Jamia Millia Islamia Sets and Relations Previous Year Question Paper and Solution with PDF

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Jamia Millia Islamia Sets And Relations PYQ

Qus : 1 Jamia Millia Islamia PYQ

Points within a set are connected by a line segment must follow the condition that points are

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

In a convex set, any line joining two points lies entirely within the set.

Qus : 2 Jamia Millia Islamia PYQ

$\text{The set }(A\cap B)'\ \cup\ (B\cap C)\ \text{ is equal to:},$

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Solution

$ (A\cap B)'=A'\cup B' ,$ (De Morgan) $\Rightarrow (A\cap B)'\cup(B\cap C)=(A'\cup B')\cup(B\cap C)=A'\cup\big(B'\cup(B\cap C)\big)$ $B'\cup(B\cap C)=(B'\cup B)\cap(B'\cup C)=U\cap(B'\cup C)=B'\cup C$ $\Rightarrow A'\cup(B'\cup C)=A'\cup B'\cup C$

Qus : 3 Jamia Millia Islamia PYQ

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 $e=6$, $2×6=12\equiv2,\ 4×6=24\equiv4,\ 8×6=48\equiv8.$

Qus : 4 Jamia Millia Islamia PYQ

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.

Qus : 5 Jamia Millia Islamia PYQ

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.

Qus : 6 Jamia Millia Islamia PYQ

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.

Qus : 7 Jamia Millia Islamia PYQ

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 $B_3 = 5$. $\boxed{\text{Answer: (D) 5}}$

Qus : 8 Jamia Millia Islamia PYQ

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,

|P(S)| = 2^{|S|} = 2^4 = 16

Qus : 9 Jamia Millia Islamia PYQ

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

Qus : 10 Jamia Millia Islamia PYQ

Suppose that $A_i = {1, 2, 3, \ldots, i}$ for $i = 1, 2, 3, \ldots$ Then find $\displaystyle \bigcup_{i=1}^{\infty} A_i = ?$ Here $Z$ denotes the set of integers.

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Solution

$A_{i} = {1, 2, 3, \dots, i} (i = 1, 2, \dots)$ $\bigcup_{i=1}^{\infty}A_i=\{1,2,3,\dots\}= \mathbb Z^{+}$ $\boxed{\mathbb Z^{+}}$

Qus : 11 Jamia Millia Islamia PYQ

$,\text{Find } \displaystyle \bigcup_{i=1}^{\infty} A_i \text{ and } \bigcap_{i=1}^{\infty} A_i,\ \text{for every positive integer } i \text{ where } A_i={-i,i}.$ (Here $\mathbb Z$ denotes the set of integers.)

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Solution

$A_1={-1,1},\ A_2={-2,2},\ldots$ so $\displaystyle \bigcup_{i=1}^{\infty}A_i={\ldots,-3,-2,-1,1,2,3,\ldots}=\mathbb Z\setminus{0}$, $\displaystyle \bigcap_{i=1}^{\infty}A_i=\varnothing$.

Qus : 12 Jamia Millia Islamia PYQ

Which of the following relations are functions? (i) ${(1,(a,b)),\ (2,(b,c)),\ (3,(c,a)),\ (4,(a,b))}$ (ii) ${(1,(a,b)),\ (2,(b,a)),\ (3,(c,a)),\ (1,(a,c))}$ (iii) ${(1,(a,b)),\ (2,(a,b)),\ (3,(a,b))}$ (iv) ${(1,(a,b)),\ (2,(b,c)),\ (1,(c,a))}$

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Solution

A relation $R$ is a function iff every first component occurs exactly once. (i) first components $1,2,3,4$ appear once each $\Rightarrow$ function. (ii) $1$ appears twice $\Rightarrow$ not a function. (iii) $1,2,3$ appear once each $\Rightarrow$ function. (iv) $1$ appears twice $\Rightarrow$ not a function.

Qus : 13 Jamia Millia Islamia PYQ

$(A\cap B')\ \cup\ (A'\cap B)\ \cup\ (A'\cap B')$ is equal to

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

$(A'\cap B)\cup(A'\cap B')=A'\cap(B\cup B')=A'$. Then $A'\cup(A\cap B')=(A'\cup A)\cap(A'\cup B')=\mathbf U\cap(A'\cup B')=A'\cup B'$.

Qus : 14 Jamia Millia Islamia PYQ

The number of students who take both the subjects’ mathematics and chemistry are 30.

