JEE Advanced (Single Correct Type): Sets, Relation & Functions

Q.1. How many elements are there in the complement of set A?
(a) 0
(b) 1
(c) All the elements of A
(d) None of these

The complement of a set A will contain the elements that are not present in set A.

Q.2. If f(x) = x3 – (1/x3), then f(x) + f(1/x) is equal to
(a) 2x3
(b) 2/x3
(c) 0
(d) 1

Given,
f(x) = x3 – (1/x3)
Now,
f(1/x) = (1/x)3 – 1/(1/x)3
= (1/x3) – x3
Thus, f(x) + f(1/x) = x3 – (1/x3) + (1/x3) – x3 = 0

Q.3. A relation R in a set A is called _______, if (a1, a2) ∈ R implies (a2, a1) ∈ R, for all a1, a2 ∈ A.
(a) symmetric
(b) transitive
(c) equivalence
(d) non-symmetric

A relation R in a set A is called symmetric, if (a1, a2) ∈ R implies (a2, a1) ∈ R, for all a1, a2 ∈ A.

Q.4. Empty set is a _______.
(a) Infinite set
(b) Finite set
(c) Unknown set
(d) Universal set

The cardinality of the empty set is zero, since it has no elements. Hence, the size of the empty set is zero.

Q.5. Let n (A) = m, and n (B) = n. Then the total number of non-empty relations that can be defined from A to B is

(a) mn
(b) nm – 1
(c) mn – 1
(d) 2mn – 1

Given,
n(A) = m and n(B) = n
We know that,
n(A x B) = n(A). n(B) = mn
Total number of relations from A to B = Number of subsets of A x B = 2mn
So, the total number of non-empty relations from A to B = 2mn – 1.

Q.6. 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

Given that n divides m, ∀ n ∈ N, R is reflexive.
Let n = 3 and m = 6
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 also divide r.
So, the given relation is reflexive, transitive, but not symmetric.

Q.7. The number of elements in the Power set P(S) of the set S = {1, 2, 3} is:
(a) 4
(b) 8
(c) 2
(d) None of these

Number of elements in the set S = 3
Number of elements in the power set of set S = {1,2,3} = 23
= 8

Q.8. If f(x) = x2 + 2, x ∈ R, then the range of f(x) is
(a) [2, ∞)
(b) (-∞, 2]
(c) (2, ∞)
(d) (-∞, 2) U (2, ∞)

Given,
f(x) = x2 + 2
We know that the square of any number is positive, i.e. greater than or equal to 0.
So, x2 ≥ 0
x. + 2 ≥ 0 + 2
f(x) ≥ 2
Therefore, f(x) range is [2, ∞).

Q.9. The maximum number of equivalence relations on the set A = {1, 2, 3} are
(a) 1
(b) 2
(c) 3
(d) 5

Given, set A = {1, 2, 3}
The equivalence relations for the given set are:
R1 = {(1, 1), (2, 2), (3, 3)}
R2 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R3 = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}
R4 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}
R5 = {(1, 2, 3) ⇔ A x A = A2}
Therefore, the maximum number of an equivalence relation is ‘5’.

Q.10. Order of the power set P(A) of a set A of order n is equal to:
(a) n
(b) 2n
(c) 2n
(d) n2

The cardinality of the power set is equal to 2n, where n is the number of elements in a given set.

Q.11. What will be the domain for which the functions f(x) = 2x2 – 1 and g(x) = 1 – 3x are equal?
(a) {-2, 1}
(b) {1/2, -2}
(c) [2, 12]
(d) (-1, 2)

Given,
f(x) = 2x2 – 1
g(x) = 1 – 3x
Now,
f(x) = g(x)
⇒ 2x2 – 1 = 1 – 3x
⇒ 2x2 + 3x – 2 = 0
⇒ 2x2 + 4x – x – 2 = 0
⇒ 2x(x + 2) – 1(x + 2) = 0
⇒ (2x – 1) (x + 2) = 0
Thus the domain for which the function f (x) = g (x) is {1/2, -2}.

Q.12. If set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is
(a) 720
(b) 120
(c) 0
(d) none of these

Given,
n(A) = 5
n(B) = 6
Each element in set B is assigned to only one element in set A for the one-one function.
Here, only ‘5’ elements of set B are assigned to ‘5’ elements of set ‘A’ and one element will be left in set B.
The range of the function must be equal to B. However, for the given sets, it is not possible.
Thus, the range of functions does not contain all ‘6’ elements of set ‘ B’. Therefore, if the function is one-one it cannot be onto.
Hence, the number of one-one and onto mappings from A to B is 0.

