CBSE Class–12 Mathematics

NCERT Exemplar
Chapter - 1
Relations and Functions - Short Answer Questions

1. Let A = {0, 1, 2, 3} and define a relation R on A as follows:

R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}.

Is R reflexive? symmetric? transitive?

Sol. R is reflexive and symmetric, but not transitive since for (1, 0) ∈ R and (0, 3) ∈ R whereas (1, 3) ∉ R.

2. For the set A = {1, 2, 3}, define a relation R in the set A as follows:

R = {(1, 1), (2, 2), (3, 3), (1, 3)}.

Write the ordered pairs to be added to R to make it the smallest equivalence relation.

Sol. (3, 1) is the single ordered pair which needs to be added to R to make it the smallest equivalence relation.

3. Let R be the equivalence relation in the set Z of integers given by R = {(a, b) : 2 divides a– b}. Write the equivalence class [0].

Sol. [0] = {0, ± 2, ± 4, ± 6, ...}

4. Let the function R → R be defined by Then, show that is one-one.

Sol. For any two elements ∈ R such that , we have

Hence is one-one.

5. If = {(5, 2), (6, 3)}, = {(2, 5), (3, 6)}, write o.

Sol. o = {(2, 2), (3, 3)}

6. Let R → R be the function defined by Then write

Sol. Given that then

Hence

7. Is the binary operation * defined on Z (set of integer) by commutative?

Sol. No. Since for while so that 1*2 ≠2 *1.

8. If = {(5, 2), (6, 3)} and = {(2, 5), (3, 6)}, write the range of and .

Sol. The range of = {2, 3} and the range of = {5, 6}.

9. If A = {1, 2, 3} and , are relations corresponding to the subset of indicated against them, which of , is a function? Why?

= {(1, 3), (2, 3), (3, 2)}

= {(1, 2), (1, 3), (3, 1)}

Sol. is a function since each element of A in the first place in the ordered pairs is related to only one element of A in the second place while g is not a function because 1 is related to more than one element of A, namely, 2 and 3.

10. If A = {a, b, c, d} and = {a, b), (b, d), (c, a), (d, c)}, show that is one-one from A onto A. Find

Sol. is one-one since each element of A is assigned to distinct element of the set A. Also, is onto since (A) = A. Moreover, = {(b, a), (d, b), (a, c), (c, d)}.

11. In the set N of natural numbers, define the binary operation *by . Is the operation * commutative and associative?

Sol. The operation is clearly commutative since

m*n= g.c.d (m, n) = g.c.d(n, m) = n * m m, n N.

It is also associative because for l, m, n N, we have

l *(m*n) = g. c. d (l, g.c.d (m, n))

= g.c.d. (g. c. d(l, m), n)

= (l*m) *n.

Long Answer Questions

12. In the set of natural numbers N, define a relation R as follows:

n, m N, nRm if on division by 5 each of the integers n and m leaves the remainder less than 5, i.e. one of the numbers 0, 1, 2, 3 and 4. Show that R is equivalence relation. Also, obtain the pairwise disjoint subsets determined by R.

Sol. R is reflexive since for each a N, aRa. R is symmetric since if aRb, then bRa for a, b N. Also, R is transitive since for a, b, c N, if aRb and bRc, then aRc.

Hence R is an equivalence relation in N which will partition the set N into the pairwise disjoint subsets. The equivalent classes are as mentioned below:

A0= {5, 10, 15, 20 ...}

A1= {1, 6, 11, 16, 21 ...}

A2= {2, 7, 12, 17, 22, ...}

A3= {3, 8, 13, 18, 23, ...}

A4= {4, 9, 14, 19, 24, ...}

It is evident that the above five sets are pairwise disjoint and

13. Show that the function R → R defined by , is neither one-one nor onto.

Sol. For x1, x2R, consider

(x1) = (x2)

⇒

We note that there are point, with and for instance, if we take and then we have and but . Hence is not one-one. Also, is not onto for if so then for such that (x) = 1 which gives . But there is no such x in the domain R, since the equation does not give any real value of x.

14. Let , : R → R be two functions defined as and x R. Then, find and .

Sol. Here which can be redefined as

Similarly, the function defined by may be redefined as

Therefore, gets defined as:

For x ≥0, ( ) (x) = ( (x) = (2x) = 0

and for x <0, ( ) (x) = ( (x) = (0) = 0.

Consequently, we have ( ) (x) = 0, x R.

Similarly, gets defined as:

For x ≥0, ( ) (x) = ( (x) = (0) = 0,

and for x <0, ( ) (x) = ( (x)) = (–2 x) = – 4x.

i.e.

15. Let R be the set of real numbers and be the function defined by (x) = 4x+ 5. Show that is invertible and find .

Sol. Here the function is defined as (say). Then

This leads to a function defined as

Therefore, =

Similarly, ( ) (y) = ( (y))

=

or .

Hence is invertible and which is given by

16. Let * be a binary operation defined on Q. Find which of the following binary operations are associative.

(i) a *b= a– b for a, b Q.

