JEE Main Sets Previous Year Questions With Solutions

Sets Previous Year Questions with Solutions are available here. What do you mean by Sets? In mathematics, Sets are an organized collection of objects. In this article you will find the solved past year questions for IIT JEE Main along with important formulas and basic concepts. Detailed solution of each question has been provided by our subject experts. All questions are very important from examination point of view. Students are advised to practice all the questions and understand the difficulty level of questions in the upcoming JEE exams.

Download Sets Previous Year Solved Questions PDF

Basic Points to Remember:

Definition: A set is a well-defined collection of objects.

Notation: Sets are usually denoted by a capital letter and elements of the group are represented by small letters.

Representations of a Set: Sets can be represented by: roster form and set builder form

Sets Operations: Union of sets, Intersection of sets, Complement of sets, Cartesian product of sets, Difference of sets.

Some of the important formulas are:

For any 3 sets X, Y and Z:

1. n ( X U Y ) = n(X) + n(Y) – n ( X ∩ Y)

2. n( Y – X) + n( X ∩ Y ) = n(Y)

3. n( X – Y) + n( X ∩ Y ) = n(X)

4. If X ∩ Y = φ, then n ( X U Y ) = n(X) + n(Y)

5. n( X – Y) + n ( X ∩ Y) + n( Y – X) = n ( X U Y )

6. n(X U Y U Z)= n(X) + n(Y) + n(Z) – n( X ∩ Y) – n( Y ∩ Z) – n( Z ∩ X) + n( X ∩ Y ∩ Z)

To Solve All Chapter wise Maths Questions: Click Here

JEE Main Maths Past Year Questions With Solutions on Sets

Question 1: If A = {x Є R : |x| < 2} and B = {x Є R : |x − 2| ≥ 3} then:

(a) A − B = [−1, 2]

(b) B − A = R − (−2, 5)

(c) A U B = R − (2, 5)

(d) A ∩ B = (−2, −1)

Answer: (b)

Solution:

A = {x:x Є (−2, 2)}

B = {x:x Є (−∞,−1] U [5, ∞)}

A ∩ B = {x:x Є (−2,−1]}

B − A = {x:x Є (−∞,−2] U [5,∞)}

A − B = {x:x Є (−1,2)}

A U B = {x:x Є (−∞,2) U [5,∞)}

Question 2: Let A and B be two finite sets having m and n elements respectively. Then the total number of mapping from A to B is

(a) mn                      (b) mn                          (c) 2mn                      (d) nm

Answer: (d)

Solution:

Consider an element a, which is belongs to A, it can be mapped to any of the n elements of B, that is, it has n images.

Similarly, each of the m elements of A can have n images in B.

Therefore, the number of mapping = n × n × n × …x m times = nm.

Question 3: Two sets A and B are as under:

A = {(a, b) ϵ R × R : |a − 5| < 1 and |b − 5| < 1};

B = {a, b) ϵ R × R : 4 (a − 6)2 + 9(b−5) 2 ≤ 36}. Then,

(a) A is subset of B

(b) A n B = φ (null set)

(c) neither A subset of B nor B is subset of A

(d) B is subset of A

Answer: (a)

Solution:

A = {(a, b) ϵ R × R : |a − 5| < 1 and |b − 5| < 1}

=> a ϵ (4, 6) and b ϵ (4, 6)

Therefore, A = {(a, b): a ϵ (4, 6) and b ϵ (4, 6)}

Now, B = {a, b) ϵ R × R : 4 (a − 6) 2 + 9(b−5) 2 ≤ 36}

using conditions of above set A, the maximum value of B is

4 (a − 6) 2 + 9(b−5) 2 = 4 (4 − 6)2 + 9(6−5)2 = 25 ≤ 36

We can check for other values as well.

Elements of the set A satisfying the conditions in Set B.

Hence, A is subset of B. Answer!!

Question 4: Let A and B be two sets containing 4 and 2 elements respectively. Then the number of subsets of the set AxB, each having at least 3 elements is

(a) 256                    (b) 275                       (c) 510                        (d) 219

Answer:(d)

Solution:

n(A) = 4 and n(B) = 2

Therefore, n(A x B) = 8

Number of subsets of A x B having atleast 3 elements = 28 – 8C0 – 8C1 – 8C2

= 256 – 1 – 8 – 28

= 219

Question 5: If A, B and C are three sets such that A n B = A n C and A U B = A U C then

(a) B = C

(b) A = B

(c) A = C

(d) A n B = 0 (empty set)

Answer: (a)

Solution:

A U B = A U C

=> (A U B) ∩ C = (A U C) ∩ C

=> (A ∩ C) U (B ∩ C) = C

Using property: (A U C) ∩ C = C

=> (A ∩ B) U (B ∩ C) = C …..(i)

We know, A ∩ C = A ∩ B

Again, A U B = A U C

(A U B) ∩ B = (A U C) ∩ B

B = (A ∩ B) U (C ∩ B)

B = (A ∩ B) U (B ∩ C) ……….(ii)

From (i) and (ii)

B = C

Question 6: Let S = {x Є R : x ≥ 0 and 2|√x − 3|+ √x (√x − 6) + 6 = 0}. Then S :

(a) is an empty set

(b) contains exactly one element

(c) contains exactly two elements

(d) contains exactly four elements

Answer: (c)

Solution:

S = {x Є R : x ≥ 0 and 2|√x − 3|+ √x (√x − 6) + 6 = 0}

Let t = √x

so, 2|t- 3| + t2 – 6t + 6 = 0

Case 1: When t < 3

-2t + 6 + t2 – 6t + 6 = 0

t = 2, 6

Since t < 3, t = 2 => x = 4

case 2:

When t ≥ 3

2t – 6 + t2 – 6t + 6 = 0

t = 0, 4

Since t ≥ 3, t = 4 => x = 16

Question 7: Let S be the set of all functions f:[0,1] -> R, which are continuous on [0, 1] and differentiable on (0, 1). Then for every f in S, there exists a C Є (0, 1), depending on f, such that:

(a) |f(1) − f(C)| = (1- C)|f′(C)|

(b) |f(C) + f(1)| < (1 + c)|f′(C)|

(c) |f(C) − f(1)| < (1 − C)|f′(C)|

(d) None of these

Answer: (d)

Solution:

Here S is set of all functions.

If we consider a constant function, then option (b) and (c) are incorrect.

For option (a):

f(1) − f(C) = (1 − C)f′(C)

This may not be true for f(x) = x2

None of the option are correct.

Question 8: If f(x) + 2 f(1/x) = 3x, x not equal to 0 and S = {x Є R : f(x) = f(- x)}; then S

(a) is an empty set

(b) contains exactly one element

(c) contains more than two elements

(d) contains exactly two elements

Answer: (d)

Solution: f(x) + 2 f(1/x) = 3x, x not equal to 0

Put x = 1/x, we have

f(1/x) + 2 f(x) = (3/x)

=> 2 f(x) + f(1/x) = 3/x

Multiply above equation by 2 and subtract given equation from it.

3f(x) = 6/x – 3x

or f(x) = 2/x – x

Again, we are given f(x) = f(-x)

2/x – x = -2/x + x

=> x = ±√2

Therefore,  S contains exactly 2 elements.

Solve More JEE Previous Year Questions: Sets Relations and Functions