Topic Question Set


Q 11 :

Let A = {1, 2, 3, ..., 100} and R be a relation on A such that R = {(a, b) : a = 2b + 1}. Let (a1,a2),(a2,a3),(a3,a4),...,(ak,ak+1) be a sequence of k elements of R such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k, for which such a sequence exists, is equal to :          [2025]

  • 8

     

  • 5

     

  • 6

     

  • 7

     

(2)

Let the smallest value is ak+1 in A = {1, 2, 3, ..., 100}.

Since, ak=2ak+1+1=(2×1)+1=3

∴       ak–1=2ak+1=(2×3)+1=7

           ak–2=2ak–1+1=(2×7)+1=15

          ak–3=2ak–2+1=(2×15)+1=31

          ak–4=2ak–3+1=(2×31)+1=63

          ak–5=2ak–4+1=(2×63)+1=127∉A

∴   The sequence is {(63, 31), (31, 15), (15, 7), (7, 3), (3, 1)}

∴   k = 5



Q 12 :

Let A = {–3, –2, –1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if 0<x2+2y≤4. Let l be the number of elements in R and m be the minimum number of elements required to be added in R to make it reflexive relation. Then l + m is equal to          [2025]

  • 18

     

  • 20

     

  • 17

     

  • 19

     

(1)

A = {–3, –2, –1, 0, 1, 2, 3}

xRy if and only if –2y≤x2≤4–2y

y = –3          6≤x2≤10               ⇒ x∈{–3,3}

y = –2          4≤x2≤8                 ⇒ x∈{–2,2}

y = –1          2≤x2≤6                 ⇒ x∈{–2,2}

y = 0            0≤x2≤4                 ⇒ x∈{–2,–1,0,1,2}

y = 1            –2≤x2≤2              ⇒ x∈{–1,0,1}

y = 2            –4≤x2≤0              ⇒ x∈{0}

y = 3            –6≤x2≤–2           ⇒ Value of x does not exist

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

∴   l = 15

To make it reflexive we will add (–1, –1), (2, 2), (3, 3) in R

∴   l + m = 15 + 3 = 18.



Q 13 :

Let A = {–2, –1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if y = max{x, 1}. Let l be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then l + m + n is equal to          [2025]

  • 13

     

  • 12

     

  • 14

     

  • 11

     

(2)

We have, A = {–2, –1, 0, 1, 2, 3} and

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

Now, number of elements in R i.e., l = 6

For R to be reflexive,

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

So, we need to add three elements to make it reflexive.

∴   m = 3

For R to be symmetric,

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

So, we need to add three elements to make it symmetric.

∴   n = 3

So, l + m + n = 6 + 3 + 3 = 12.



Q 14 :

Let A = {–3, –2, –1, 0, 1, 2, 3} and R be a relation on A defined by xRy if and only if 2x – y ∈ {0, 1}. Let l be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then l + m + n is equal to :          [2025]

  • 17

     

  • 16

     

  • 18

     

  • 15

     

(1)

We have, A = {–3, –2, –1, 0, 1, 2, 3}, R is defined on A as xRy such that 2x – y ∈ {0, 1}.

i.e., 2x – y = 0 or 2x – y = 1

∴   R = {(0, 0), (–1, –2), (1, 2), (0, –1), (2,3), (1, 1), (–1, –3)} i.e., l = 7

For R to be reflexive, i.e., we need 5 more elements {(2, 2), (–1, –1), (3, 3), (–3, –3), (–2, –2)} so m = 5 and for R to be symmetric, we need 5 more elements {(–2, –1), (2, 1), (–1, 0), (3, 2), (–3, –1)}, so n = 5.

∴   l + m + n = 7 + 5 + 5 = 17.



Q 15 :

Let A = {0, 1, 2, 3, 4, 5}. Let R be a relation on A defined by (x, y) ∈ R if and only if max {x, y} ∈ {3, 4}. Then among the statements

(S1) : The number of elements in R is 18, and

(S2) : The relation R is symmetric but neither reflexive nor transitive.           [2025]

  • both are true

     

  • only (S2) is true

     

  • both are false

     

  • only (S1) is true

     

(2)

R = {(0, 3), (3, 0), (0, 4), (4, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)}

∴   Number of elements in R = 16

∴   S1 is false.

Now, (0, 0), (1, 1), (2, 2), (5, 5) ∉ R   ∴ R is not reflexive.