This represents 10% of the enrolment in mathematics and 12% of enrolment in chemistry.

How many students take at least one of these two subjects?

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

Let total mathematics students be $M$, chemistry students be $C$, and both be $30$. \[ 0.1M=30 \Rightarrow M=300,\qquad 0.12C=30 \Rightarrow C=250. \] At least one $= M+C-\text{both}=300+250-30=520$.

Qus : 15 Jamia Millia Islamia PYQ

Let $A$ and $B$ be two disjoint subsets of a universal set $E$. Then $(A \cup B)\cap B'$ is equal to

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

Since $A$ and $B$ are disjoint, elements common with $B'$ are only from $A$. Hence $(A \cup B)\cap B' = A.$

Qus : 16 Jamia Millia Islamia PYQ

$(A - B) - A$ is equal to

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Solution

$(A-B)$ means elements in $A$ but not in $B$. Subtracting $A$ again gives an empty set.

Qus : 17 Jamia Millia Islamia PYQ

Let 10 is the cardinality of set A. The number of bijective mappings from set A to itself is

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Solution

Bijections on a 10-element set = number of permutations = 10. So, 10 = 3628800.

Qus : 18 Jamia Millia Islamia PYQ

Let cardinality of set A and B are 2 and 5 respectively. The number of relations from A to B is

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Solution

$|A\times B| = 2\cdot5=10$. A relation is any subset of $A\times B$. Count = $2^{10}=1024$.

Qus : 19 Jamia Millia Islamia PYQ

Let cardinality of A and B are 3 and 10 respectively. The number of one-to-one functions from A to B is

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Solution

Injective maps from 3 to 10 = permutations $P(10,3)=10\cdot9\cdot8=720$.

Qus : 20 Jamia Millia Islamia PYQ

Let $A=\{1,2,3,4\}$ and $B=\{a,b\}$. The number of surjective mappings from A to B is

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Solution

Onto functions to a 2-element codomain: $2^{4}-\binom{2}{1}1^{4}=16-2=14$ (I.E. principle).

Qus : 21 Jamia Millia Islamia PYQ

The relation represented by $R = \{(1,1), (2,2), (3,3), (1,3), (3,2), (1,2)\}$ on the set $A = \{1, 2, 3\}$ is what kind of relation?

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Solution

R includes $(1,1), (2,2), (3,3)$ → Reflexive. Check symmetry: $(1,2)$ exists but $(2,1)$ does not → Not symmetric. Check transitivity: $(1,2)$ and $(2,2)$ imply $(1,2)$ already → transitive holds. Hence, relation is reflexive and transitive but not symmetric.

Qus : 22 Jamia Millia Islamia PYQ

For $a, b \in \mathbb{R}$ define $a = b$ to mean that $|x| = |y|$. If $[x]$ is an equivalence relation in $R$, then the equivalence relation for $[17]$ is...

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Solution

Given relation $a = b \iff |a| = |b|$. Then equivalence class of $17$ is $[17] = \{x \in \mathbb{R} : |x| = |17|\} = \{-17, 17\}.$

Qus : 23 Jamia Millia Islamia PYQ

The sets A and B have the same cardinality if and only if there is a ..... correspondence from A to B.

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

Two sets have the same cardinality if there exists a one-to-one and onto (bijective) correspondence between them.

Qus : 24 Jamia Millia Islamia PYQ

The relation $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ on $A=\{1,2,3\}$ is:

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Solution

Reflexive — yes (all $(a,a)$ are present). Symmetric — no, since $(1,2)\in R$ but $(2,1)\notin R$. Transitive — yes, because $(1,2)$ and $(2,3)$ imply $(1,3)$ (which exists).

Qus : 25 Jamia Millia Islamia PYQ

For real numbers $x,y$, define $x\,R\,y$ iff $x-y+\sqrt{2}$ is irrational. Then the relation $R$ is:

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Solution

For reflexivity: $xRx$ means $x-x+\sqrt{2}=\sqrt{2}$ (irrational) ⇒ true. For symmetry: $xRy⇒x-y+\sqrt{2}$ irrational, but $y-x+\sqrt{2}=-(x-y)+\sqrt{2}$ may be rational. Not always true ⇒ not symmetric. For transitivity: fails similarly.

