Sets, Relations, and Functions Solved Examples | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


Solved Examples on Sets, Relations and 
Functions 
JEE Mains 
 
Sets and Relations 
 
Q1: Let A = {(x, y) : x
2
 + y
2
 = 9} and B = {(x, y) : x
2
 + 2y
2
 = 16}.  
Determine the number of distinct points in the intersection A n B.  
Sol: To find the points in the intersection A n B, we need to solve the system of equations given 
by the conditions of both sets A and B:  
1. The equation for set A is x 2 + y 2 = 9.  
2. The equation for set B is x 2 + 2y 2 = 16.  
Now, let’s solve this system to find the points (x, y) that satisfy both equations. Subtracting the 
first equation from the second: (x
2
 + 2y
2
) - (x
2
 + y
2
) = 16 - 9  
Simplifying: y
2
 = 7  
Taking the square root of both sides:  
y = ± v 7  
Now, substitute these values of y back into the equation for set A or B to find the corresponding 
x values. Using the equation for set A:  
x
2
 + (v7)
2
 = 9  
x
2
 + 7 = 9  
x
2
 = 2  
x = ± v2  
So, the points of intersection are (v2, v7), (- v2, v7), (v2, - v7), and (- v2, - v7). There are 
four distinct points in the intersection A n B. 
 
 
Q2: Let R be the relation on the set of real numbers defined by aRb if and only if a
3
-b
3
 is 
divisible by 7. Determine whether R is an equivalence relation or not.  
Sol: Let’s analyze the properties of R:  
Page 2


Solved Examples on Sets, Relations and 
Functions 
JEE Mains 
 
Sets and Relations 
 
Q1: Let A = {(x, y) : x
2
 + y
2
 = 9} and B = {(x, y) : x
2
 + 2y
2
 = 16}.  
Determine the number of distinct points in the intersection A n B.  
Sol: To find the points in the intersection A n B, we need to solve the system of equations given 
by the conditions of both sets A and B:  
1. The equation for set A is x 2 + y 2 = 9.  
2. The equation for set B is x 2 + 2y 2 = 16.  
Now, let’s solve this system to find the points (x, y) that satisfy both equations. Subtracting the 
first equation from the second: (x
2
 + 2y
2
) - (x
2
 + y
2
) = 16 - 9  
Simplifying: y
2
 = 7  
Taking the square root of both sides:  
y = ± v 7  
Now, substitute these values of y back into the equation for set A or B to find the corresponding 
x values. Using the equation for set A:  
x
2
 + (v7)
2
 = 9  
x
2
 + 7 = 9  
x
2
 = 2  
x = ± v2  
So, the points of intersection are (v2, v7), (- v2, v7), (v2, - v7), and (- v2, - v7). There are 
four distinct points in the intersection A n B. 
 
 
Q2: Let R be the relation on the set of real numbers defined by aRb if and only if a
3
-b
3
 is 
divisible by 7. Determine whether R is an equivalence relation or not.  
Sol: Let’s analyze the properties of R:  
1. Reflexive: For aRa, a
3
 - a
3
 = 0, which is divisible by 7. Therefore, R is reflexive. 
2. Symmetric: If aRb, then a
3
 - b
3
 is divisible by 7. This implies b
3
 - a
3 
is also divisible by 7, so 
bRa. Therefore, R is symmetric.  
3. Transitive: If aRb and bRc, then a
3
 - b
3
 and b
3
 - c
3
 are divisible by 7.  
This implies a
3
 - c
3
 is also divisible by 7. Therefore, R is transitive. 
Since R is reflexive, symmetric, and transitive, it is an equivalence relation. 
 
 
Q3: Let R be a relation in N defined by 
R = {(1 + x, 1 + x
2
) : x = 5, x ? N}. 
Which of the following is false 
(a) R = {(2, 2), (3, 5), (4, 10), (5, 17), (6, 25)} 
(b) Domain of R = {2, 3, 4, 5, 6} 
(c) Range of R = {2, 5, 10, 17, 26} 
(d) None of these 
Ans: (a) 
R = {(1 + x, 1 + x
2
): x = 5, x ? N}  
(A) R = {(2, 2), (3, 5), (4, 10), (5, 17), (6, 25)} 
 
Option A is false.  
 
