NCERT Solutions for Class 12 Mathematics Chapter 1: Relations and Functions

Answer

(v) Relation R in the set A of human beings in a town at a particular time given by (a) R = {(x, y) : x and y work at the same place} (b) R = {(x, y) : x and y live in the same locality} (c) R = {(x, y) : x is exactly 7 cm taller than y} (d) R = {(x, y) : x is wife of y} (e) R = {(x, y) : x is father of y}

Ans. (i) 3x – y = 0 means y = 3x. With x, y in A: R = {(1, 3), (2, 6), (3, 9), (4, 12)}.

Reflexive? (1, 1) ∉ R, so not reflexive. Symmetric? (1, 3) ∈ R but (3, 1) ∉ R, so not symmetric. Transitive? (1, 3) ∈ R and (3, 9) ∈ R but (1, 9) ∉ R, so not transitive.

R is neither reflexive, nor symmetric, nor transitive.

(ii) x ∈ N and x < 4 gives x = 1, 2, 3, so R = {(1, 6), (2, 7), (3, 8)}.

(1, 1) ∉ R
⇒ not reflexive. (1, 6) ∈ R but (6, 1) ∉ R
⇒ not symmetric. There is no pair (x, y) and (y, z) both in R (the second components 6, 7, 8 never appear as first components), so the condition for transitivity is never violated
⇒ R is transitive.

R is transitive but neither reflexive nor symmetric.

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

Every x divides itself, so (x, x) ∈ R
⇒ reflexive. (1, 2) ∈ R but 1 is not divisible by 2, so (2, 1) ∉ R
⇒ not symmetric. If x divides y and y divides z, then x divides z
⇒ transitive.

R is reflexive and transitive but not symmetric.

(iv) For any x ∈ Z, x – x = 0 is an integer
⇒ reflexive. If x – y is an integer, then y – x = –(x – y) is also an integer
⇒ symmetric. If x – y and y – z are integers, then x – z = (x – y) + (y – z) is an integer
⇒ transitive.

R is reflexive, symmetric and transitive (an equivalence relation).

(v)(a) Every person works at the same place as himself
⇒ reflexive. If x and y work at the same place, so do y and x
⇒ symmetric. If x, y work at the same place and y, z work at the same place, then x, z work at the same place
⇒ transitive. R is reflexive, symmetric and transitive.

(v)(b) Exactly the same reasoning with "live in the same locality": R is reflexive, symmetric and transitive.

(v)(c) No one is 7 cm taller than himself
⇒ not reflexive. If x is 7 cm taller than y, then y is 7 cm shorter than x
⇒ not symmetric. If x is 7 cm taller than y and y is 7 cm taller than z, x is 14 cm taller than z
⇒ not transitive. R is neither reflexive, nor symmetric, nor transitive.

(v)(d) No one is his/her own wife
⇒ not reflexive. If x is wife of y, then y is husband (not wife) of x
⇒ not symmetric. For (x, y) and (y, z) both to be in R, y would have to be a wife (female) and also have a wife, which is impossible, so the transitivity condition is never violated
⇒ transitive. R is transitive but neither reflexive nor symmetric.

(v)(e) No one is his own father
⇒ not reflexive. If x is father of y, y cannot be father of x
⇒ not symmetric. If x is father of y and y is father of z, then x is grandfather of z, not father
⇒ not transitive. R is neither reflexive, nor symmetric, nor transitive.

Answer

Not reflexive: take a = 1/2. Then a² = 1/4 and 1/2 ≤ 1/4 is false, so (1/2, 1/2) ∉ R.

Not symmetric: 1 ≤ 4² = 16, so (1, 4) ∈ R; but 4 ≤ 1² = 1 is false, so (4, 1) ∉ R.

Not transitive: 3 ≤ 2² = 4, so (3, 2) ∈ R; 2 ≤ (1.5)² = 2.25, so (2, 1.5) ∈ R; but 3 ≤ (1.5)² = 2.25 is false, so (3, 1.5) ∉ R.

Hence R is neither reflexive, nor symmetric, nor transitive.