(ii)

(iii) a *b= a– b+ ab for a, b Q.

(iv) for a, b Q.

Sol. (i) * is not associative for if we take a= 1, b= 2 and c= 3, then

(a*b) *c= (1 *2) * 3 = (1 – 2) * 3 = – 1 – 3 = – 4 and
a *(b*c) = 1 *(2 *3) = 1 *(2 – 3) = 1 – (– 1) = 2.
Thus (a*b) *c ≠ a*(b*c) and hence * is not associative.
(ii) * is associative since Q is associative with respect to multiplication.
(iii) * is not associative for if we take a= 2, b= 3 and c = 4, then
(a*b) *c= (2 *3) * 4 = (2 – 3 + 6) *4 = 5*4 = 5 – 4 + 20 = 21, and
a*(b*c) = 2 *(3*4) = 2*(3 – 4 + 12) = 2 *11 = 2 – 11 + 22 = 13
Thus (a*b) * c ≠ a*(b*c) and hence * is not associative.
(iv) * is not associative for if we take a= 1, b= 2 and c= 3, then (a*b) * c=(1 * 2) * 3 = 4 * 3 = 4 × 9 = 36 and a*(b*c) = 1 * (2 * 3) = 1*18 =
Thus (a * b) * c ≠ a * (b * c) and hence*is not associative.

Objective Questions

Choose the correct answer from the given four options in each of the Examples 17 to 25.

17. Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is

(A) Reflexive and symmetric
(B) Transitive and symmetric
(C) Equivalence
(D) Reflexive, transitive but not symmetric

Sol. The correct choice is (D).
Since n divides n, n N, R is reflexive. R is not symmetric since for 3, 6 N, 3 R 6 ≠6 R 3. R is transitive since for n, m, r whenever n/m and m/r ⇒ n/r, i.e., n divides m and m divides r, then n will divide r.

18. Let L denote the set of all straight lines in a plane. Let a relation R be defined by lRm if and only if l is perpendicular to m l, m L. Then R is

(A) reflexive
(B) symmetric
(C) transitive
(D) none of these

Sol. The correct choice is (B).

19. Let N be the set of natural numbers and the function : N → N be defined by (n) = 2n+ 3 n N. Then is

(A) surjective
(B) injective
(C) bijective
(D) none of these

Sol. (B) is the correct option.

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

(A) 144
(B) 12
(C) 24
(D) 64

Sol. The correct choice is (C). The total number of injective mappings from the set containing 3 elements into the set containing 4 elements is

21. Let : R → R be defined by (x) = sin x and : R → R be defined by, then is

(A)

(B) (sin x)2

(C)

(D)

Sol. (C) is the correct choice.

22. Let : R → R be defined by. Then –1(x) is given by

(A)

(B)

(C)

(D) None of these

Sol. (A) is the correct choice.

23. Let : R → R be defined by (x) = x2 + 1. Then, pre-images of 17 and – 3, respectively, are

(A)

(B)

(C)

(D) {4, – 4, {2, – 2}

Sol. (C) is the correct choice since for or ⇒ x= ± 4 or f –1( 17) = {4, – 4} and for and hence.

24. For real numbers x and y, define xRy if and only if is an irrational number. Then the relation R is

(A) reflexive

(B) symmetric

(D) none of these

Sol. (A) is the correct choice.

Fill in the blanks in each of the Examples 25 to 30.

25. Consider the set A = {1, 2, 3} and R be the smallest equivalence relation on A, then R = ________

Sol. R = {(1, 1), (2, 2), (3, 3)}.

26. The domain of the function : R → R defined by is ________.

Sol. Here x2– 3x+ 2 ≥0

⇒ (x– 1) (x– 2) ≥0

⇒ x≤1 or x ≥2

Hence the domain of

27. Consider the set A containing n elements. Then, the total number of injective functions from A onto itself is ________.

Sol. n!

28. Let Z be the set of integers and R be the relation defined in Z such that aRb if a– b is divisible by 3. Then R partitions the set Z into ________ pairwise disjoint subsets.

Sol. Three.

29. Let R be the set of real numbers and *be the binary operation defined on R as a * b= a + b– ab a, b R. Then, the identity element with respect to the binary operation * is _______.

Sol. 0 is the identity element with respect to the binary operation *.

State True or False for the statements in each of the Examples 30 to 34.

30. Consider the set A = {1, 2, 3} and the relation R = {(1, 2), (1, 3)}. R is a transitive relation.

Sol. True.

31. Let A be a finite set. Then, each injective function from A into itself is not surjective.

Sol. False.

32. For sets A, B and C, let : A →B, : B →C be functions such that is injective. Then both and are injective functions.

Sol. False.

33. For sets A, B and C, let : A →B, : B →C be functions such that is surjective. Then is surjective.

Sol. True.

34. Let N be the set of natural numbers. Then, the binary operation * in N defined as has identity element.

Sol. False.

Relations and Functions - Exemplar Solutions

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