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
Correct Answer is option (a)
The complement of a set A will contain the elements that are not present in set A.
Q.2. If f(x) = x^{3} – (1/x^{3}), then f(x) + f(1/x) is equal to
(a) 2x^{3}
(b) 2/x^{3}
(c) 0
(d) 1
Correct Answer is option (c)
Given,
f(x) = x^{3} – (1/x^{3})
Now,
f(1/x) = (1/x)^{3} – 1/(1/x)^{3}
= (1/x^{3}) – x^{3}
Thus, f(x) + f(1/x) = x^{3} – (1/x^{3}) + (1/x^{3}) – x^{3} = 0
Q.3. A relation R in a set A is called _______, if (a_{1}, a_{2}) ∈ R implies (a_{2}, a_{1}) ∈ R, for all a_{1}, a_{2} ∈ A.
(a) symmetric
(b) transitive
(c) equivalence
(d) nonsymmetric
Correct Answer is option (a)
A relation R in a set A is called symmetric, if (a_{1}, a_{2}) ∈ R implies (a_{2}, a_{1}) ∈ R, for all a_{1}, a_{2} ∈ A.
Q.4. Empty set is a _______.
(a) Infinite set
(b) Finite set
(c) Unknown set
(d) Universal set
Correct Answer is option (b)
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 nonempty relations that can be defined from A to B is
(a) m^{n}
(b) n^{m} – 1
(c) mn – 1
(d) 2^{mn} – 1
Correct Answer is option (d)
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 = 2^{mn}
So, the total number of nonempty relations from A to B = 2^{mn} – 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
Correct Answer is option (d)
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
Correct Answer is option (b)
Number of elements in the set S = 3
Number of elements in the power set of set S = {1,2,3} = 2^{3}
= 8
Q.8. If f(x) = x^{2} + 2, x ∈ R, then the range of f(x) is
(a) [2, ∞)
(b) (∞, 2]
(c) (2, ∞)
(d) (∞, 2) U (2, ∞)
Correct Answer is option (a)
Given,
f(x) = x^{2} + 2
We know that the square of any number is positive, i.e. greater than or equal to 0.
So, x^{2} ≥ 0
Adding 2 on both sides,
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
Correct Answer is option (d)
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) 2^{n}
(d) n^{2}
Correct Answer is option (c)
The cardinality of the power set is equal to 2^{n}, where n is the number of elements in a given set.
Q.11. What will be the domain for which the functions f(x) = 2x^{2} – 1 and g(x) = 1 – 3x are equal?
(a) {2, 1}
(b) {1/2, 2}
(c) [2, 12]
(d) (1, 2)
Correct Answer is option (b)
Given,
f(x) = 2x^{2} – 1
g(x) = 1 – 3x
Now,
f(x) = g(x)
⇒ 2x^{2} – 1 = 1 – 3x
⇒ 2x^{2} + 3x – 2 = 0
⇒ 2x^{2} + 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 oneone and onto mappings from A to B is
(a) 720
(b) 120
(c) 0
(d) none of these
Correct Answer is option (c)
Given,
n(A) = 5
n(B) = 6
Each element in set B is assigned to only one element in set A for the oneone 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 oneone it cannot be onto.
Hence, the number of oneone 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
Correct Answer is option (a)
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)
Correct Answer is option (d)
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) = x^{2} – 4x + 5, then the range of f is
(a) R
(b) [1, ∞)
(c) [4, ∞)
(d) [5, ∞)
Correct Answer is option (b)
Given,
f : [2, ∞) → R
f(x) = x^{2} – 4x + 5
Let f(x) = y
x^{2} – 4x + 5 = y
x^{2} – 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 realvalued 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
Correct Answer is option (a)
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) = 2^{10} = 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
Correct Answer is option (b)
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) oneone
(b) onto
(c) bijective
(d) f is not defined
Correct Answer is option (d)
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}
Correct Answer is option (d)
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/(x^{2} + 3x + 2) is
(a) [2, 1]
(b) R – {1, 2}
(c) R – {1, 2}
(d) R – {2}
Correct Answer is option (c)
Given f(x) is a rational function of the form g(x)/h(x), where g(x) = x and h(x) = x^{2} + 3x + 2.
Now h(x) ≠ 0
⇒ x^{2} + 3x + 2 ≠ 0
⇒ x^{2} + 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}.
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