Q4: The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on the set A = {1, 2, 3} is 
(a) Reflexive but not symmetric 
(b) Reflexive but not transitive 
Page 3


Solved Examples on Sets, Relations and 
Functions 
JEE Mains 
 
Sets and Relations 
 
Q1: Let A = {(x, y) : x
2
 + y
2
 = 9} and B = {(x, y) : x
2
 + 2y
2
 = 16}.  
Determine the number of distinct points in the intersection A n B.  
Sol: To find the points in the intersection A n B, we need to solve the system of equations given 
by the conditions of both sets A and B:  
1. The equation for set A is x 2 + y 2 = 9.  
2. The equation for set B is x 2 + 2y 2 = 16.  
Now, let’s solve this system to find the points (x, y) that satisfy both equations. Subtracting the 
first equation from the second: (x
2
 + 2y
2
) - (x
2
 + y
2
) = 16 - 9  
Simplifying: y
2
 = 7  
Taking the square root of both sides:  
y = ± v 7  
Now, substitute these values of y back into the equation for set A or B to find the corresponding 
x values. Using the equation for set A:  
x
2
 + (v7)
2
 = 9  
x
2
 + 7 = 9  
x
2
 = 2  
x = ± v2  
So, the points of intersection are (v2, v7), (- v2, v7), (v2, - v7), and (- v2, - v7). There are 
four distinct points in the intersection A n B. 
 
 
Q2: Let R be the relation on the set of real numbers defined by aRb if and only if a
3
-b
3
 is 
divisible by 7. Determine whether R is an equivalence relation or not.  
Sol: Let’s analyze the properties of R:  
1. Reflexive: For aRa, a
3
 - a
3
 = 0, which is divisible by 7. Therefore, R is reflexive. 
2. Symmetric: If aRb, then a
3
 - b
3
 is divisible by 7. This implies b
3
 - a
3 
is also divisible by 7, so 
bRa. Therefore, R is symmetric.  
3. Transitive: If aRb and bRc, then a
3
 - b
3
 and b
3
 - c
3
 are divisible by 7.  
This implies a
3
 - c
3
 is also divisible by 7. Therefore, R is transitive. 
Since R is reflexive, symmetric, and transitive, it is an equivalence relation. 
 
 
Q3: Let R be a relation in N defined by 
R = {(1 + x, 1 + x
2
) : x = 5, x ? N}. 
Which of the following is false 
(a) R = {(2, 2), (3, 5), (4, 10), (5, 17), (6, 25)} 
(b) Domain of R = {2, 3, 4, 5, 6} 
(c) Range of R = {2, 5, 10, 17, 26} 
(d) None of these 
Ans: (a) 
R = {(1 + x, 1 + x
2
): x = 5, x ? N}  
(A) R = {(2, 2), (3, 5), (4, 10), (5, 17), (6, 25)} 
 
Option A is false.  
 
Q4: The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on the set A = {1, 2, 3} is 
(a) Reflexive but not symmetric 
(b) Reflexive but not transitive 
(c) Symmetric and transitive 
(d) Neither symmetric nor transitive 
Ans: (a) 
R = {(1, 1), (2, 2), (3, 3),(1, 2), (2, 3), (1, 3)} on the set A = {1, 2, 3} ? (1, 1), (2, 2), (3, 3) ? R 
R is reflexive  
? (1, 2) ? R but (2, 1) ? R 
R is not symmetric  
? (1, 2) ? R and (2, 3) and also (1, 3) ? R  
So R is transitive.  
 
Q5: If A, B and C are these sets such that A nB = A n C and A ? B = A ? C, then 
(a) A = B  
(b) A = C 
(c) B = C  
(d) A n B = f 
Ans: (c)  
 
Given A n B = A n C … (i)  
and A ? B = A ? C … (ii)  
we know that A ? B = A + B – A n B  
from (i) and (ii)  
A ? C = A + B – A n C … (iii)  
but A ? C = A + C – A n C … (iv)  
From (iii) – (iv) O = B – C 
B = C  
 
Q6: Given the relation R = {(2, 3), (3, 4)} on the set {2, 3, 4}. The number of minimum 
number of ordered pair to be added to R so that R is reflexive and symmetric 
(a) 4  
(b) 5  
Page 4