Again, let (a, b) ∈ R then (b, a) ∈ R

As max {a, b} = max {b, a}   ∴ R is symmetric.

Now, R is not transitive as (0, 3), (3, 1) ∈ R but (0, 1) ∉ R.

∴   S2 is true.



Q 16 :

The number of non-empty equivalence relations on the set {1, 2, 3} is:          [2025]

  • 5

     

  • 6

     

  • 7

     

  • 4

     

(1)

Partitions of set {1, 2, 3} is {{1}}, {2}, {3}}, {{1, 2}, {3}}, {{1, 3}, {2}}, {{2, 3}, {1}}, {{1, 2, 3}}

∴   Number of non-empty equivalence relations on the set {1, 2, 3} = 5.



Q 17 :

Let R = {(1, 2), (2, 3), (3, 3)} be a relation defined on the set {1, 2, 3, 4}. Then the minimum number of elements, needed to be added in R so that R becomes an equivalence relation, is:          [2025]

  • 7

     

  • 10

     

  • 9

     

  • 8

     

(1)

Let A = {1, 2, 3, 4}

R = {(1, 2), (2, 3), (3, 3)}          [Given]

For an equivalence relation 'R' should be reflexive, symmetric and transitive.

∴   Minimum elements that should be added are

(2, 1), (3, 2), (1, 1), (2, 2), (4, 4), (1, 3), (3, 1)

∴   Minimum number of elements required = 7.



Q 18 :

Let A={(x,y)∈R×R:|x+y|≥3} and B={(x,y)∈R×R:|x|+|y|≤3}

If C={(x,y)∈A∩B:x=0 or y=0}, then ∑(x,y)∈C|x+y| is:          [2025]

  • 18

     

  • 12

     

  • 24

     

  • 15

     

(2)

We have, A={(x,y)∈R×R:|x+y|≥3} and B={(x,y)∈R×R:|x|+|y|≤3}

From given figure, we get C = {(3, 0), (–3, 0), (0, 3), (0, –3)}

∑(x,y)∈C|x+y|=12



Q 19 :

Let X = R × R. Define a relation R on X as:

(a1,b1)R(a2,b2)⇔b1=b2.

Statement I: R is an equivalence relation.

Statement II: For some (a, b) ∈ X, the set

S = {(x, y) ∈ X :(x, y) R (a, b)} represents a line parallel to y = x.

In the light of the above statements, choose the correct answer from the options given below:          [2025]

  • Statement I is true but Statement II is false.

     

  • Statement I is false but State II is true.

     

  • Both Statement I and Statement II are false.

     

  • Both Statement I and Statement II is true.

     

(1)

Statement I:

Reflexive : (a1,b1)R(a1,b1)⇒b1=b1

∴   R is reflexive.

Symmetric : (a1,b1)R(a2,b2)⇒b1=b2

                    (a2,b2)R(a1,b1)⇒b2=b1

∴   R is symmetric.

Transitive : (a1,b1)R(a2,b2)⇒b1=b2

and (a2,b2)R(a3,b3)⇒b2=b3⇒b1=b3

⇒(a1,b1)R(a3,b3)

Hence, relation R is equivalence relation.

∴  Statement I is true.

For Statement II: (x, y) R (a, b), for some (a,b)∈X⇒y=b

So, Statement II is false.



Q 20 :

The relation R = {(x, y) : x, y ∈ Z and x + y is even} is:          [2025]

  • reflexive and transitive but not symmetric.

     

  • reflexive and symmetric but not transitive.

     

  • symmetric and transitive but not reflexive.

     

  • an equivalence relation.

     

(4)

R = {(x, y) : x, y ∈ Z and x + y is even}

For reflexive : x + x = 2x is even

For symmetric : x + y = y + x is even

For transitive : x + y is even and y + z is even then z + x is also even.

So, the relation is an equivalence.



Q 21 :

Define a relation R on the interval [0,π2) by xRy if and only if sec2x–tan2y=1. Then R is :          [2025]

  • an equivalence relation.

     

  • reflexive but neither symmetric not transitive.

     

  • both reflexive and symmetric but not transitive.

     

  • both reflexive and transitive but not symmetric.

