Sets And Relations – JEE Main MathematicsPractice Questions & PYQs

The definition of reflexivity requires every element $A \in M$ to satisfy $A \cap A \neq \phi$, but the empty set $\phi \in M$ has $\phi \cap \phi = \phi$, failing this condition. Symmetry is satisfied because the intersection operation is commutative, meaning $A \cap B = B \cap A$, so $A \cap B \neq \phi$ implies $B \cap A \neq \phi$. Transitivity fails because one can construct sets $A, B, C$ such that $A \cap B \neq \phi$ and $B \cap C \neq \phi$ while $A \cap C = \phi$ by choosing $A = \{1\}, B = \{1, 2\}, C = \{2\}$.

An equivalence relation must satisfy the properties of reflexivity, symmetry, and transitivity. The first relation $R_1$ fails reflexivity because $a^{2+}a^2$ does not equal 1 for all real numbers, and it fails symmetry and transitivity as well. The second relation $R_2$ is equivalent to $a-b=c-d$, which inherently satisfies all three properties, making it the only equivalence relation among the two.

Applying the Principle of Inclusion-Exclusion, the union of the student sets plus the 10 students who studied none equals 210, establishing an equation for the triple intersection $x$. Substituting the given bounds for the individual subjects and intersection values leads to the inequality $15 \leq x \leq 45$, while the constraint that the number of students studying exactly two subjects must be non-negative further restricts the upper bound to 30, resulting in $m+n=15+30=45$.

The relation $R$ is defined by $2x=3y$, implying pairs of the form $(3k, 2k)$, which generates 33 distinct elements for $k \leq 33$. Because $R_1$ must be a symmetric relation containing $R$, every pair $(x, y)$ in $R$ necessitates its inverse $(y, x)$ to be included in $R_1$. Consequently, $R_1$ must contain all 33 original pairs from $R$ as well as their 33 reversed counterparts, resulting in a minimum of 66 elements.

A relation R is reflexive if $(a, b) R (a, b)$ holds for all $(a, b) imes A imes B$. Evaluating the condition $3ad-7bc$ for $(a, b) R (a, b)$ yields $3ab-7ba=-4ab$, which is an even integer, confirming reflexivity. Symmetry holds because $3ad-7bc ext{ is even} ext{ if and only if } ad+bc ext{ is even}$, and testing for transitivity with specific values such as $(3,4) R (6,4)$ and $(6,4) R (3,1)$ fails, as $(3,4) R (3,1)$ results in an odd value $3(3)(1)-7(4)(3)=-75$.

The number of subsets of a set with $k$ elements is $2^k$. Given the condition $2^m - 2^n = 56$, we factor it as $2^n(2^{m-n} - 1) = 8 \times 7$. Comparing terms, $2^n = 8$ gives $n=3$, and $2^{m-n} - 1 = 7$ implies $2^{m-n} = 8$, so $m-n = 3$, meaning $m = 6$. The distance between points $P(6, 3)$ and $Q(-2, -3)$ is calculated using the distance formula $\sqrt{(6 - (-2))^2 + (3 - (-3))^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$.

A relation is reflexive if $(a, b) R (a, b)$ holds, which is true here since $ab - ba = 0$ is divisible by $5$. It is symmetric because if $ad - bc$ is divisible by $5$, then $bc - ad$ is also divisible by $5$, satisfying $(c, d) R (a, b)$. However, it is not transitive because while $(3, 1) R (10, 5)$ and $(10, 5) R (1, 1)$ are valid as $3(5) - 1(10) = 5$ and $10(1) - 5(1) = 5$, the relation $(3, 1) R (1, 1)$ fails since $3(1) - 1(1) = 2$ is not divisible by $5$.

The inclusion-exclusion principle is used to find the number of elements meeting specific criteria by evaluating the sets of multiples. Counting the multiples of 3, 4, and their common multiple 12 in the range $[100, 700]$ gives 200, 151, and 50 elements respectively. Applying $n(3 imes 4) = n(3) + n(4) - n(3 imes 4)$ results in 301 integers divisible by either, which when subtracted from the total of 601 integers, leaves 300 that are not divisible by 3 or 4.