Answer

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), (2, 3) ∈ R but (1, 3) ∉ R
⇒ not transitive.

R is neither reflexive, nor symmetric, nor transitive.

Answer

Reflexive: a ≤ a for every real a, so (a, a) ∈ R.

Not symmetric: 1 ≤ 2, so (1, 2) ∈ R, but 2 ≤ 1 is false, so (2, 1) ∉ R.

Transitive: if a ≤ b and b ≤ c then a ≤ c, so (a, b), (b, c) ∈ R
⇒ (a, c) ∈ R.

Hence R is reflexive and transitive but not symmetric.

Answer

Not reflexive: (1/2)³ = 1/8 and 1/2 ≤ 1/8 is false, so (1/2, 1/2) ∉ R.

Not symmetric: 1 ≤ 2³ = 8, so (1, 2) ∈ R; but 2 ≤ 1³ = 1 is false, so (2, 1) ∉ R.

Not transitive: 3 ≤ (3/2)³ = 27/8 = 3.375, so (3, 3/2) ∈ R; 3/2 ≤ (4/3)³ = 64/27 ≈ 2.37, so (3/2, 4/3) ∈ R; but 3 ≤ 64/27 is false, so (3, 4/3) ∉ R.

R is neither reflexive, nor symmetric, nor transitive.

Answer

Symmetric: (1, 2) ∈ R and (2, 1) ∈ R; these are the only pairs, so for every (a, b) ∈ R, (b, a) ∈ R.

Not reflexive: (1, 1) ∉ R (also (2, 2), (3, 3) ∉ R).

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

Answer

Reflexive: every book has the same number of pages as itself, so (x, x) ∈ R.

Symmetric: if x and y have the same number of pages, then so do y and x, so (x, y) ∈ R
⇒ (y, x) ∈ R.

Transitive: if x, y have the same number of pages and y, z have the same number of pages, then x and z have the same number of pages, so (x, z) ∈ R.

Hence R is an equivalence relation.

Answer

Reflexive: |a – a| = 0, which is even, so (a, a) ∈ R.

Symmetric: |a – b| = |b – a|, so if |a – b| is even then |b – a| is even.

Transitive: if |a – b| and |b – c| are even, then a – b and b – c are even, so a – c = (a – b) + (b – c) is even, i.e. |a – c| is even.

So R is an equivalence relation.

Note that |a – b| is even exactly when a and b are both odd or both even. Elements of {1, 3, 5} are all odd, so the difference of any two is even
⇒ all related to each other. Elements of {2, 4} are both even
⇒ related to each other. An element of {1, 3, 5} (odd) and an element of {2, 4} (even) have an odd difference, so no element of {1, 3, 5} is related to any element of {2, 4}.

Answer

A = {0, 1, 2, ..., 12}.

(i) Reflexive: |a – a| = 0 = 4 × 0, a multiple of 4. Symmetric: |a – b| = |b – a|. Transitive: if a – b = 4m and b – c = 4n (m, n integers), then a – c = 4(m + n), a multiple of 4. So R is an equivalence relation.

Elements related to 1: b with |1 – b| a multiple of 4
⇒ b = 1, 5, 9. Set = {1, 5, 9}.

(ii) Reflexive: a = a. Symmetric: a = b
⇒ b = a. Transitive: a = b and b = c
⇒ a = c. So R is an equivalence relation.

Elements related to 1: only b = 1. Set = {1}.

Answer

(i) In A = {1, 2, 3}, R = {(1, 2), (2, 1)}. Symmetric; (1, 1) ∉ R so not reflexive; (1, 2), (2, 1) ∈ R but (1, 1) ∉ R so not transitive.

(ii) In R (real numbers), R = {(a, b) : a < b}. a < a is false
⇒ not reflexive; 1 < 2 but not 2 < 1
⇒ not symmetric; a < b and b < c
⇒ a < c
⇒ transitive.

(iii) In A = {1, 2, 3}, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}. Reflexive and symmetric; (1, 2), (2, 3) ∈ R but (1, 3) ∉ R
⇒ not transitive.