     

(1)

sec2x–tan2x=1, ∀x∈[0,π2)

∴   R is reflexive

Consider, sec2x–tan2y=1

⇒ 1+tan2x–(sec2y–1)=1

⇒ 1+tan2x–sec2y+1=1 ⇒ sec2y–tan2x=1

∴   R is symmetric.

Now, if sec2x–tan2y=1 and sec2y–tan2z=1

Adding both equation, sec2x–tan2y+sec2y–tan2z=2

⇒ sec2x–tan2z=1        [∵ sec2y–tan2y=1]

∴   R is transitive

Thus R is an equivalence relation.



Q 22 :

Let S=N∪{0}. Define a relation R from S to R by:

R={(x,y):logey=xloge(25), x∈S, y∈R}

Then, the sum of all the elements in the range of R is equal to:          [2025]

  • 52

     

  • 53

     

  • 32

     

  • 109

     

(2)

We have, S=N∪{0}={0,1,2,3,......}

Also, logey=xloge(25)⇒y=(25)x

∴   Required sum = (25)0+(25)1+(25)2+........

                              =11–25=53.



Q 23 :

The number of relation on the set A = {1, 2, 3}, containing at most 6 elements including (1, 2), which are reflexive and transitive but not symmetric, is __________.          [2025]



(5)

Given, A = {1, 2, 3}

Let the relation be R on A, which is reflexive and transitive but not symmetric, then

(1, 1), (2, 2), (3, 3), (1, 2) ∈ R

Remaining elements are

(2, 1), (2, 3), (1, 3), (3, 1), (3, 2)

Case I : If relation contains exactly 4 elements ⇒ 1 way

Case II : If relation contains exactly 5 elements, so we can add (1, 3) or (3, 2) ⇒ 2 ways

Case III : If relation contains exactly 6 elements, so we can add (2, 3), (1, 3) or (1, 3), (3, 2) or (3, 1), (3, 2) ⇒ 3 ways

∴   Total number of relations is 6.



Q 24 :

Let A = {1, 2, 3}. The number of relations on A, containing (1, 2) and (2, 3), which are reflexive and transitive but not symmetric, is __________.          [2025]



(3)

For transitive : (1, 2) and (2, 3) ∈ R ⇒ (1, 3) ∈ R

For reflexive : (1, 1), (2, 2), (3, 3) ∈ R

Now, for (2, 1), (3, 2), (3, 1); (3, 1) cannot be taken for not symmetric relation.

Case I : (2, 1) taken and (3, 2) not taken

Case II : (3, 2) taken and (2, 1) not taken

Case III : (2, 1) and (3, 2) are not taken

Therefore, 3 relations are possible.



Q 25 :

Let A={-2,-1,0,1,2,3,4}. Let R be a relation on A defined by xRy if and only if 2x+y≤2. Let l be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations respectively. Then l+m+n is equal to:                 [2026]

  • 33

     

  • 32

     

  • 35

     

  • 34

     

(1)

R = {(-2, a), (-1, b), (0, c), (1, d), (2, e)}

a = {-2, -1, 0, 1, 2, 3, 4};  
b = {-2, -1, 0, 1, 2, 3, 4}  
c = {-2, -1, 0, 1, 2};  
d = {-2, -1, 0};  
e = {-2}

∴ No. of elements in R  
= 7 + 7 + 5 + 3 + 1 = 23 = ℓ

Minimum number of element to be added to make it reflexive = m = 4  
⇒ {(1, 1), (2, 2), (3, 3), (4, 4)}

Minimum number of element to be added to make it symmetric = n = 6  

⇒ R = {(3, -2), (4, -2), (2, -1), (2, 0), (3, -1), (4, -1)}

∴ ℓ + m + n = 23 + 4 + 6 = 33



Q 26 :

Let R be a relation defined on the set {1,2,3,4}×{1,2,3,4} by R={((a,b),(c,d)):2a+3b=3c+4d}.

Then the number of elements in R  is                       [2026]

  • 15

     

  • 6

     

  • 18

     

  • 12

     

(4)

(a, b)      (c, d)

(1, 1)         x
(1, 2)         x
(1, 3)      (1, 2)
(1, 4)      (2, 2)
(2, 1)      (1, 1)
(2, 2)      (2, 1)
(2, 3)      (3, 1)
(2, 4)      (4, 1)
(3, 1)         x
(3, 2)         x
(3, 3)      (1, 3)
(3, 4)      (2, 3)
(4, 1)      (1, 2)
(4, 2)      (2, 2)
(4, 3)      (3, 2)
(4, 4)      (4, 2)



Q 27 :

Let the relation R on the set M={1,2,3,…,16} be given by R={(x,y):4y=5x-3, x,y∈M}.

