NCERT Solutions›Class 12 Mathematics›Chapter 1

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Chapter 1 · Class 12 Mathematics

Relations and Functions

3 exercises38 questions solved

Exercise 1.1Reflexive, Symmetric and Transitive Relations

Determine whether each of the following relations are reflexive, symmetric and transitive: (i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0} (ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer} (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}

Solution

(i) A = {1, 2, ..., 14}, R = {(x,y) : 3x – y = 0} Listing R: {(1,3), (2,6), (3,9), (4,12)} Reflexive: (1,1) ∉ R since 3(1) – 1 = 2 ≠ 0. Not reflexive. Symmetric: (1,3) ∈ R but (3,1) ∉ R since 3(3) – 1 = 8 ≠ 0. Not symmetric. Transitive: (1,3) ∈ R and (3,9) ∈ R but (1,9) ∉ R since 3(1) – 9 = –6 ≠ 0. Not transitive. (ii) N, R = {(x,y) : y = x+5, x < 4} Listing R: {(1,6), (2,7), (3,8)} Reflexive: (1,1) ∉ R since 1 ≠ 1+5. Not reflexive. Symmetric: (1,6) ∈ R but (6,1) ∉ R since 6 ≠ 1+5. Not symmetric. Transitive: No pair (a,b),(b,c) both exist in R (6, 7, 8 appear only as second elements). Vacuously transitive. ✓ (iii) A = {1,2,3,4,5,6}, R = {(x,y) : y is divisible by x} Reflexive: x is always divisible by itself, so (x,x) ∈ R for all x ∈ A. Reflexive ✓ Symmetric: (1,2) ∈ R (2 divisible by 1) but (2,1) ∉ R (1 not divisible by 2). Not symmetric. Transitive: If x|y and y|z, then x|z. Transitive ✓ (iv) Z, R = {(x,y) : x–y is an integer} Reflexive: x–x = 0 ∈ Z for all x ∈ Z. Reflexive ✓ Symmetric: If x–y ∈ Z, then y–x = –(x–y) ∈ Z. Symmetric ✓ Transitive: If x–y ∈ Z and y–z ∈ Z, then x–z = (x–y)+(y–z) ∈ Z. Transitive ✓ ∴ R is an equivalence relation. (v)(a) x and y work at the same place: Reflexive: x works at the same place as x. ✓ Symmetric: If x works with y, then y works with x. ✓ Transitive: If x works with y and y works with z, all work at same place. ✓ ∴ Equivalence relation. (v)(b) x and y live in same locality: Reflexive ✓, Symmetric ✓, Transitive ✓ → Equivalence relation. (v)(c) x is exactly 7 cm taller than y: Not Reflexive: x cannot be 7 cm taller than itself. Not Symmetric: If x is 7 cm taller than y, then y is 7 cm shorter than x, not taller. Not Transitive: If x is 7 cm taller than y and y is 7 cm taller than z, then x is 14 cm taller than z — not 7 cm. (v)(d) x is wife of y: Not Reflexive: No one is their own wife. Not Symmetric: If x is wife of y, then y is husband (not wife) of x. Not Transitive: x is wife of y means x is female; y (male) cannot be wife of anyone. (v)(e) x is father of y: Not Reflexive: No one is their own father. Not Symmetric: If x is father of y, y is the child (not father) of x. Not Transitive: If x is father of y and y is father of z, then x is grandfather of z, not father.

Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b²} is neither reflexive nor symmetric nor transitive.

Solution

Not Reflexive: Take a = 1/2. We need (1/2) ≤ (1/2)² = 1/4. But 1/2 > 1/4. So (1/2, 1/2) ∉ R. Hence R is not reflexive. Not Symmetric: (1, 4) ∈ R since 1 ≤ 4² = 16. ✓ But (4, 1): Is 4 ≤ 1² = 1? No. So (4, 1) ∉ R. Hence R is not symmetric. Not Transitive: (3, 2) ∈ R since 3 ≤ 2² = 4. ✓ (2, 3/2) ∈ R since 2 ≤ (3/2)² = 9/4 = 2.25. ✓ But (3, 3/2): Is 3 ≤ (3/2)² = 2.25? No. So (3, 3/2) ∉ R. Hence R is not transitive.

