CBSE Class 12th Math 1 – Relation and Functions MCQs

A function f : A → B is a bijection if it is ____.

Correct! Wrong!

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A function f: A → B is a bijection if it is one and onto

If f: R → R; f(x) = x2 and g : R → R, g(x) = sin 3x, f o g = ____ .

Correct! Wrong!

Given f(x) = x2; g(x) = sin 3x f o g = f(g(x)) = f(sin 3x) = (sin 3x)2 = s in2 3x

Let f: A → B be one to one function such that range of f is {b}. Then, the number of elements in A is _______.

Correct! Wrong!

Let f: A → B be one to one function such that range of f is {b}. Then, the number of elements in A is 1. Since f: A → B is one to one function having {b} as its range, then A should also have 1 element.

Let f : N − {1} → N be defined as f (n) = highest prime factor of n. Then f is ____.

Correct! Wrong!

Given, f (n) = highest prime factor of n. f (10) = 5 ; f (20) = 5 ; f (30) = 5 Therefore, f is many one. Range of f = Set of prime numbers (because f (n) is the largest prime) Therefore, range of f ≠ N. Therefore, f is not onto too.

Let R be the relation on (N × N) defined by (a, b) R (c, d) ⇒ ad = bc. Then R is ____.

Correct! Wrong!

Consider (a, b) R (a, b) Because, ad = bc, we get ab = ba ⇒ (a, b) R (a, b) Therefore, R is reflexive. (a, b) R (c, d) ⇒ ad = bc ⇒ da = cb ⇒ cb = da ⇒ (c, d) R (a, b) Therefore, R is symmetric. (a, b) R (c, d) and (c, d) R (e, f) ⇒ ad = bc and cf = de ⇒ adcf = bcde ⇒ (a, b) R (e, f)

Let R be a relation on the set of natural numbers such that R = {(p, q) ∈ N × N, p is divisible by q}. Then, ____.

Correct! Wrong!

Given R = {(p, q) ∈ N × N and p is divisibility by q} p R p for all p ∈ N, as every natural number is divisible by itself. Therefore, R is reflexive. p R q ⇏ q R p (for e.g. 8 is divisible by 4. but 4 is not divisible by 8) Therefore, R is not symmetric p R q, q R r ⇒ p R r (because p is divisible by q, and q is divisible by r) ⇒ p is divisible by r. Therefore, R is transitive.

Let A = {a, b}. Then, the number of one-one function from A to A is ___________.

Correct! Wrong!

the number of one-one function from A to A is 2. Let f : A → A be a one-one function. Then f(a), has 2 choices; f(a) = a or f(a) = b. If f(a) = a, then f(b) = b means f is one-one. If f(a) = b, then f(b) = a, again f is one-one. Thus, there are 2 one-one functions.

Let f: A → B be a real valued one-one and onto function such that f(a) = 2a - 3 : a ∈ A and set A = {2, 3, 4, 5}. Then, set B is __________________.

Correct! Wrong!

If f: A → B be a real valued one-one and onto function such that f(a) = 2a - 3 : a ∈ A and set A = {2, 3, 4, 5}, then set B is {1, 3, 5, 7}. Since f is a one-one function and onto set, B = {1, 3, 5, 7}, the same elements as in A.

Given f (x) = |x| and g (x) = |2x - 1|, g o f is ____.

Correct! Wrong!

g o f (x) = g (f (x)) = g (|x|) = |2|x| - 1| = |2x - 1| if x ≥ 0. = |-2x - 1| if x < 0.

The function f : N → N defined by f(x) = 7x is ______________.

Correct! Wrong!

Since the co domain is set of natural numbers, every number cannot be mapped. For example, we cannot find ax such that 7x = 2 (since x is a natural number). Therefore, 2, 3, 4, 5 … numbers are not connected Since every image is unique, it is a one-one function. But, it is an into function. The function f: N → N defined by f(x) = 7x is one-one and into.