Qus : 26 Jamia Millia Islamia PYQ

If $A$ be a set of cardinality $n$, then number of one-to-one onto functions from set $A$ to $A$ is …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

Number of bijective (one-one and onto) functions from a set of $n$ elements to itself $= n!$.

Qus : 27 Jamia Millia Islamia PYQ

If $R$ is the largest equivalence relation on a set $A$ and $S$ is any relation on $A$, then:

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Solution

Largest equivalence on $A$ is $A\times A$ (universal relation), so every relation $S$ on $A$ satisfies $S\subseteq A\times A=R$.

Qus : 28 Jamia Millia Islamia PYQ

If $f$ is a function from a finite set $A$ having $10$ elements to a finite set $B$ having $5$ elements, then number of functions from $A$ to $B$ is …

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Solution

Each of $10$ elements in $A$ can be mapped to any of $5$ elements in $B$. Hence total functions $= 5^{10}$.

Qus : 29 Jamia Millia Islamia PYQ

If $A$ and $B$ are two sets, then $(A \cup B)' \cap B$ is equal to …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

$(A \cup B)'$ means elements not in $A$ or $B$. Hence, $(A \cup B)' \cap B$ contains elements that are both in $B$ and not in $B$, i.e., none. So result is $\phi$.

Qus : 30 Jamia Millia Islamia PYQ

If $A = \{a, b, c\}$ and $B = \{a, b, d, e, f\}$ are two sets, then number of elements in $(A - B) \times (A \cap B)$ is …

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Solution

$A - B = \{c\}$ (only $c$ not in $B$). $A \cap B = \{a, b\}$. Then $(A - B) \times (A \cap B) = \{(c, a), (c, b)\}$ → 2 elements.

Qus : 31 Jamia Millia Islamia PYQ

If $A$ and $B$ are two disjoint sets having $3$ and $5$ elements respectively, then power-set of $A \times (B - A)$ contains … elements.

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Solution

Since $A$ and $B$ are disjoint, $B - A = B$. Then $A \times B$ has $3 \times 5 = 15$ ordered pairs. Power-set of it will have $2^{15}$ elements.

Qus : 32 Jamia Millia Islamia PYQ

Let A={1,2,3} and R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}. Then R is

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Solution

$A={1,2,3},; R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}$

Reflexive: $(1,1),(2,2),(3,3)\in R\Rightarrow$ reflexive.

Symmetric: $(1,2)\in R$ but $(2,1)\notin R\Rightarrow$ not symmetric.

Transitive: check the only nontrivial chain: $(1,2)$ and $(2,3)\Rightarrow (1,3)$, which is in $R$. Pairs with $(x,x)$ keep the second pair; $(1,3)$ can only compose with $(3,3)$ giving $(1,3)$; $(2,3)$ with $(3,3)$ gives $(2,3)$. All required compositions are in $R$. $\boxed{\text{$R$ is reflexive and transitive, but not symmetric.}}$

Qus : 33 Jamia Millia Islamia PYQ

On A={1,2,3} let R={(1,2)}. Then R is

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Solution

Not reflexive (missing $(1,1),(2,2),(3,3)$). Not symmetric (since $(2,1)\notin R$). Transitive holds vacuously because there is no pair starting at $2$ to trigger $(1,2)$∘$(2,\cdot)$.

Qus : 34 Jamia Millia Islamia PYQ

A={1,2,3}. Which $f:A\to A$ does not have an inverse?

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Solution

A, C, and D are bijections (permutations), hence invertible. In B, both $2$ and $3$ map to $1$ (not one-to-one), so not bijective $\Rightarrow$ no inverse.

Qus : 35 Jamia Millia Islamia PYQ

The maximum number of equivalence relations on the set $A = {1, 2, 3}$ are

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Solution

Number of equivalence relations = Number of partitions of the set = Bell number $B_3 = 5$.

Qus : 36 Jamia Millia Islamia PYQ

A relation $R$ in a set $A$ is called ________, if $(a_1, a_2) \in R$ implies $(a_2, a_1) \in R$, for all $a_1, a_2 \in A$.

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Solution

Qus : 37 Jamia Millia Islamia PYQ

Power set of empty set has exactly _____ subset.

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Solution

Power set of $\phi$ = ${\phi}$ has 1 subset.

Qus : 38 Jamia Millia Islamia PYQ

What is the cardinality of the power set of ${0,1,2}$?

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Solution

For a set of $n$ elements, power set size $= 2^n = 2^3 = 8.$

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