Solved Examples on Sets, Relations and 
Functions 
JEE Mains 
 
Sets and Relations 
 
Q1: Let A = {(x, y) : x
2
 + y
2
 = 9} and B = {(x, y) : x
2
 + 2y
2
 = 16}.  
Determine the number of distinct points in the intersection A n B.  
Sol: To find the points in the intersection A n B, we need to solve the system of equations given 
by the conditions of both sets A and B:  
1. The equation for set A is x 2 + y 2 = 9.  
2. The equation for set B is x 2 + 2y 2 = 16.  
Now, let’s solve this system to find the points (x, y) that satisfy both equations. Subtracting the 
first equation from the second: (x
2
 + 2y
2
) - (x
2
 + y
2
) = 16 - 9  
Simplifying: y
2
 = 7  
Taking the square root of both sides:  
y = ± v 7  
Now, substitute these values of y back into the equation for set A or B to find the corresponding 
x values. Using the equation for set A:  
x
2
 + (v7)
2
 = 9  
x
2
 + 7 = 9  
x
2
 = 2  
x = ± v2  
So, the points of intersection are (v2, v7), (- v2, v7), (v2, - v7), and (- v2, - v7). There are 
four distinct points in the intersection A n B. 
 
 
Q2: Let R be the relation on the set of real numbers defined by aRb if and only if a
3
-b
3
 is 
divisible by 7. Determine whether R is an equivalence relation or not.  
Sol: Let’s analyze the properties of R:  
1. Reflexive: For aRa, a
3
 - a
3
 = 0, which is divisible by 7. Therefore, R is reflexive. 
2. Symmetric: If aRb, then a
3
 - b
3
 is divisible by 7. This implies b
3
 - a
3 
is also divisible by 7, so 
bRa. Therefore, R is symmetric.  
3. Transitive: If aRb and bRc, then a
3
 - b
3
 and b
3
 - c
3
 are divisible by 7.  
This implies a
3
 - c
3
 is also divisible by 7. Therefore, R is transitive. 
Since R is reflexive, symmetric, and transitive, it is an equivalence relation. 
 
 
Q3: Let R be a relation in N defined by 
R = {(1 + x, 1 + x
2
) : x = 5, x ? N}. 
Which of the following is false 
(a) R = {(2, 2), (3, 5), (4, 10), (5, 17), (6, 25)} 
(b) Domain of R = {2, 3, 4, 5, 6} 
(c) Range of R = {2, 5, 10, 17, 26} 
(d) None of these 
Ans: (a) 
R = {(1 + x, 1 + x
2
): x = 5, x ? N}  
(A) R = {(2, 2), (3, 5), (4, 10), (5, 17), (6, 25)} 
 
Option A is false.  
 
Q4: The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on the set A = {1, 2, 3} is 
(a) Reflexive but not symmetric 
(b) Reflexive but not transitive 
(c) Symmetric and transitive 
(d) Neither symmetric nor transitive 
Ans: (a) 
R = {(1, 1), (2, 2), (3, 3),(1, 2), (2, 3), (1, 3)} on the set A = {1, 2, 3} ? (1, 1), (2, 2), (3, 3) ? R 
R is reflexive  
? (1, 2) ? R but (2, 1) ? R 
R is not symmetric  
? (1, 2) ? R and (2, 3) and also (1, 3) ? R  
So R is transitive.  
 
Q5: If A, B and C are these sets such that A nB = A n C and A ? B = A ? C, then 
(a) A = B  
(b) A = C 
(c) B = C  
(d) A n B = f 
Ans: (c)  
 
Given A n B = A n C … (i)  
and A ? B = A ? C … (ii)  
we know that A ? B = A + B – A n B  
from (i) and (ii)  
A ? C = A + B – A n C … (iii)  
but A ? C = A + C – A n C … (iv)  
From (iii) – (iv) O = B – C 
B = C  
 
Q6: Given the relation R = {(2, 3), (3, 4)} on the set {2, 3, 4}. The number of minimum 
number of ordered pair to be added to R so that R is reflexive and symmetric 
(a) 4  
(b) 5  
(c) 7  
(d) 6 
Ans: (b) 
Relation R = {(2, 3), (3, 4)} on the set assume A = {2, 3, 4}  
? For reflexive ? (x, x) ? R, x ? A 
 
? For symmetric ? given (2, 3), (3, 4) ? R to be R ? symmetric (3, 2) and (4, 3) should be pair of 
R  
Total added pair is 5{(2, 2),(3, 3),(4, 4),(3, 2),(4,3)} 
 
 
 
Functions 
 
Q7: Period of the function  
(a)p/2  
(b) p 
(c) 2p  
(d) 4p 
Ans: (c)  
 