Given A = {a, b, c} and R is {(a, a), (a, b), (a, c), (b, b), (c, c)}, then R is ____ .

Correct! Wrong!

∀a ∈ A, (a, a) ∈ R Therefore, R is reflexive. Let us consider (a, b) ∈ A, but (b, a) ∉ A. Therefore, R is not symmetric. Let us consider (a, b) ∈ A, (b, c) ∈ A ⇏a, c ∈ A Therefore, R is not transitive.

If f is a bijection from z to z, then it may be defined as __________.

Correct! Wrong!

Here, we know that modulus function is not one one and onto, for, f(-2) = |2(-2)| + 3 = 7and f(2) = |2(2)| +3 = 7 Hence, f is not one-one. Consider f(x) = x2 + x Here, f(3) = 32 + 3 = 12 and f(-3) = (-3)2+ x = 12 Hence, it is not one–one. Consider f(x) = x3 Here, in the co-domain which is a set of integers, we cannot find pre-images of 2, 3, 4, since they are not perfect cubes. Therefore, all three functions are not bijective.

A is a finite set. f: A → A is an onto function. f is ____.

Correct! Wrong!

Given A is finite. Let A = { a1, a2 ....., an }. Range of f = { f (a1), f (a2) ....., f (an)}, as f (a1), f (a2)....... f (an) are distinct elements of A (f is one one). A is finite with n elements. Therefore, f (a1), f (a2)..... f (an) are distinct. Therefore, f: A → A is one one.

Given R = {(a, b) ∈ N × N such that a < b}, then R is ____.

Correct! Wrong!

Given R = {(a, b) ∈ N × N, a < b} a R a (because a = a and not a < a) Therefore, R is not reflexive. a R b ⇏ b R a (because a < b ⇏ b < a) a R b and b R c ⇒ a < b, b < c ⇒ a < c ∀a, b, c ∈ N Therefore, R is transitive.

Let R be the relation on the set A = {2, 4, 6, 8, 10} given by R = {(a, b) ∈ A × A / a - b is an odd number}. Then, R is ____.

Correct! Wrong!

Consider 2 elements and find their difference. For e.g. 4 - 2 = 2; 6 - 4 = 2...... (i.e.) even - even = even Therefore, R has no element. Therefore, R is an empty relation.

Let A be the set of all real numbers. If R is the relation defined by R = {(a, b) / a ≤ b2}, then R is ____.

Correct! Wrong!

Given (a, b), a ≤ b2. Consider a and a2. E.g. 1/2 and (1/2)2 1/2 ≰ 1/2 ⇒ R is not reflexive. Given a ≤ b2 Consider 2 < 4 ⇏ 4 < 2 Therefore, R is not symmetric. Consider 4 and 16. 4 < 16 and 16 < 256 ⇒ 4 < 256 a R b and b R c ⇒ a R c R is transitive.

If f: R → R be defined as f (x) = sin x, then f is ____.

Correct! Wrong!

Given f (x) = sin x ⇒ f (0) = sin 0 and f (Π) = sin Π = 0 ⇒ f (0) = f (Π) As, 0 and Π have the same image, it is many one and not one one. Range f = [−1, 1] ⊂ R. Therefore, f is not onto too. Hence, f is neither one one nor onto.

Let R be the relation given by R = {a, b / |a - b|}. If R is even in the set of whole numbers (N), then ____.

Correct! Wrong!

Let us consider a ∈ W. Then, a - a = 0 ⇒ |a - a| = 0 is an even number. ⇒ (a, a)∈R Therefore, R is reflexive. Let us consider a, b ∈ W such that a R b. Then, |a - b| = even ⇒ |- (a - b)| is also even ⇒ |b - a| is even ⇒ b R a ⇒ R is symmetric. Let a, b, c ∈ N such that a R b, b R c. Then, (a, b) ∈ R and (b, c) ∈ R ⇒ |a - b| is even and |b - c| is also even. Adding both we get, a - b + b - c = a - c is even (∵ even- even = even) ⇒ (a, c)∈R ∴a R c ⇒R is transitive As R is reflexive, symmetric and transitive, R is an equivalent relation.