(iv) In R, R = {(a, b) : a ≤ b}. Reflexive and transitive, but (1, 2) ∈ R, (2, 1) ∉ R
⇒ not symmetric.

(v) In A = {1, 2, 3}, R = {(1, 1), (1, 2), (2, 1), (2, 2)}. Symmetric; every chain (a, b), (b, c) in R gives (a, c) ∈ R
⇒ transitive; (3, 3) ∉ R
⇒ not reflexive.

Answer

Let O be the origin. (P, Q) ∈ R ⇔ OP = OQ.

Reflexive: OP = OP, so (P, P) ∈ R. Symmetric: OP = OQ
⇒ OQ = OP. Transitive: OP = OQ and OQ = OS
⇒ OP = OS. Hence R is an equivalence relation.

Points related to P (P ≠ O): all points Q with OQ = OP = r (say), r > 0. The locus of points at a fixed distance r from O is the circle with centre O and radius r, and it passes through P. So the required set is the circle through P with the origin as centre.

Answer

Reflexive: every triangle is similar to itself. Symmetric: if T₁ ~ T₂ then T₂ ~ T₁. Transitive: if T₁ ~ T₂ and T₂ ~ T₃ then T₁ ~ T₃. So R is an equivalence relation.

Ratios of sides: T₁ and T₃
⇒ 3/6 = 4/8 = 5/10 = 1/2, so the sides are proportional and T₁ ~ T₃.

T₁ and T₂
⇒ 3/5, 4/12, 5/13 are not equal; T₂ and T₃
⇒ 5/6, 12/8, 13/10 are not equal.

Hence only T₁ and T₃ are related.

Answer

Reflexive: a polygon has the same number of sides as itself. Symmetric: if P₁ and P₂ have the same number of sides, so do P₂ and P₁. Transitive: if P₁, P₂ have equal numbers of sides and P₂, P₃ have equal numbers of sides, then P₁ and P₃ do too. So R is an equivalence relation.

T has 3 sides, so the polygons related to T are exactly those with 3 sides: the set of all triangles in A.

Answer

Reflexive: every line is parallel to itself (taking coincident lines as parallel). Symmetric: L₁ ∥ L₂
⇒ L₂ ∥ L₁. Transitive: L₁ ∥ L₂ and L₂ ∥ L₃
⇒ L₁ ∥ L₃. So R is an equivalence relation.

Lines parallel to y = 2x + 4 have the same slope 2. Required set = {lines y = 2x + c : c ∈ R}.

Answer

(1, 1), (2, 2), (3, 3), (4, 4) ∈ R
⇒ reflexive. (1, 2) ∈ R but (2, 1) ∉ R
⇒ not symmetric.

Transitivity check: (1, 3), (3, 2)
⇒ (1, 2) ∈ R; (1, 2), (2, 2)
⇒ (1, 2) ∈ R; (3, 2), (2, 2)
⇒ (3, 2) ∈ R; pairs with (a, a) give nothing new
⇒ transitive.

Correct answer: (B).

Answer

(2, 4): b = 4 is not > 6. (3, 8): 8 – 2 = 6 ≠ 3. (6, 8): 8 > 6 and 8 – 2 = 6 ✓. (8, 7): 7 – 2 = 5 ≠ 8.

Correct answer: (C).

Answer

One-one: f(x₁) = f(x₂)
⇒ 1/x₁ = 1/x₂
⇒ x₁ = x₂.

Onto: for any y ∈ R∗, take x = 1/y ∈ R∗; then f(x) = 1/(1/y) = y. So f is one-one and onto.

With domain N: g : N
⇒ R∗, g(x) = 1/x. It is still one-one (same argument). But it is not onto: for y = 2 ∈ R∗, 1/x = 2 gives x = 1/2 ∉ N. So the result is not fully true — g is one-one but not onto.

Answer

(i) For x₁, x₂ ∈ N, x₁² = x₂²
⇒ x₁ = x₂ (both positive)
⇒ injective. 2 ∈ N has no natural x with x² = 2
⇒ not surjective.

