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Free PDF Download of CBSE Maths Multiple Choice Questions for Class 12 with Answers Chapter 1 Relations and Functions. Maths MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Maths Relations and Functions MCQs Pdf with Answers to know their preparation level.

1. Let R be a relation on the set L of lines defined by l_{1} R l_{2} if l_{1} is perpendicular to l_{2}, then relation R is

(a) reflexive and symmetric

(b) symmetric and transitive

(c) equivalence relation

(d) symmetric

Answer: d

Explaination: (d), not reflexive, as l_{1} R l_{2}

⇒ l_{1} ⊥ l_{1} Not true

Symmetric, true as l_{1} R l_{2} ⇒ l2R h

Transitive, false as l_{1} R l_{2}, l_{2} R l_{3}

⇒ l_{1} || l_{3} . l_{1} R l_{2}.

2. Given triangles with sides T_{1} : 3, 4, 5; T_{2} : 5, 12, 13; T_{3} : 6, 8, 10; T_{4} : 4, 7, 9 and a relation R in set of triangles defined as R = {(Δ_{1}, Δ_{2}) : Δ_{1} is similar to Δ_{2}}. Which triangles belong to the same equivalence class?

(a) T_{1} and T_{2}

(b) T_{2} and T_{3}

(c) T_{1} and T_{3}

(d) T_{1} and T_{4}

Answer: c

Explaination: (c), T_{1} and T_{3} are similar as their sides are proportional.

3. Given set A ={1, 2, 3} and a relation R = {(1, 2), (2, 1)}, the relation R will be

(a) reflexive if (1, 1) is added

(b) symmetric if (2, 3) is added

(c) transitive if (1, 1) is added

(d) symmetric if (3, 2) is added

Answer: c

Explaination: (c), here (1,2) e R, (2,1) € R, if transitive (1,1) should belong to R.

4. Given set A = {a, b, c). An identity relation in set A is

(a) R = {(a, b), (a, c)}

(b) R = {(a, a), (b, b), (c, c)}

(c) R = {(a, a), (b, b), (c, c), (a, c)}

(d) R= {(c, a), (b, a), (a, a)}

Answer: b

Explaination: (b), A relation R is an identity relation in set A if for all a ∈ A, (a, a) ∈ R.

5. A relation S in the set of real numbers is defined as xSy ⇒ x – y+ √3 is an irrational number, then relation S is

(a) reflexive

(b) reflexive and symmetric

(c) transitive

(d) symmetric and transitive

Answer: a

Explaination:

6. Set A has 3 elements and the set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is

(a) 144

(b) 12

(c) 24

(d) 64

Answer: c

Explaination: (c), total injective mappings/functions

= ^{4} P_{3} = 4! = 24.

7. Given a function lf as f(x) = 5x + 4, x ∈ R. If g : R → R is inverse of function ‘f then

(a) g(x) = 4x + 5

(b) g(x) = \(\frac{5}{4 x-5}\)

(c) g(x) = \(\frac{x-4}{5}\)

(d) g(x) = 5x – 4

Answer: c

Explaination:

8. Let Z be the set of integers and R be a relation defined in Z such that aRb if (a – b) is divisible by 5. Then R partitions the set Z into ______ pairwise disjoint subsets.

Answer:

Explaination: Five, as remainder can be 0, 1, 2, 3, 4.

9. Consider set A = {1, 2, 3 } and the relation R= {(1, 2)}, then? is a transitive relation. State true or false.

Answer:

Explaination: True, as there is no situation

(a, b) ∈ R, (b, c) ∈ R Hence, transitive. We can also say, a relation containing only one element is transitive.

10. Every relation which is symmetric and transitive is reflexive also. State true or false.

Answer:

Explaination: False,e.g.if R is arelationinset A = {2,3,4} defined as {(2, 3), (3, 2), (2, 2)} is symmetric and transitive but not reflexive.

11. Let R be a relation in set N, given by R = {(a, b): a = b – 2, b > 6} then (3, 8) ∈ R. State true or false with reason.

Answer:

Explaination: False, as in (3, 8), b = 8

⇒ a = 8 – 2

⇒ a = 6, but here a = 3.

12. Let R be a relation defined as R = {(x, x), (y, y), (z, z), (x, z)} in set A = {x, y, z} then R is (reflexive/symmetric) relation.