Then the minimum number of elements required to be added in R, in order to make the relation symmetric, is equal to         [2026]

  • 1

     

  • 2

     

  • 4

     

  • 3

     

(2)

R = {(3, 3), (7, 8), (11, 13)}

to make it symmetric (8, 7), (13, 11) must be added.



Q 28 :

Let A={0,1,2,…,9} Let R be a relation on A defined by (x,y)∈R if and only if ∣x−y∣ is a multiple of 3.

Given below are two statements:

Statement I: n(R)=36.

Statement II: R is an equivalence relation.

In the light of the above statements, choose the correct answer from the options given below. [2026]

  • Statement I is correct but Statement II is incorrect

     

  • Statement I is incorrect but Statement II is correct

     

  • Both Statement I and Statement II are incorrect

     

  • Both Statement I and Statement II are correct

     

(2)

Number of form 3K=4

Number of form 3K+1=3

Number of form 3K+2=4

4×4+3×3+3×3=34 relations

⇒xRy⇒yRx

⇒(x-y)=3λ, (y-z)=3μ

⇒(x-z)=3(λ+μ)

R is reflexive, symmetric and transitive S2 is true

Ans. S1 is false but S2 is true



Q 29 :

Let A = {2,3,5,7,9} Let R be the relation on A defined by xRy if and only if 2x≤3y. Let l be the number of elements in R, and m be the minimum number of elements required to be added in R to make it a symmetric relation. Then l+m is equal to :         [2026]

  • 25

     

  • 23

     

  • 27

     

  • 21

     

(1)

A={2,3,5,7,9}

y≥2x3

x=2,y=2,3,5,7,9x=3,y=2,3,5,7,9x=5,y=5,7,9x=7,y=5,7,9x=9,y=7,9]⇒ℓ=18

To make it symmetric the elements to be added are {(5,2),(7,2),(9,2),(5,3),(7,3),(9,3),(9,5)}

m=7

∴ ℓ+m=25



Q 30 :

The number of elements in the relation R={(x,y):4x2+y2<52, x,y∈ℤ} is    [2026]

  • 86

     

  • 89

     

  • 67

     

  • 77

     

(4)

4x2+y2<52       ,    x,y∈ℤ

↓     ↓

0    0,±1,±2,±3,±4,±5,±6,±7  →  1×15=15

±1           0,±1,±2,±3,….....,±6  →  2×13=26

±2             0,±1,±2,±3,.....…,±5   →  2×11=22      

±3                                 0,±1,±2,±3   →  2×7=14

Number of elements =15+26+22+14=77



Q 31 :

Let R1 and R2 be two relations defined on ℝ by aR1b⇔ab≥0 and aR2b⇔a≥b. Then

  • R1 is an equivalence relation but not R2

     

  • R2 is an equivalence relation but not R1

     

  • Both R1 and R2 are equivalence relations

     

  • Neither R1 nor R2 is an equivalence relation

     

(4)

a=1, b=0, c=-2 then transitive fails for R1



Q 32 :

Let R be the relation on N defined by R={(x,y):x+2y=8}. Then, the domain of R is:

  • {2,4,6,8}

     

  • {2,4,8}

     

  • {2,4,6}

     

  • {1,2,3,4}

     

(3)

Ans.   {2, 4, 6}

Explanation:

Since R is defined as x+2y=8.

Now, x and y are natural numbers.

∴                 x=2,  y=3

Because if we put x=3, then y will not be a natural number.

                      x=4,  y=2

                      x=6,  y=1

∴    Domain of the given relation is {2, 4, 6}



Q 33 :

The relation R in the set of natural numbers N defined as R={(x,y):y=x+5 and x<4} is:

  • reflexive

     

  • symmetric

     

  • transitive

     

  • None of these

     

(4)

Ans.    None of these

Explanation:

R={(1,6),(2,7),(3,8)}

∵   (6,6)∉R

⇒R is not reflexive.

For   (2,7)∈R

⇒    (2,7)∈R but (7,2)∉R

⇒R is not symmetric.

Clearly, R is not transitive.