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Solution

R = {(1,2), (2,3), (3,4), (4,5), (5,6)} Reflexive: (1,1) ∉ R since 1 ≠ 1+1. Not reflexive. Symmetric: (1,2) ∈ R but (2,1) ∉ R since 1 ≠ 2+1. Not symmetric. Transitive: (1,2) ∈ R and (2,3) ∈ R but (1,3) ∉ R since 3 ≠ 1+1. Not transitive. ∴ R is neither reflexive, nor symmetric, nor transitive.

Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.

Solution

Reflexive: For every a ∈ R, a ≤ a. So (a, a) ∈ R for all a. ✓ Transitive: Let (a,b) ∈ R and (b,c) ∈ R. Then a ≤ b and b ≤ c. So a ≤ c, meaning (a,c) ∈ R. ✓ Not Symmetric: (1, 2) ∈ R since 1 ≤ 2. But (2, 1) ∉ R since 2 > 1. ∴ R is reflexive and transitive but not symmetric.

Check whether the relation R in R defined by R = {(a, b) : a ≤ b³} is reflexive, symmetric or transitive.

Solution

Not Reflexive: Take a = 1/2. Need 1/2 ≤ (1/2)³ = 1/8. But 1/2 > 1/8. So (1/2, 1/2) ∉ R. Not reflexive. Not Symmetric: (1, 2) ∈ R since 1 ≤ 8 ✓. But (2, 1): Is 2 ≤ 1³ = 1? No. Not symmetric. Not Transitive: (3, 3/2) ∈ R since 3 ≤ (3/2)³ = 27/8 = 3.375 ✓ (3/2, 6/5) ∈ R since 1.5 ≤ (6/5)³ = 216/125 = 1.728 ✓ But (3, 6/5): Is 3 ≤ (6/5)³ = 1.728? No. Not transitive. ∴ R is neither reflexive, symmetric, nor transitive.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Solution

Not Reflexive: (1,1) ∉ R, (2,2) ∉ R, (3,3) ∉ R. So R is not reflexive. Symmetric: (1,2) ∈ R and (2,1) ∈ R. These are the only pairs and both directions are present. ✓ Not Transitive: (1,2) ∈ R and (2,1) ∈ R but (1,1) ∉ R. Not transitive. ∴ R is symmetric but neither reflexive nor transitive.

Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have the same number of pages} is an equivalence relation.

Solution

Reflexive: Every book x has the same number of pages as itself. So (x,x) ∈ R for all x. ✓ Symmetric: If x and y have the same number of pages, then y and x also have the same number of pages. So (x,y) ∈ R ⟹ (y,x) ∈ R. ✓ Transitive: If x has the same pages as y, and y has the same pages as z, then x has the same pages as z. So (x,y) ∈ R and (y,z) ∈ R ⟹ (x,z) ∈ R. ✓ Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

Solution

Reflexive: |a – a| = 0, which is even. So (a,a) ∈ R for all a ∈ A. ✓ Symmetric: If |a–b| is even, then |b–a| = |a–b| is also even. So (a,b) ∈ R ⟹ (b,a) ∈ R. ✓ Transitive: If |a–b| is even and |b–c| is even, then (a–b) and (b–c) are both even. So a–c = (a–b) + (b–c) is the sum of two even integers = even. Hence |a–c| is even. ✓ ∴ R is an equivalence relation. Elements of {1,3,5}: |1–3| = 2 (even) ✓, |3–5| = 2 (even) ✓, |1–5| = 4 (even) ✓ So all elements of {1,3,5} are related to each other. Elements of {2,4}: |2–4| = 2 (even) ✓. So 2 and 4 are related. Cross check: |1–2| = 1 (odd) ✗. No element of {1,3,5} is related to any element of {2,4}.

Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by (i) R = {(a, b) : |a – b| is a multiple of 4} (ii) R = {(a, b) : a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.

