Jamia Millia Islamia MCA Sets And Relations Previous Year Questions (PYQs) – Page 1 of 5
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Showing 41 questions
Points within a set are connected by a line segment must follow the condition that points are
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Solution
In a convex set, any line joining two points lies entirely within the set.
Order of the power set $ P(A) $ of a set $ A $ of order $ n $ is:
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Solution
If
$ |A| = n $
Then
$ |P(A)| = 2^n $
Number of subsets $ = 2^n $
$\text{The set }(A\cap B)'\ \cup\ (B\cap C)\ \text{ is equal to:},$
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Solution
$ (A\cap B)'=A'\cup B' ,$ (De Morgan)
$\Rightarrow (A\cap B)'\cup(B\cap C)=(A'\cup B')\cup(B\cap C)=A'\cup\big(B'\cup(B\cap C)\big)$
$B'\cup(B\cap C)=(B'\cup B)\cap(B'\cup C)=U\cap(B'\cup C)=B'\cup C$
$\Rightarrow A'\cup(B'\cup C)=A'\cup B'\cup C$
In the group G = {2, 4, 6, 8} under multiplication modulo 10, the identity element is
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Solution
$(a × e) \bmod 10 = a.$
For $e=6$, $2×6=12\equiv2,\ 4×6=24\equiv4,\ 8×6=48\equiv8.$
A partition of {1, 2, 3, 4, 5} is the family
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Solution
A partition divides a set into disjoint non-empty subsets covering all elements.
{{1,2},{3,4,5}} satisfies it.
Let P(S) denote the power set of set S. Which of the following is always TRUE?
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Solution
A partition divides a set into disjoint non-empty subsets covering all elements.
{{1,2},{3,4,5}} satisfies it.
G = {a, b, c} is an abelian group with ‘e’ as identity element. The order of other elements is
A. 2, 2, 3 B. 3, 3, 3 C. 2, 2, 4 D. 2, 2, 2
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Solution
With 3 elements, it must be cyclic of order 3.
The maximum number of equivalence relations on the set $A = \{1,2,3\}$ is:
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Solution
Number of equivalence relations = Number of partitions of set $A$.
For 3 elements, Bell number $B_3 = 5$.
$\boxed{\text{Answer: (D) 5}}$
How many elements does the set
$P\big({\varnothing, a, {a}, {{a}}}\big)$ has;
where $a$ and $b$ are distinct elements and $P$ denotes the power set?
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Solution
The given set ${\varnothing, a, {a}, {{a}}}$ has $4$ elements.
Therefore,
What is the cardinality of these sets in the order of their serial number?
(i) ${a}$
(ii) ${{a}}$
(iii) ${a, {a}}$
(iv) ${a, {a}, {{a}}}$
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Solution
(i) ${a}$ → 1 element
(ii) ${{a}}$ → 1 element
(iii) ${a, {a}}$ → 2 elements
(iv) ${a, {a}, {{a}}}$ → 3 elements
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