Answer:

Explaination: Reflexive, as for all a ∈ A, (a, a) ∈ R.

13. Let R be a relation in the set of natural numbers N defined by R = {(a, b) ∈ N × N: a < b}. Is relation R reflexive? Give a reason.

Answer:

Explaination:

Given R = {(a, b) ∈ N × N: a < b}.

Not reflexive, as for (a, a) × R

⇒ a< a, not true.

14. Let A be any non-empty set and P(A) be the power set of A. A relation R defined on P(A) by X R Y ⇔ X ∩ Y = X, X, Y ∈ P(A). Examine whether ? is symmetric.

Answer:

Explaination: X R Y ⇔ X ∩ Y = X ⇒ Y ∩ X = X ⇒ Y R X.

Hence, symmetric.

15. State the reason for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} not to be transitive. [NCERT; Delhi 2011]

Answer:

Explaination: (1, 2) ∈ R, (2, 1) ∈ R, but (1, 1) ∉ R.

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

Answer:

Explaination:

Given R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} defined on R: {1, 2, 3} → {1, 2, 3}

For reflexive: As (1, 1), (2,2), (3, 3) ∈ R. Hence, reflexive

For symmetric: (1, 2) ∈ R but (2, 1) ∉ R. Hence, not symmetric.

For transitive: (1, 2) ∈ R and (2, 3) ∈R but (1, 3) ∉ R. Hence, not transitive.

17. Let A = {3, 4, 5} and relation R on set A is defined as R = {(a, b) e A x A : a – b – 10). Is relation an empty relation?

Answer:

Explaination: We notice for no value of a, b s A, a-b = 10. Hence, (a, b) £ R for a, b e A. Hence, empty relation.

18. Given set A = {a, b} and relation R on A is defined as R = {(a, a), (b, b)}. Is relation an identity relation?

Answer:

Explaination: Yes, as (a, a) ∈ R, for all a ∈ A..

19. Let set A represents the set of all the girls of a particular class. Relation R on A is defined as R = {(a, b) ∈ A × A : difference between weights of a and b is less than 30 kg}. Show that relation R is a universal relation.

Answer:

Explaination: Let a, b ∈ A then a – b < 30 kg, always true for students of a particular class, i.e. aRb ∀ a, b ∈ A. Hence, universal relation.

20. If A = {1, 2, 3} and relation R = {(2, 3)} in A. Check whether relation R is reflexive, symmetric and transitive.

Answer:

Explaination:

Not reflexive, as (1, 1) ∉ R.

Not symmetric, as (2, 3) ∈ R but (3,2) ∉ R.

Transitive, as relation R in a non empty set containing one element is transitive.

21. State the reason for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} not to be transitive. [Delhi]

Answer:

Explaination: As (1, 2) ∈ R, (2, 1) ∈ R, but (1, 1) ∉ R.

22. Consider the set A containing n elements, then the total number of injective functions from set A onto itself is _____ .

Answer:

Explaination: Total number of injective functions from set containing n elements to a set containing n elements is ^{n} P_{n} = n!

23. The domain of the function *f* : R → R defined by f(x) = \(\sqrt{4-x^{2}}\) is ______ .

Answer:

Explaination:

[-2, 2]. For domain 4 – x² ≥ 0

⇒ 4 ≥ x²

⇒ x² ≤ 4

⇒ x² ≤ (2)²

⇒ -2 ≤ x ≤ 2, i.e. [-2, 2].

24. Let A = {a, b }. Then number of one-one functions from A to A possible are

(a) 2

(b) 4

(c) 1

(d) 3

Answer:

Explaination: (a), as if n(A) = m, then possible one-one functions from A to A are m!

25. Let A = {1, 2, 3, 4} and B = {a, b, c}. Then number of one-one functions from A to B are ______.

Answer:

Explaination: 0, as n(A) > n(B)

26. If n(A) = p, then number of bijective functions from set A to A are ______ ..

Answer:

Explaination: p!, as for bijective functions from A to B, n(A) = n(B) and function is one-one onto.

27. The function *f *: R → R defined as *f*(x) = [x], where [x] is greatest integer ≤ x, is onto function. State true or false.

Answer:

Explaination: False, as range of *f* is set of integers, i.e.