Solution

(i) R = {(a,b) : |a–b| is a multiple of 4}, A = {0,1,2,...,12} Reflexive: |a–a| = 0 = 4×0. Reflexive ✓ Symmetric: If 4|( a–b), then 4|(b–a). Symmetric ✓ Transitive: If 4|(a–b) and 4|(b–c), then a–c = (a–b)+(b–c) is divisible by 4. Transitive ✓ ∴ R is an equivalence relation. Elements related to 1: {a ∈ A : |a–1| is multiple of 4} = {1, 5, 9} (ii) R = {(a,b) : a = b} Reflexive: a = a ✓; Symmetric: a=b ⟹ b=a ✓; Transitive: a=b, b=c ⟹ a=c ✓ ∴ R is an equivalence relation (identity relation). Elements related to 1: {a ∈ A : a = 1} = {1}

Q10

Give an example of a relation which is: (i) Symmetric but neither reflexive nor transitive (ii) Transitive but neither reflexive nor symmetric (iii) Reflexive and symmetric but not transitive (iv) Reflexive and transitive but not symmetric (v) Symmetric and transitive but not reflexive

Solution

Let A = {1, 2, 3} (i) Symmetric but not reflexive/transitive: R = {(1,2), (2,1)} Symmetric ✓; Not reflexive: (1,1) ∉ R; Not transitive: (1,2),(2,1) ∈ R but (1,1) ∉ R. (ii) Transitive but not reflexive/symmetric: R = {(1,2), (2,3), (1,3)} Transitive ✓ (check all pairs); Not reflexive: (1,1) ∉ R; Not symmetric: (1,2) ∈ R but (2,1) ∉ R. (iii) Reflexive and symmetric but not transitive: R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)} Reflexive ✓; Symmetric ✓; Not transitive: (1,2),(2,3) ∈ R but (1,3) ∉ R. (iv) Reflexive and transitive but not symmetric: R = {(a,b) : a ≤ b} in {1,2,3} Reflexive ✓; Transitive ✓; Not symmetric: (1,2) ∈ R but (2,1) ∉ R. (v) Symmetric and transitive but not reflexive: R = {(1,1),(2,2),(1,2),(2,1)} on A = {1,2,3} Symmetric ✓; Transitive ✓; Not reflexive: (3,3) ∉ R.

Q11

Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Solution

Let O = (0,0) be the origin, and d(P) = distance of P from O. Reflexive: d(P) = d(P) for every point P. So (P,P) ∈ R. ✓ Symmetric: If d(P) = d(Q), then d(Q) = d(P). So (P,Q) ∈ R ⟹ (Q,P) ∈ R. ✓ Transitive: If d(P) = d(Q) and d(Q) = d(R), then d(P) = d(R). So (P,Q),(Q,R) ∈ R ⟹ (P,R) ∈ R. ✓ ∴ R is an equivalence relation. Set of all points related to P ≠ (0,0): {Q : d(Q,O) = d(P,O)} = circle centred at origin with radius d(P,O). This is the circle passing through P with centre at origin.

Q12

Show that the relation R defined in the set A of all triangles as R = {(T₁, T₂) : T₁ is similar to T₂}, is equivalence relation. Consider three right angle triangles T₁ with sides 3, 4, 5; T₂ with sides 5, 12, 13; and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂ and T₃ are related?

Solution

Reflexive: Every triangle is similar to itself (AAA with same angles). ✓ Symmetric: If T₁ ~ T₂, then T₂ ~ T₁ (similarity is mutual). ✓ Transitive: If T₁ ~ T₂ and T₂ ~ T₃, then T₁ ~ T₃ (corresponding angles are equal). ✓ ∴ R is an equivalence relation. Checking which triangles are related: T₁ has sides 3:4:5. T₃ has sides 6:8:10 = 3:4:5 (ratio 2:1). Since corresponding sides are proportional, T₁ ~ T₃. So T₁ and T₃ are related. T₂ has sides 5:12:13, which is a different ratio. T₂ is not similar to T₁ or T₃. ∴ T₁ and T₃ are related to each other.

Q13

Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂) : P₁ and P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Solution

Reflexive: Every polygon has the same number of sides as itself. ✓ Symmetric: If P₁ and P₂ have the same sides, then P₂ and P₁ also do. ✓ Transitive: If P₁ and P₂ have same sides, and P₂ and P₃ have same sides, then P₁ and P₃ have same sides. ✓ ∴ R is an equivalence relation. Triangle T has 3 sides. Set of all elements related to T = {P ∈ A : P has 3 sides} = set of all triangles in A.