(ii) f(–1) = f(1) = 1
⇒ not injective. –1 ∈ Z has no preimage since x² ≥ 0
⇒ not surjective.

(iii) f(–1) = f(1)
⇒ not injective. –1 ∈ R has no preimage
⇒ not surjective.

(iv) x₁³ = x₂³
⇒ x₁ = x₂
⇒ injective. 2 ∈ N is not a cube of any natural number
⇒ not surjective.

(v) x₁³ = x₂³
⇒ x₁ = x₂
⇒ injective. 2 ∈ Z is not the cube of any integer
⇒ not surjective.

Answer

Not one-one: f(1.2) = 1 and f(1.9) = 1, but 1.2 ≠ 1.9.

Not onto: [x] is always an integer, so a non-integer like 0.7 in the co-domain R has no preimage.

Answer

Not one-one: f(–1) = |–1| = 1 = f(1), but –1 ≠ 1.

Not onto: |x| ≥ 0 for all x, so –1 ∈ R has no preimage.

Answer

Not one-one: f(1) = f(2) = 1 but 1 ≠ 2.

Not onto: the range of f is {–1, 0, 1}; for example 2 ∈ R has no preimage.

Answer

f(1) = 4, f(2) = 5, f(3) = 6. Distinct elements of A have distinct images, so f is one-one. (It is not onto, since 7 has no preimage.)

Answer

(i) One-one: 3 – 4x₁ = 3 – 4x₂
⇒ x₁ = x₂. Onto: for y ∈ R, put x = (3 – y)/4 ∈ R
⇒ f(x) = 3 – (3 – y) = y. So f is bijective.

(ii) f(1) = f(–1) = 2
⇒ not one-one. 1 + x² ≥ 1, so e.g. –2 has no preimage
⇒ not onto. f is neither one-one nor onto.

Answer

One-one: f(a₁, b₁) = f(a₂, b₂)
⇒ (b₁, a₁) = (b₂, a₂)
⇒ b₁ = b₂ and a₁ = a₂
⇒ (a₁, b₁) = (a₂, b₂).

Onto: any element of B × A is (b, a) with b ∈ B, a ∈ A; then (a, b) ∈ A × B and f(a, b) = (b, a).

Hence f is bijective.

Answer

f(1) = (1 + 1)/2 = 1 and f(2) = 2/2 = 1. So f(1) = f(2) with 1 ≠ 2
⇒ f is not one-one.

(f is onto: for any m ∈ N, f(2m) = m.)

Since f is not one-one, f is not bijective.

Answer

One-one: (x₁ – 2)/(x₁ – 3) = (x₂ – 2)/(x₂ – 3)
⇒ (x₁ – 2)(x₂ – 3) = (x₂ – 2)(x₁ – 3)
⇒ x₁x₂ – 3x₁ – 2x₂ + 6 = x₁x₂ – 3x₂ – 2x₁ + 6
⇒ –x₁ = –x₂
⇒ x₁ = x₂.

Onto: let y ∈ B (y ≠ 1). Solve y = (x – 2)/(x – 3)
⇒ xy – 3y = x – 2
⇒ x(y – 1) = 3y – 2
⇒ x = (3y – 2)/(y – 1).

This x is defined since y ≠ 1, and x ≠ 3 (otherwise 3y – 2 = 3y – 3, impossible), so x ∈ A. Check: f(x) = y.

Hence f is one-one and onto (bijective).

Answer

f(1) = f(–1) = 1
⇒ not one-one. x⁴ ≥ 0, so –2 has no preimage
⇒ not onto.

Correct answer: (D).

Answer

3x₁ = 3x₂
⇒ x₁ = x₂
⇒ one-one. For y ∈ R, x = y/3 gives f(x) = y
⇒ onto.

Correct answer: (A).

Answer

For x ≥ 0, f(x) = x/(1 + x) ≥ 0; for x < 0, f(x) = x/(1 – x) < 0.

One-one: Case 1, x₁, x₂ ≥ 0: x₁/(1 + x₁) = x₂/(1 + x₂)
⇒ x₁ + x₁x₂ = x₂ + x₁x₂
⇒ x₁ = x₂.

