Sets Previous Year JEE Questions with Solutions are available here. 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 required formulas and basic concepts. Our subject experts have provided a detailed solution to each question. All questions are very important from an examination point of view. Students are advised to practice all the questions and understand the difficulty level of questions for the upcoming JEE exams.
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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, the intersection of sets, the complement of sets, Cartesian product of sets, the difference of sets, etc.
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) |
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JEE Main Maths Past Year Questions With Solutions on Sets
Question 1: 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 belongs to A and can be mapped to any of the n elements of B, which 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 2: Two sets, A and B, are as under:
A = {(a, b) ∈ R × R : |a − 5| < 1 and |b − 5| < 1};
B = {a, b) ∈ R × R : 4 (a − 6)2 + 9(b−5) 2 ≤ 36}. Then,
(a) A is a subset of B
(b) A n B = φ (null set)
(c) neither A subset of B nor B is a subset of A
(d) B is a subset of A
Answer: (a)
Solution:
A = {(a, b) ∈ R × R : |a − 5| < 1 and |b − 5| < 1}
⇒ a ∈ (4, 6) and b ∈ (4, 6)
Therefore, A = {(a, b): a ∈ (4, 6) and b ∈ (4, 6)}
Now, B = {a, b) ∈ R × R : 4 (a − 6) 2 + 9(b−5) 2 ≤ 36}
From the conditions above for 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 set A satisfy the conditions in Set B.
Hence, A is a subset of B.
Question 3: 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 4: If A, B and C are three sets such that A ∩ B = A ∩ 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) {from the given 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 5: Let S = {x ∈ 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 ∈ 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 6: 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 ∈ (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: (c)
Solution:
By LMVT, f'(C) = [(f(1) – f(C)]/(1 – C)
|(f(1) – f(C)|/(1- C) < |f'(C)|
|f(C)- f(1) |< (1-C)|f'(C)|
Question 7: If f(x) + 2 f(1/x) = 3x, x not equal to 0 and S = {x ∈ 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 the above equation by 2 and subtract the given equation from it.
3f(x) = 6/x – 3x
⇒ 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
Also, Check:
JEE Advanced Maths Chapter-wise Previous Year Questions With Solutions
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