 
Page 5


Solved Examples on Sets, Relations and 
Functions 
JEE Mains 
 
Sets and Relations 
 
Q1: Let A = {(x, y) : x
2
 + y
2
 = 9} and B = {(x, y) : x
2
 + 2y
2
 = 16}.  
Determine the number of distinct points in the intersection A n B.  
Sol: To find the points in the intersection A n B, we need to solve the system of equations given 
by the conditions of both sets A and B:  
1. The equation for set A is x 2 + y 2 = 9.  
2. The equation for set B is x 2 + 2y 2 = 16.  
Now, let’s solve this system to find the points (x, y) that satisfy both equations. Subtracting the 
first equation from the second: (x
2
 + 2y
2
) - (x
2
 + y
2
) = 16 - 9  
Simplifying: y
2
 = 7  
Taking the square root of both sides:  
y = ± v 7  
Now, substitute these values of y back into the equation for set A or B to find the corresponding 
x values. Using the equation for set A:  
x
2
 + (v7)
2
 = 9  
x
2
 + 7 = 9  
x
2
 = 2  
x = ± v2  
So, the points of intersection are (v2, v7), (- v2, v7), (v2, - v7), and (- v2, - v7). There are 
four distinct points in the intersection A n B. 
 
 
Q2: Let R be the relation on the set of real numbers defined by aRb if and only if a
3
-b
3
 is 
divisible by 7. Determine whether R is an equivalence relation or not.  
Sol: Let’s analyze the properties of R:  
1. Reflexive: For aRa, a
3
 - a
3
 = 0, which is divisible by 7. Therefore, R is reflexive. 
2. Symmetric: If aRb, then a
3
 - b
3
 is divisible by 7. This implies b
3
 - a
3 
is also divisible by 7, so 
bRa. Therefore, R is symmetric.  
3. Transitive: If aRb and bRc, then a
3
 - b
3
 and b
3
 - c
3
 are divisible by 7.  
This implies a
3
 - c
3
 is also divisible by 7. Therefore, R is transitive. 
Since R is reflexive, symmetric, and transitive, it is an equivalence relation. 
 
 
Q3: Let R be a relation in N defined by 
R = {(1 + x, 1 + x
2
) : x = 5, x ? N}. 
Which of the following is false 
(a) R = {(2, 2), (3, 5), (4, 10), (5, 17), (6, 25)} 
(b) Domain of R = {2, 3, 4, 5, 6} 
(c) Range of R = {2, 5, 10, 17, 26} 
(d) None of these 
Ans: (a) 
R = {(1 + x, 1 + x
2
): x = 5, x ? N}  
(A) R = {(2, 2), (3, 5), (4, 10), (5, 17), (6, 25)} 
 
Option A is false.  
 
Q4: The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on the set A = {1, 2, 3} is 
(a) Reflexive but not symmetric 
(b) Reflexive but not transitive 
(c) Symmetric and transitive 
(d) Neither symmetric nor transitive 
Ans: (a) 
R = {(1, 1), (2, 2), (3, 3),(1, 2), (2, 3), (1, 3)} on the set A = {1, 2, 3} ? (1, 1), (2, 2), (3, 3) ? R 
R is reflexive  
? (1, 2) ? R but (2, 1) ? R 
R is not symmetric  
? (1, 2) ? R and (2, 3) and also (1, 3) ? R  
So R is transitive.  
 
Q5: If A, B and C are these sets such that A nB = A n C and A ? B = A ? C, then 
(a) A = B  
(b) A = C 
(c) B = C  
(d) A n B = f 
Ans: (c)  
 
Given A n B = A n C … (i)  
and A ? B = A ? C … (ii)  
we know that A ? B = A + B – A n B  
from (i) and (ii)  
A ? C = A + B – A n C … (iii)  
but A ? C = A + C – A n C … (iv)  
From (iii) – (iv) O = B – C 
B = C  
 
Q6: Given the relation R = {(2, 3), (3, 4)} on the set {2, 3, 4}. The number of minimum 
number of ordered pair to be added to R so that R is reflexive and symmetric 
(a) 4  
(b) 5  
(c) 7  
(d) 6 
Ans: (b) 
Relation R = {(2, 3), (3, 4)} on the set assume A = {2, 3, 4}  
? For reflexive ? (x, x) ? R, x ? A 
 
? For symmetric ? given (2, 3), (3, 4) ? R to be R ? symmetric (3, 2) and (4, 3) should be pair of 
R  
Total added pair is 5{(2, 2),(3, 3),(4, 4),(3, 2),(4,3)} 
 
 
 
Functions 
 
Q7: Period of the function  
(a)p/2  
(b) p 
(c) 2p  
(d) 4p 
Ans: (c)  
 
 
 
 
 
Q8: If f(x) =  denoting the greatest integer function, then 
(a) f(0) = 0 
(b)  
(c)  
(d) f(p) = 0 
Ans: (c)