive: |a - a| = 0, which is even. (a,a) ∈ R.
- Symmetric: |a - b| is even → |b - a| is also even. (b,a) ∈ R.
- Transitive: If |a - b| is even and |b - c| is even, then (a-b) and (b-c) are both even or both odd. Their sum a-c is even. So |a-c| is even. (a,c) ∈ R.

Subsets:
- {1, 3, 5}: The difference between any two odd numbers is always even. Thus, they are related to each other.
- {2, 4}: The difference between any two even numbers is even. They are related.
- {1, 3, 5} and {2, 4}: The difference between an odd and an even number is always odd. Hence, no element from the first set is related to the second.

Solution:

(i) |a - b| is a multiple of 4:
- Ref: |a - a| = 0 (multiple of 4). Sym: |a - b| multiple of 4 → |b - a| is too. Trans: a-b = 4k, b-c = 4m → a-c = 4(k+m). Equivalence.
- Elements related to 1: |x - 1| is a multiple of 4. x ∈ A.
|x - 1| = 0, 4, 8, 12 → x = 1, 5, 9. Set = {1, 5, 9}.

(ii) a = b:
- Ref: a = a. Sym: a = b → b = a. Trans: a = b and b = c → a = c. Equivalence.
- Elements related to 1: x = 1. Set = {1}.

Solution:

Let A = {1, 2, 3}.

• (i): R = {(1, 2), (2, 1)}
• (ii): R = {(1, 2), (2, 3), (1, 3)}
• (iii): R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
• (iv): R = {(1, 1), (2, 2), (3, 3), (1, 2)}
• (v): R = {(1, 1), (2, 2), (1, 2), (2, 1)} (Notice (3,3) is missing, so not reflexive on A)

Solution:

Let O be the origin. OP = OQ.

- Reflexive: OP = OP. True.
- Symmetric: If OP = OQ, then OQ = OP. True.
- Transitive: If OP = OQ and OQ = OS, then OP = OS. True.
Since it satisfies all three, it's an equivalence relation.

Further: The set of all points related to P form a circle passing through P with origin as center.

Solution:

Similarity is Reflexive (T1 ~ T1), Symmetric (T1 ~ T2 → T2 ~ T1), and Transitive (T1 ~ T2 & T2 ~ T3 → T1 ~ T3). Hence, Equivalence.

Comparing ratios of sides:
T1/T3 = 3/6 = 4/8 = 5/10 = 1/2.
Therefore, T1 and T3 have proportional sides and are similar. T1 is related to T3.

Solution:

Same number of sides is clearly Reflexive (n=n), Symmetric (n1=n2 → n2=n1), and Transitive (n1=n2 & n2=n3 → n1=n3). Hence, Equivalence.

The triangle T has 3 sides. Elements related to T will be all polygons with 3 sides. Set = The set of all triangles in a plane.

Solution:

- A line is parallel to itself (Reflexive).
- L1 || L2 → L2 || L1 (Symmetric).
- L1 || L2 & L2 || L3 → L1 || L3 (Transitive). Hence, Equivalence.

Lines related to y = 2x + 4 must have the same slope (m = 2).
Set of lines: y = 2x + c, where c is any real number.

Solution:

Reflexive? Yes, contains (1,1), (2,2), (3,3), (4,4).
Symmetric? No, contains (1,2) but not (2,1).
Transitive? (1,3) ∈ R and (3,2) ∈ R → (1,2) ∈ R. Yes, it satisfies all transitive cases.

Correct Answer: (B) R is reflexive and transitive but not symmetric.

Solution:

Given: a = b - 2, and b > 6.
Check options:
(A) (2, 4) ∈ R → False, because b=4 is not > 6.
(B) (3, 8) ∈ R → False, 3 ≠ 8 - 2.
(C) (6, 8) ∈ R → True, 6 = 8 - 2, and 8 > 6.
(D) (8, 7) ∈ R → False.

Correct Answer: (C) (6, 8) ∈ R