Case 2, x₁, x₂ < 0: x₁/(1 – x₁) = x₂/(1 – x₂)
⇒ x₁ – x₁x₂ = x₂ – x₁x₂
⇒ x₁ = x₂.

Case 3, x₁ ≥ 0 > x₂: f(x₁) ≥ 0 > f(x₂), so f(x₁) ≠ f(x₂). Hence f is one-one.

Onto: let y ∈ (–1, 1).

If 0 ≤ y < 1, take x = y/(1 – y) ≥ 0
⇒ f(x) = [y/(1 – y)] ÷ [1 + y/(1 – y)] = [y/(1 – y)] ÷ [1/(1 – y)] = y.

If –1 < y < 0, take x = y/(1 + y) < 0
⇒ f(x) = x/(1 – x) = [y/(1 + y)] ÷ [1/(1 + y)] = y.

Every y has a preimage, so f is onto. Hence f is one-one and onto.

Answer

Let f(x₁) = f(x₂)
⇒ x₁³ – x₂³ = 0
⇒ (x₁ – x₂)(x₁² + x₁x₂ + x₂²) = 0.

x₁² + x₁x₂ + x₂² = (x₁ + x₂/2)² + 3x₂²/4, which is 0 only when x₁ = x₂ = 0. In every case we get x₁ = x₂.

So f is injective (one-one).

Answer

Reflexive: every set is a subset of itself, A ⊂ A, so ARA.

Transitive: A ⊂ B and B ⊂ C ⇒ A ⊂ C.

Not symmetric: φ ⊂ X, so φ R X; but X ⊄ φ (X is non-empty), so X is not related to φ.

Since R is not symmetric, R is not an equivalence relation on P(X).

Answer

For a finite set, a function from the set to itself is onto if and only if it is one-one, i.e. a bijection. The number of bijections of an n-element set onto itself equals the number of arrangements of n elements = n!.

Number of onto functions = n!.

Answer

f(–1) = 1 + 1 = 2; g(–1) = 2|–3/2| – 1 = 3 – 1 = 2.

f(0) = 0; g(0) = 2(1/2) – 1 = 0.

f(1) = 1 – 1 = 0; g(1) = 2(1/2) – 1 = 0.

f(2) = 4 – 2 = 2; g(2) = 2(3/2) – 1 = 2.

f(a) = g(a) for every a ∈ A, and both have the same domain and co-domain. Hence f and g are equal.

Answer

Reflexive and symmetric relation containing (1, 2), (1, 3) must contain R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)}.

R is not transitive: (2, 1), (1, 3) ∈ R but (2, 3) ∉ R.

The only remaining pairs are (2, 3), (3, 2); symmetry forces adding both together, which gives A × A, the universal relation, which is transitive. So exactly one such relation exists.

Correct answer: (A).

Answer

The smallest equivalence relation containing (1, 2) is R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}.

Adding any pair involving 3 with 1 or 2 (with symmetry and transitivity) forces all pairs, giving the universal relation R₂ = A × A.

So there are exactly 2 such equivalence relations. Correct answer: (B).

Answer

Yes. a + a = 2a is even (reflexive); a + b = b + a (symmetric); if a + b and b + c are even, a + c = (a + b) + (b + c) – 2b is even (transitive). It splits Z into two classes: even integers and odd integers.

Answer

[0] = {b ∈ A : b is a multiple of 3} = {0, 3, 6, 9}.

Answer

Each of the 3 elements can go to 2 images
⇒ 2³ = 8 functions. A one-one function needs at least 3 distinct images, so there are 0 one-one functions.

Answer

2x₁ = 2x₂
⇒ x₁ = x₂, so f is one-one. Odd numbers such as 1 have no preimage in N (2x = 1 gives x = 1/2), so f is not onto.

Answer

No. f(0) = 3 and f(–2) = 4 – 4 + 3 = 3, so two different inputs give the same image. (It is also not onto, since f(x) = (x + 1)² + 2 ≥ 2.)