Q14

Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L₁, L₂) : L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Solution

Reflexive: Every line is parallel to itself. So (L,L) ∈ R for all L ∈ L. ✓ Symmetric: If L₁ ∥ L₂, then L₂ ∥ L₁. So (L₁,L₂) ∈ R ⟹ (L₂,L₁) ∈ R. ✓ Transitive: If L₁ ∥ L₂ and L₂ ∥ L₃, then L₁ ∥ L₃. ✓ ∴ R is an equivalence relation. The line y = 2x + 4 has slope 2. All lines parallel to it also have slope 2. Set of all lines related to y = 2x + 4: {y = 2x + c : c ∈ R}, i.e., all lines with slope 2.

Q15

Let R be the relation in the set {1, 2, 3, 4} given by R = {(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)}. Choose the correct answer. (A) R is reflexive and symmetric but not transitive. (B) R is reflexive and transitive but not symmetric. (C) R is symmetric and transitive but not reflexive. (D) R is an equivalence relation.

Solution

Check Reflexive: (1,1) ✓, (2,2) ✓, (3,3) ✓, (4,4) ✓. All elements have (a,a) ∈ R. Reflexive ✓ Check Symmetric: (1,2) ∈ R but (2,1) ∉ R. Not symmetric. Check Transitive: (1,2),(2,2) → need (1,2) ∈ R ✓ (1,3),(3,3) → need (1,3) ∈ R ✓ (1,3),(3,2) → need (1,2) ∈ R ✓ (3,2),(2,2) → need (3,2) ∈ R ✓ All cases satisfied. Transitive ✓ Answer: (B) R is reflexive and transitive but not symmetric.

Q16

Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer. (A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R

Solution

For (a,b) ∈ R, we need a = b–2 and b > 6. (A) (2,4): b = 4, but 4 > 6 is false. (2,4) ∉ R. (B) (3,8): b = 8 > 6 ✓. But a = b–2 = 8–2 = 6 ≠ 3. (3,8) ∉ R. (C) (6,8): b = 8 > 6 ✓. a = b–2 = 8–2 = 6 ✓. (6,8) ∈ R. ✓ (D) (8,7): b = 7 > 6 ✓. a = b–2 = 7–2 = 5 ≠ 8. (8,7) ∉ R. Answer: (C)

Exercise 1.2Types of Functions — One-One, Onto, Bijective

Show that the function f : R* → R* defined by f(x) = 1/x is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true if the domain R* is replaced by N with co-domain being same as R*?

Solution

f : R* → R*, f(x) = 1/x One-one: Suppose f(x₁) = f(x₂). Then 1/x₁ = 1/x₂ ⟹ x₁ = x₂. So f is one-one. ✓ Onto: For any y ∈ R* (y ≠ 0), take x = 1/y ∈ R*. Then f(x) = f(1/y) = 1/(1/y) = y. So every y has a pre-image. f is onto. ✓ ∴ f is bijective on R*. If domain is N, co-domain R*: Range of f = {1/n : n ∈ N} = {1, 1/2, 1/3, ...} ⊂ R* but ≠ R*. For example, 2 ∈ R* has no pre-image in N (since f(1/2) = 2 but 1/2 ∉ N). So f is one-one but not onto. Result is NOT true for N → R*.

Check the injectivity and surjectivity of the following functions: (i) f : N → N given by f(x) = x² (ii) f : Z → Z given by f(x) = x² (iii) f : R → R given by f(x) = x² (iv) f : N → N given by f(x) = x³ (v) f : Z → Z given by f(x) = x³

Solution

(i) f : N → N, f(x) = x² Injective: f(x₁)=f(x₂) ⟹ x₁²=x₂² ⟹ x₁=x₂ (both positive in N). Injective ✓ Not surjective: 2 ∈ N has no pre-image (no natural number n with n²=2). (ii) f : Z → Z, f(x) = x² Not injective: f(1)=f(–1)=1 but 1≠–1. Not surjective: –1 ∈ Z has no pre-image (x²≥0 for all x). (iii) f : R → R, f(x) = x² Not injective: f(1)=f(–1)=1. Not surjective: –1 ∈ R has no pre-image. (iv) f : N → N, f(x) = x³ Injective: x₁³=x₂³ ⟹ x₁=x₂. Injective ✓ Not surjective: 2 ∈ N has no pre-image (no n ∈ N with n³=2). (v) f : Z → Z, f(x) = x³ Injective: x₁³=x₂³ ⟹ x₁=x₂. Injective ✓ Not surjective: 2 ∈ Z has no integer pre-image.

Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Solution

Not one-one: f(1.2) = [1.2] = 1 and f(1.8) = [1.8] = 1, but 1.2 ≠ 1.8. So two different elements have the same image. Not one-one. Not onto: The range of f is Z (only integers). For example, 0.5 ∈ R has no pre-image x with [x] = 0.5 since [x] is always an integer. ∴ f is neither one-one nor onto.

Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither one-one nor onto.

Solution

Not one-one: f(1) = |1| = 1 and f(–1) = |–1| = 1, but 1 ≠ –1. Two different inputs give the same output. Not one-one. Not onto: f(x) = |x| ≥ 0 for all x ∈ R. So the range is [0, ∞) ≠ R. For example, –1 ∈ R has no pre-image. Not onto.

Show that the Signum Function f : R → R, given by f(x) = 1 if x > 0; 0 if x = 0; –1 if x < 0, is neither one-one nor onto.

Solution

Not one-one: f(2) = 1 and f(3) = 1, but 2 ≠ 3. Two different inputs map to the same value. Not one-one. Not onto: Range of f = {–1, 0, 1} ≠ R. For example, 2 ∈ R has no pre-image x with f(x) = 2. Not onto.

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Solution

The function f maps: f(1)=4, f(2)=5, f(3)=6. All three images (4, 5, 6) are distinct. f(x₁) = f(x₂) ⟹ the image is the same ⟹ x₁ = x₂ (since no two distinct elements share an image). ∴ f is one-one.

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) f : R → R defined by f(x) = 3 – 4x (ii) f : R → R defined by f(x) = 1 + x²

Solution

(i) f(x) = 3 – 4x One-one: f(x₁) = f(x₂) ⟹ 3–4x₁ = 3–4x₂ ⟹ x₁ = x₂. One-one ✓ Onto: For any y ∈ R, take x = (3–y)/4 ∈ R. Then f(x) = 3–4·(3–y)/4 = 3–(3–y) = y. Onto ✓ ∴ f is bijective. (ii) f(x) = 1 + x² Not one-one: f(1) = 1+1 = 2 and f(–1) = 1+1 = 2, but 1 ≠ –1. Not onto: f(x) = 1+x² ≥ 1 for all x ∈ R. Range = [1,∞) ≠ R. For example, 0 ∈ R has no pre-image. ∴ f is neither one-one nor onto.

Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.

Solution

One-one: Suppose f(a₁, b₁) = f(a₂, b₂). Then (b₁, a₁) = (b₂, a₂) ⟹ b₁=b₂ and a₁=a₂ ⟹ (a₁,b₁) = (a₂,b₂). So f is one-one. ✓ Onto: For any (b, a) ∈ B × A, there exists (a, b) ∈ A × B such that f(a,b) = (b,a). So f is onto. ✓ ∴ f is bijective.

Let f : N → N be defined by f(n) = (n+1)/2 if n is odd; n/2 if n is even, for all n ∈ N. State whether the function f is bijective. Justify your answer.

Solution

Not injective: f(1) = (1+1)/2 = 1 (1 is odd) f(2) = 2/2 = 1 (2 is even) f(1) = f(2) = 1 but 1 ≠ 2. So f is not one-one. Surjective: For any k ∈ N, we can find n = 2k–1 (odd): f(2k–1) = (2k–1+1)/2 = k. So every k ∈ N has a pre-image. f is onto. ✓ ∴ f is surjective but not injective. f is NOT bijective.

Q10

Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x–2)/(x–3). Is f one-one and onto? Justify your answer.

