RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Maths · 11 questions · 26 marks
This chapter formalises how elements of sets relate to one another and builds towards the precise conditions under which a function can be reversed. Reflexive, symmetric and transitive properties combine to define equivalence relations, which quietly partition a set into disjoint classes. Injectivity and surjectivity then decide whether a function is invertible, a theme that resurfaces in inverse trigonometric functions.
Let R = {(a,b) : a,b ∈ Z, a − b is divisible by 5}. Then R is:
Answer
R is an equivalence relation. Reflexive: a − a = 0, divisible by 5, so (a,a)∈R for all a. Symmetric: if a − b is divisible by 5, then b − a = −(a−b) is also divisible by 5. Transitive: if a − b and b − c are both divisible by 5, then (a−b)+(b−c) = a−c is divisible by 5. All three properties hold, so R is an equivalence relation.
If f:R→R is defined by f(x) = 3x − 4, then f is invertible and f⁻¹(x) equals:
Answer
f⁻¹(x) = (x+4)/3. Let y = 3x − 4. Solving for x: 3x = y + 4, so x = (y+4)/3. Interchanging the roles of x and y gives f⁻¹(x) = (x+4)/3. Check: f(f⁻¹(x)) = 3·((x+4)/3) − 4 = x + 4 − 4 = x. Correct.
The number of one-one functions from A = {1,2,3} to B = {1,2,3,4} is:
Answer
The number of one-one functions is 24. With |A| = 3 and |B| = 4, the count of one-one functions is 4!/(4−3)! = 4!/1! = 24. Each element of A must map to a distinct element of B, giving 4 choices for the first, 3 for the second, 2 for the third: 4×3×2 = 24.
Let f:N→N be defined by f(n) = n+1 if n is odd, and f(n) = n−1 if n is even. Then f is:
Answer
f is bijective. f pairs up consecutive natural numbers: f(1)=2, f(2)=1, f(3)=4, f(4)=3, and so on, swapping each odd number with the next even number. Every natural number is hit exactly once as an image (onto), and distinct inputs always give distinct outputs (one-one). Since f is both one-one and onto, it is bijective.
Assertion (A): The relation R = {(1,1),(2,2),(3,3),(1,2),(2,1)} on the set {1,2,3} is an equivalence relation. Reason (R): Every relation that is both symmetric and reflexive is automatically an equivalence relation.
Answer
A is true but R is false. For A: R is reflexive since (1,1),(2,2),(3,3) are all present. It is symmetric since (1,2) and (2,1) both appear. For transitivity, the only non-trivial chain is (1,2) and (2,1) giving (1,1), which is present; no other chains arise. So R is indeed an equivalence relation, and A is true. But R (the reason) is false as a general statement: a symmetric and reflexive relation need not be transitive, so it need not be an equivalence relation in general — this particular R only happens to work out.
Show that the relation R on the set of real numbers defined by R = {(a,b) : a ≤ b²} is neither reflexive nor symmetric.
Answer
Not reflexive: take a = 1/2. Then a ≤ a² becomes 1/2 ≤ 1/4, which is false. So (1/2, 1/2) ∉ R, and R is not reflexive. Not symmetric: take a = 1, b = 2. Then a ≤ b² is 1 ≤ 4, true, so (1,2) ∈ R. But b ≤ a² is 2 ≤ 1, false, so (2,1) ∉ R. Since (1,2) ∈ R but (2,1) ∉ R, R is not symmetric.
Let A = {1,2,3,4} and R = {(1,1),(2,2),(3,3),(4,4),(1,2),(2,1)}. Show that R is an equivalence relation on A, and write the equivalence class of 1.
Answer
Reflexive: (1,1),(2,2),(3,3),(4,4) are all in R, so every element is related to itself. Symmetric: the only non-diagonal pairs are (1,2) and (2,1), and both are present together. Transitive: the only chain to check is (1,2) and (2,1) giving (1,1), which is present. No other combinations of non-diagonal pairs exist. So R is reflexive, symmetric and transitive, hence an equivalence relation. The equivalence class of 1 is the set of all elements related to 1: [1] = {1, 2}.
Let f:R→R be given by f(x) = x² + 1. Show that f is not one-one.
Answer
To show f is not one-one, it is enough to find two distinct inputs with the same output. Take x1 = 2 and x2 = −2. Then f(2) = 4 + 1 = 5 and f(−2) = 4 + 1 = 5. Since x1 ≠ x2 but f(x1) = f(x2), the function f is not one-one.
Show that f:R→R defined by f(x) = (4x+3)/(6x−4), x ≠ 2/3, is invertible, and find f⁻¹. Verify that f⁻¹ equals f itself.
Answer
To find the inverse, set y = (4x+3)/(6x−4) and solve for x in terms of y. y(6x − 4) = 4x + 3 6xy − 4y = 4x + 3 6xy − 4x = 4y + 3 x(6y − 4) = 4y + 3 x = (4y+3)/(6y−4). Since every y (other than 2/3) gives exactly one x, f is bijective on its domain and hence invertible. Interchanging x and y: f⁻¹(x) = (4x+3)/(6x−4), which is identical in form to f(x). So f⁻¹ = f, meaning f is an involution. Verification: f(f(x)) should equal x for any valid x; substituting f(x) back into the formula and simplifying algebraically returns x, confirming the result.
Define the relation R on the set of integers Z by aRb if and only if a ≡ b (mod 4), i.e. 4 divides (a−b). Prove that R is an equivalence relation and list the equivalence classes.
Answer
Reflexive: a − a = 0, and 4 divides 0, so aRa for every integer a. Symmetric: if 4 divides (a−b), then a−b = 4k for some integer k, so b−a = 4(−k), which means 4 divides (b−a). So aRb ⇒ bRa. Transitive: if aRb and bRc, then a−b = 4k1 and b−c = 4k2 for integers k1, k2. Adding: a−c = 4(k1+k2), so 4 divides (a−c), giving aRc. Since R is reflexive, symmetric and transitive, it is an equivalence relation. Every integer leaves remainder 0, 1, 2 or 3 on division by 4, so there are exactly four equivalence classes: [0] = {…,−8,−4,0,4,8,…}, [1] = {…,−7,−3,1,5,9,…}, [2] = {…,−6,−2,2,6,10,…}, [3] = {…,−5,−1,3,7,11,…}.
Read the following and answer the questions that follow: A class teacher records the month of birth of six students with roll numbers A = {1,2,3,4,5,6}. She defines a relation R on A by R = {(a,b) : a and b were born in the same month}. (a) Is R reflexive? Justify. (b) Is R symmetric? Justify. (c) Is R transitive? Justify. (d) Is R an equivalence relation? What real-world sets do its equivalence classes correspond to?
Answer
(a) Yes, R is reflexive. Every student is trivially born in the same month as themselves, so (a,a) ∈ R for all a ∈ A. (b) Yes, R is symmetric. If student a was born in the same month as student b, then student b was equally born in the same month as student a. (c) Yes, R is transitive. If a and b share a birth month, and b and c share a birth month, then a, b and c all share that same birth month, so a and c also share it. (d) Since R is reflexive, symmetric and transitive, it is an equivalence relation. Its equivalence classes correspond to groups of students who share the same birth month — each class is exactly the set of students born in one particular month.
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