(Since, there is no pair in R such that (x,y) and (y,z)∈R, then (x,z)∈R)

⇒   R is not an equivalence relation.



Q 34 :

If A={x∈Z:0≤x≤12} and R is the relation in A given by R={(a,b):a=b}. Then, the set of all elements related to 1 is:

  • {1, 2}

     

  • {2, 3}

     

  • {1}

     

  • {2}

     

(3)

Ans.    {1}

Explanation:

The set of all elements related to 1 is

{a∈A:a=1}

Hence,   {1}



Q 35 :

The range of the relation R = {(x,x2):x is a prime number less than 13} is:

  • {2, 3, 5, 7}

     

  • {4, 9, 25, 49, 121}

     

  • {2, 3, 5, 7, 11}

     

  • {1, 4, 9, 25, 49, 121}

     

(2)

Ans.    {4, 9, 25, 49, 121}

Explanation:

R = {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121)}

Range = {4, 9, 25, 49, 121}



Q 36 :

R is a relation from {11, 12, 13} to {8, 10, 12} defined by y=x−3. Then R-1 is:

  • {(8, 11), (10, 13)}

     

  • {(11, 8), (13, 10)}

     

  • {(10, 13), (8, 11), (8, 10)}

     

  • None of the above

     

(1)

Ans.    {(8,11), (10,13)}

Explanation:

Let               A={11,12,13},

                    B={8,10,12}

RB'A          y=x-3

                    R={(11,8),(13,10)}

               R-1={(8,11),(10,13)}



Q 37 :

Let T be the set of all triangles in the Euclidean plane and let a relation R on T be defined as aRb if a is congruent to b for all a,b∈T, then R is:

  • reflexive but not symmetric

     

  • transitive but not symmetric

     

  • equivalence

     

  • neither symmetric nor transitive

     

(3)

Ans.     Equivalence

Explanation:

Each triangle is congruent to itself.

∴   Relation is reflexive.

        Suppose  ∆ABC≅∆PQR

        then         ∆PQR≅∆ABC

∴  Relation is symmetric.

       Suppose   ∆ABC≅∆PQR,

       and           ∆PQR≅∆XYZ

       then          ∆ABC≅∆XYZ

∴  Relation is transitive.

∴  Relation is an equivalence relation.



Q 38 :

Let us define a relation R in R as aRb if a≥b. Then R is: 

  • an equivalence relation

     

  • reflexive, transitive but not symmetric

     

  • symmetric, transitive but not reflexive

     

  • neither transitive nor reflexive but symmetric

     

(2)

Ans.      Reflexive, transitive but not symmetric

Explanation:

Given that, aRb if a≥b

⇒aRa⇒a≥a, which is true.

Hence, R is reflexive.

Let aRb, i.e., a≥b.

For symmetry, we require bRa, i.e., b≥a, which is not true in general.

Hence, R is not symmetric.

But if  aRb and  bRc,

⇒        a≥b  and  b≥c

⇒        a≥c

Hence, R is transitive.



Q 39 :

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: 

  • R is reflexive and symmetric but not transitive

     

  • R is reflexive and transitive but not symmetric

     

  • R is symmetric and transitive but not reflexive

     

  • R is an equivalence relation

     

(2)

Ans.    R is reflexive and transitive but not symmetric.

Explanation:

R={(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)}

It is seen that (a,a)∈R, for every a∈{1,2,3,4}

∴  R is reflexive.

It is seen that (1,2)∈R,but (2,1)∉R.

∴  R is not symmetric.

Also, it is observed that (a,b), (b,c)∈R ⇒(a,c)∈R, for all a,b,c∈{1,2,3,4}.

∴  R is transitive.

Hence,  R is reflexive and transitive but not symmetric.



Q 40 :

If R={(x,y):x+2y=8} is a relation on N, then the range of R is:

  • {2, 3, 4}

     

  • R

     

  • {1, 2, 3}

     

  • None of these

     

(3)

Ans.     {1, 2, 3}

Explanation:

The given relation on N is  R={(x,y):x+2y=8}

Since both x,y∈N,

x can take the values (2, 4, 6) for all x,y∈N

For x=2,              2+2y=8

⇒                                   y=3

For x=4,              4+2y=8

⇒                                   y=2

For x=6,              6+2y=8

⇒                                   y=1

Hence,                             R={(2,3),(4,2),(6,1)}

Range of R = Set of second elements ={1,2,3}