Q.13. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, P = {1, 2, 5}, Q = {6, 7}. Then P ∩ Q’ is :
(a) P
(b) Q
(c) Q’
(d) None

Given,
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
P = {1, 2, 5}
Q = {6, 7}
Q’ = {1, 2, 3, 4, 5, 8, 9, 10}
Hence,
P ∩ Q’ = {1, 2, 5} = P

Q.14. If [x]2 – 5 [x] + 6 = 0, where [ . ] denotes the greatest integer function, then
(a) x ∈ [3, 4]
(b) x ∈ (2, 3]
(c) x ∈ [2, 3]
(d) x ∈ [2, 4)

Given,
[x]2 – 5 [x] + 6 = 0, where [ . ] denotes the greatest integer function.
[x]2 – 5[x] + 6 = 0
[x]2 – 2[x] – 3[x] + 6 = 0
[x]([x – 2) – 3([x] – 2) = 0
([x] – 2)([x] – 3) = 0
When [x] = 2, 2 ≤ x < 3
When [x] = 3, 3 ≤ x < 4
From the above, x ∈ [2, 4).

Q.15. Let f : [2, ∞) → R be the function defined by f(x) = x2 – 4x + 5, then the range of f is
(a) R
(b) [1, ∞)
(c) [4, ∞)
(d) [5, ∞)

Given,
f : [2, ∞) → R
f(x) = x2 – 4x + 5
Let f(x) = y
x2 – 4x + 5 = y
x2 – 4x + 4 + 1 = y
(x – 2)2 + 1 = y
(x – 2)2 = y – 1
x – 2 = √(y – 1)
x = 2 + √(y – 1)
If the function is a real-valued function, then y – 1 ≥ 0
So, y ≥ 1.
Therefore, the range is [1, ∞).

Q.16. The cardinality of the power set of {x : x ∈ N, x ≤ 10} is ______.
(a) 1024
(b) 1023
(c) 2048
(d) 2043

Given,
Set X = {x: x ∈ N, x ≤ 10}
X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Number of elements of power set of X, P(X) = 210 = 1024

Q.17. If f(x) = ax + b, where a and b are integers, f(–1) = – 5 and f(3) = 3, then a and b are equal to
(a) a = – 3, b = –1
(b) a = 2, b = – 3
(c) a = 0, b = 2
(d) a = 2, b = 3

Given,
f(x) = ax + b
And
f(-1) = -5
a(-1) + b = -5
-a + b = -5….(i)
Also, f(3) = 3
a(3) + b = 3
3a + b = 3….(ii)
From (i) and (ii),
a = 2, b = -3

Q.18. Let f : R → R be defined by f(x) = 1/x ∀ x ∈ R. Then f is
(a) one-one
(b) onto
(c) bijective
(d) f is not defined

f(x) = 1/x ∀ x ∈ R
Suppose x = 0, then f is not defined.
i.e. f(0) = 1/0 = undefined
So, the function f is not defined.

Q.19. Write X = {1, 4, 9, 16, 25,…} in set builder form.
(a) X = {x: x is a set of prime numbers}
(b) X = {x: x is a set of whole numbers}
(c) X = {x: x is a set of natural numbers}
(d) X = {x: x is a set of square numbers}

Given,
X = {1, 4, 9, 16, 25,…}
X = {12, 22, 32, 42, 52, …}
Therefore,
X = {x: x is a set of square numbers}

Q.20. The domain of the function f(x) = x/(x2 + 3x + 2) is
(a) [-2, -1]
(b) R – {1, 2}
(c) R – {-1, -2}
(d) R – {2}

Given f(x) is a rational function of the form g(x)/h(x), where g(x) = x and h(x) = x2 + 3x + 2.
Now h(x) ≠ 0
⇒ x2 + 3x + 2 ≠ 0
⇒ x2 + x + 2x + 2 ≠ 0
⇒ x(x + 2) + 2(x + 1)
⇒ (x + 1) (x + 2) ≠ 0
⇒ x ≠ -1, x ≠ -2
Therefore, the domain of the given function is R – {– 1, – 2}.

The document JEE Advanced (Single Correct Type): Sets, Relation & Functions | Chapter-wise Tests for JEE Main & Advanced is a part of the JEE Course Chapter-wise Tests for JEE Main & Advanced.
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## Chapter-wise Tests for JEE Main & Advanced

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## Chapter-wise Tests for JEE Main & Advanced

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