Let A be the set of all triangles in a plane with a relation R given by, R = {(A, B) / A= B}. Then R is ____.

Correct! Wrong!

Consider triangle A. A R A, since every triangle is congruent to itself. Therefore, R is reflexive. Consider triangles A and B. A R B ⇒ B R A, as A is congruent to B ⇒ B is congruent to A. Therefore, R is symmetric. Consider triangles A, B and C. A R B and B R C ⇒A is congruent to B and B is congruent to C ⇒ A is congruent to C ⇒ A R C. Therefore, R is transitive. Therefore, R is an equivalence relation.

Let L be the set of all lines in a plane with the relation R defined by R = {L1, L2/L1 ⊥ L2}. Then R is ____.

Correct! Wrong!

L1 is not perpendicular to itself. So R is not reflexive. L1 is perpendicular toL2 impliesL2 is perpendicular to L1. Therefore, R is symmetric. L1is perpendicular to L2 and L2 is perpendicular to L3 implies L1is parallel toL3 ⇒ L1is not perpendicular toL3. Therefore, R is not transitive. Hence, R is symmetric but neither reflexive nor transitive.

Let P be the set of all points in a plane. 'O' represents the origin and let R be the relation defined by R = {(A, B) / OA = OB}. Then R is ____.

Correct! Wrong!

Let A be a point on the plane. Then, OA = OA ⇒ A R A. Therefore, R is reflexive. Let A and B be two points in the plane such that, A R B ⇒ OA = OB ⇒ OB = OA ⇒ B R A. Therefore, R is symmetric. Let A, B, C be three points in the plane such that, A R B and B R C ⇒ OA = OB and OB = OC ⇒ OA = OC ⇒ A R C. Therefore, R is transitive. As R is reflexive, symmetric and transitive, R is an equivalence relation.

Given [x] denotes the greatest integer less than or equal to x, the greatest integer function f: R → R given by f (x) = [x] is ____.

Correct! Wrong!

[x] denotes the greatest integer equal to or less than x. ⇒ f (2.3) = [2.3] = 2 f (2.7) = [2.7] = 2 ⇒ two real numbers have the same image. Therefore, f is not one one. As there is no real number x such that [x] = 2.4, f is not onto. Therefore, f is neither one one nor onto.

If f(x) = x2 - 3x + 2 be a real valued function, then f(f(2)) is __________.

Correct! Wrong!

If f(x) = x2 - 3x + 2 be a real valued function, then f(f(2)) is 2. f(2) = 22 - 3(2) + 2 = 0. Then, f(f(2)) = 0 - 3(0) + 2 = 2

Let f: N - {1} → N be defined by f(n) = the highest prime factor of n. Then, the range of f is ____________.

Correct! Wrong!

The range of f is the set of all prime numbers. Here, f(2) = 2, f(3) = 3, f(4) = 2, f(6) = 3. The function is not one-one. The range is the set of all prime numbers, which is a subset of N.

P and Q are two non-empty sets. f: (P × Q) → (Q × P) : f (x, y) = (y, x) is ____.

Correct! Wrong!

Consider f(x, y), f (x1, y1) = f (x2, y2) ⇒ (y1, x1) = (y2, x2) ⇒ x1 = x2 and y1 = y2 ⇒ (x1, y1) = (x2, y2) Therefore, f is one one. Let (y, x) ∈ (Q × P) ⇒ y ∈ Q and x ∈ P ⇒(x, y) ← (P × Q) ⇒ for every (y, x) ∈ (Q × P), there exists a pre-image (x, y) in (P × Q) such that f (a, b) = (b, a). Therefore, f is onto. As f is one one and onto, f is bijective.

CBSE Class 12th Math 1 - Relation and Functions MCQs
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