Solution

One-one: Let f(x₁) = f(x₂). (x₁–2)/(x₁–3) = (x₂–2)/(x₂–3) (x₁–2)(x₂–3) = (x₂–2)(x₁–3) x₁x₂ – 3x₁ – 2x₂ + 6 = x₁x₂ – 3x₂ – 2x₁ + 6 –3x₁ – 2x₂ = –3x₂ – 2x₁ –x₁ = –x₂ ⟹ x₁ = x₂. One-one ✓ Onto: Let y ∈ B = R–{1}. Solve y = (x–2)/(x–3): y(x–3) = x–2 ⟹ xy–3y = x–2 ⟹ x(y–1) = 3y–2 ⟹ x = (3y–2)/(y–1) Since y ≠ 1, x is defined. Check x ≠ 3: 3y–2 ≠ 3(y–1) ⟹ 3y–2 ≠ 3y–3 ⟹ –2 ≠ –3 ✓ So x ∈ A for every y ∈ B. Onto ✓ ∴ f is bijective.

Q11

Let f : R → R be defined as f(x) = x⁴. Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto

Solution

Not one-one: f(1) = 1⁴ = 1 and f(–1) = (–1)⁴ = 1. Two inputs give same output. Not onto: f(x) = x⁴ ≥ 0 for all x ∈ R. Range = [0,∞) ≠ R. For example, –1 ∈ R has no pre-image. Answer: (D) f is neither one-one nor onto.

Q12

Let f : R → R be defined as f(x) = 3x. Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto

Solution

One-one: f(x₁) = f(x₂) ⟹ 3x₁ = 3x₂ ⟹ x₁ = x₂. One-one ✓ Onto: For any y ∈ R, take x = y/3 ∈ R. Then f(y/3) = 3·(y/3) = y. Onto ✓ Answer: (A) f is one-one onto (bijective).

Exercise MiscMiscellaneous Exercise

Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = f o g = I_R.

Solution

We need g such that g(f(x)) = x for all x ∈ R. f(x) = 10x + 7. Let y = f(x) = 10x+7. Solving for x: x = (y–7)/10. So g(y) = (y–7)/10, i.e., g(x) = (x–7)/10. Verify: g(f(x)) = g(10x+7) = (10x+7–7)/10 = 10x/10 = x ✓ f(g(x)) = f((x–7)/10) = 10·(x–7)/10 + 7 = (x–7)+7 = x ✓ ∴ g(x) = (x – 7)/10

Let f : W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Solution

One-one: Case 1: Both n₁, n₂ odd. f(n₁)=f(n₂) ⟹ n₁–1=n₂–1 ⟹ n₁=n₂. Case 2: Both n₁, n₂ even. f(n₁)=f(n₂) ⟹ n₁+1=n₂+1 ⟹ n₁=n₂. Case 3: n₁ odd, n₂ even. f(n₁)=n₁–1 (even), f(n₂)=n₂+1 (odd). They cannot be equal. ∴ f is one-one. Onto: For any m ∈ W: if m is even, n = m+1 (odd) gives f(n)=m+1–1=m ✓ If m is odd, n = m–1 (even) gives f(n)=m–1+1=m ✓ ∴ f is onto. Since f is bijective, f is invertible. Inverse: f⁻¹(m) = m+1 if m is even; m–1 if m is odd. Note: f⁻¹ = f (the function is its own inverse).

If f : R → R is defined by f(x) = x² – 3x + 2, find f(f(x)).

Solution

f(x) = x² – 3x + 2 f(f(x)) = f(x² – 3x + 2) Let t = x² – 3x + 2. f(t) = t² – 3t + 2 = (x²–3x+2)² – 3(x²–3x+2) + 2 Expand (x²–3x+2)²: = x⁴ – 6x³ + 13x² – 12x + 4 –3(x²–3x+2) = –3x² + 9x – 6 Combining: f(f(x)) = x⁴ – 6x³ + 13x² – 12x + 4 – 3x² + 9x – 6 + 2 = x⁴ – 6x³ + 10x² – 3x

Show that function f : R → {x ∈ R : –1 < x < 1} defined by f(x) = x / (1 + |x|), x ∈ R is one one and onto function.