Z and range of f ⊆ co-domain R. Hence,not onto e.g. for \(\frac{1}{2}\) ∈ R (co-domain) there is no x ∈ R (domain) such that y = *f*(x) or \(\frac{1}{2}\) e∈ R has no pre-image.

28. If \(*f*(x)=\frac{x-1}{|x-1|}, x(\neq 1) \in R\) then range of ‘f’ is _______ .

Answer:

Explaination:

29. If *f* : R → R be defined by *f*(x) = (3 – x^{3})^{1/3}, then find *fof*(x). [NCERT]

Answer:

Explaination:

30. If f is an invertible function defined as *f*(x) = \(\frac{3x-4}{5}\), write *f*^{-1}(x).

Answer:

Explaination:

31. 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. State whether *f* is one-one or not. [AI 2011] *f*

Answer:

Explaination: One-one, as for x_{1} ≠ x_{2}

⇒ *f*(x_{1}) ≠ *f*(x_{2}).

32. Let *f* : R → R is defined by *f* (x) = | x |. Is function f onto? Give a reason. [HOTS]

Answer:

Explaination: *f* is not onto, as for some y ∈ R from co-domain, there is no x ∈ R from domain such that y = f(x), e.g. for -2 ∈ R (co-domain) there is no x ∈ R (domain) such that *f*(x) = -2, i.e. |x| = -2. Hence, not onto.

33. If f : R → R and g : R → R are given by f (x) = sin x and g(x) = 5x², find gof(x).

Answer:

Explaination: gof(x) = g(f(x)) = g(sin x) = 5 sin² x.

34. Let *f* : {1, 3,4} → {1,2, 5} and g: {1,2, 5} → {1, 3} be given by *f* = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof. [NCERT]

Answer:

Explaination:

go*f*: {1, 3, 4} → {1, 3}.

go*f*(1) = g(*f*(1)) = g(2) = 3.

go*f*(3) = g(K3)) = g(5) = 1

go*f*(4) = g(*f*(4)) = g(1) = 3

go*f* = {(1, 3), (3, 1), (4, 3)}.

35. Prove that *f* : R → R given by *f*(x) = x^{3} + 1 is one-one function.

Answer:

Explaination:

Given *f*(x) = x^{3} + 1

For x_{1} ≠ x_{2}

⇒ x_{1}^{3} ≠ x_{2}^{3}

⇒ x_{1}^{3} + 1 ≠ x_{2}^{3} + 1

⇒ *f*(x_{1}) ≠ *f*(x_{2}). Hence, one-one

36. Show that the Signum Function *f* : R → R,

one-one nor onto.

Answer:

Explaination:

Range of function is {-1, 0, 1} and co-domain is set of real numbers R.

⇒ Range ⊆ co-domain.

There is at least one element in R(codomain) which is not image of any element of the domain, e.g. for 2 e R(co-domain), there is no x in domain such that *f*(x) = 2, x ∈ R.

Hence, function is not onto.

Also, let x_{1} = 2 and x_{2} = 3 then *f*(x_{1}) = 1 and *f*(x_{2}) = 1

i.e., x_{1} ≠ x_{2} ⇒ *f*(x_{1}) = *f*(x_{2}).

So, function is not one-one.

37. Given *f*(x) = sin x check if function f is one-one for (i) (0, π) (ii) (-\(\frac{π}{2}\), \(\frac{π}{2}\)).

Answer:

Explaination:

38. If *f* : R → R is defined by *f*(x) = 3x + 2, define *f* (*f*(x)). [Foreign]

Answer:

Explaination: *f*(*f*(x)) = *f*(3x + 2) = 3(3x + 2) + 2

= 9x+ 8.

39. Write fog, if *f* : R → R and g : R → R are given by *f*(x) = |x| and *g*(x) = |5x – 2|. [Foreign]

Answer:

Explaination: (*fog*)(x) =*f*(*g*(x))

=*f*(|5x-2|) = ||5x-2||.

40. Write fog, if *f* : R → R and g : R → R are given by *f*(x) = 8x^{3} and g(x) = x^{1/3}. [Foreign]

Answer:

Explaination: (*fog*)(x) = *f*(*g*(x)) = *f*(x^{1/3}) = 8(x^{1/3})^{3} = 8x.

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