Solution

One-one: Case 1: x₁, x₂ ≥ 0. f(x₁)=x₁/(1+x₁), f(x₂)=x₂/(1+x₂). f(x₁)=f(x₂) ⟹ x₁(1+x₂)=x₂(1+x₁) ⟹ x₁=x₂. ✓ Case 2: x₁, x₂ < 0. f(x)=x/(1–x). Similarly x₁=x₂. ✓ Case 3: x₁≥0, x₂<0. f(x₁)≥0, f(x₂)<0. Cannot be equal. ✓ ∴ f is one-one. Onto: Let y ∈ (–1,1). If y ≥ 0: x = y/(1–y) ≥ 0 and f(x) = [y/(1–y)] / [1 + y/(1–y)] = y ✓ If y < 0: x = y/(1+y) < 0 and f(x) = y ✓ ∴ f is onto. ∴ f is bijective.

Show that the function f : R → R given by f(x) = x³ is injective.

Solution

Let f(x₁) = f(x₂). x₁³ = x₂³ x₁³ – x₂³ = 0 (x₁ – x₂)(x₁² + x₁x₂ + x₂²) = 0 Now x₁² + x₁x₂ + x₂² = (x₁ + x₂/2)² + 3x₂²/4 ≥ 0, and equals 0 only when x₁=x₂=0. In general: x³ is strictly increasing, so x₁³=x₂³ ⟹ x₁=x₂. ∴ f(x) = x³ is injective.

Give examples of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective. [Hint: Consider f(x) = x and g(x) = |x|]

Solution

Let f : N → Z defined by f(x) = x (inclusion map). Let g : Z → Z defined by g(x) = |x|. g is not injective: g(1) = g(–1) = 1 but 1 ≠ –1. g o f : N → Z defined by (g o f)(x) = g(f(x)) = |x| = x (since x ∈ N, x > 0). So (g o f)(x) = x, which is the identity on N. (g o f)(x₁) = (g o f)(x₂) ⟹ x₁ = x₂. Injective ✓ ∴ g o f is injective but g is not injective.

Give examples of two functions f : N → N and g : N → N such that g o f is onto but f is not onto. [Hint: Consider f(x) = x + 1 and g(x) = x–1 if x > 1; 1 if x = 1]

Solution

Let f : N → N, f(x) = x + 1. Let g : N → N, g(x) = x–1 if x > 1; 1 if x = 1. f is not onto: 1 ∈ N has no pre-image (f(x)=x+1 ≥ 2 for all x ∈ N). (g o f)(x) = g(f(x)) = g(x+1) = (x+1)–1 = x (since x+1 ≥ 2 > 1). So g o f = identity on N, which is onto. ✓ ∴ g o f is onto but f is not onto.

Given a non-empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.

Solution

R is defined as ARB iff A ⊆ B. Reflexive: A ⊆ A for every A ∈ P(X). So ARA. Reflexive ✓ Symmetric: Let A ⊆ B. Does B ⊆ A follow? Not necessarily. Example: Let X = {1,2}. A = {1}, B = {1,2}. A ⊆ B but B ⊄ A. Not symmetric. ∴ R is NOT an equivalence relation (fails symmetry).

Find the number of all onto functions from the set {1, 2, 3, ..., n} to itself.

Solution

An onto function from a set of n elements to itself is a bijection (since domain and co-domain have the same size). The number of bijections from {1,2,...,n} to itself = number of permutations of n elements. ∴ Number of onto functions = n!

Q10

Let S = {a, b, c} and T = {1, 2, 3}. Find F⁻¹ of the following functions F from S to T, if it exists. (i) F = {(a, 3), (b, 2), (c, 1)} (ii) F = {(a, 2), (b, 1), (c, 1)}

Solution

(i) F = {(a,3),(b,2),(c,1)} F maps a→3, b→2, c→1. All images are distinct → F is one-one. Range = {1,2,3} = T → F is onto. So F is bijective and F⁻¹ exists. F⁻¹ = {(3,a),(2,b),(1,c)} (ii) F = {(a,2),(b,1),(c,1)} F(b) = F(c) = 1 but b ≠ c → F is not one-one. Since F is not injective, it is not bijective and F⁻¹ does not exist.

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