Reflexivity is satisfied because for any element in $N \times N$, the condition $x \leq x$ or $y \leq y$ is inherently true. Symmetry fails because while $(1, 1) R (2, 3)$ is true since $1 \leq 2$, the reverse pair $(2, 3) R (1, 1)$ is false as $2 \not\leq 1$ and $3 \not\leq 1$. Transitivity fails because the sequence $(2, 4) R (3, 3)$ and $(3, 3) R (1, 3)$ does not imply $(2, 4) R (1, 3)$, as $2 \not\leq 1$ and $4 \not\leq 3$.

The relation $R$ contains all pairs $(x,y)$ such that $4x imes 5y$ for $x,y imes {1, 2, 3, 4, 5}$, resulting in 16 ordered pairs. Symmetry requires that for every $(x,y) imes R$, the inverse $(y,x)$ must also be in $R$ if it is not already included. By evaluating each pair, we identify 9 inverses, such as $(2,1)$ and $(3,1)$, that must be added, yielding $m=16$ and $n=9$, so the sum $m+n=25$.

The number of elements in $R_1$ is determined by summing $\lfloor 20/a \rfloor$ for every $a$ from 1 to 20, yielding $20+10+6+5+4+3+2+2+2+2+1+1+1+1+1+1+1+1+1+1 = 66$. The intersection $R_1 \cap R_2$ comprises pairs where $b$ is a multiple of $a$ and $a$ is a multiple of $b$, which only occurs when $a=b$, resulting in 20 such pairs. Subtracting these 20 intersection pairs from 66 leaves 46 pairs that exist in $R_1$ but not in $R_2$.

The number of elements in the union of three sets is related to the individual set counts, pairwise intersections, and the triple intersection via the principle of inclusion-exclusion, where the sum of individual sets minus the sum of pairwise intersections plus the triple intersection equals the total union. Given $|A|=48, |B|=25, |C|=18$, and $|A \cup B \cup C|=60$, with $|A \cap B \cap C|=5$, the sum of pairwise intersections $\sum |A \cap B|$ is $48+25+18+5-60 = 36$. The number of people receiving medals in exactly two events is the sum of these pairwise intersections minus three times the number of people receiving all three medals, calculated as $36 - 3(5) = 21$.

Setting $2a+3b = 4c+5d = \alpha$ for $a,b,c,d \in \{1,2,3,4\}$, we evaluate possible values of $\alpha$ by calculating $2a+3b$ and $4c+5d$ for all combinations. The values of $2a+3b$ yield the set $\{5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20\}$ and $4c+5d$ yield $\{9, 13, 14, 17, 18, 19, 21, 22, 23, 24, 26, 27, 28, 31, 32, 36\}$. Comparing these sets reveals common values of $\alpha$ at $\{9, 13, 14, 17, 18\}$, and mapping these to specific pairs $(a,b)$ and $(c,d)$ identifies 6 valid instances where both expressions are equal.

A symmetric relation must contain the pair $(y, x)$ if $(x, y)$ is already in the relation. First, identify pairs in $R$: the odd positive differences occur between the 3 odd numbers from the set and the 5 even numbers, yielding 15 pairs where $x-y$ is a positive odd integer. Additionally, the condition $x-y=2$ provides 4 distinct pairs: $(6,4), (8,6), (10,8)$, and $(9,7)$, noting that $(9,7)$ is also covered by the odd difference rule, but the other three are not. Since none of these $15 + 4 = 19$ ordered pairs are symmetric within the existing set $R$, all 19 reverse pairs must be added to satisfy the symmetry requirement.

Classifying the elements of the set by their remainder when divided by 3 yields one element of the form $3k$, three of the form $3k+1$, and three of the form $3k+2$. By systematically counting subsets of size one through seven that satisfy the condition, we aggregate the valid combinations based on the distribution of these remainders. This combinatorial approach accounts for all nonempty subsets by grouping cases according to the number of elements chosen from each remainder category, confirming there are 43 such subsets.

For a relation to be reflexive and transitive on $A = \{1,2,3\}$ while containing $(1,2)$ and $(2,3)$, it must minimally contain the reflexive pairs $(1,1), (2,2), (3,3)$ and the transitive closure of the given elements, which is $\{(1,2), (2,3), (1,3)\}$. Starting with this base set $R_0 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}$, we can maintain these properties without achieving symmetry by adding only $(2,1)$, only $(3,2)$, or neither, as any further additions would eventually make the relation symmetric. Thus, the three distinct non-symmetric relations are $R_0$, $R_0 \cup \{(2,1)\}$, and $R_0 \cup \{(3,2)\}$.

An equivalence relation must satisfy reflexive, symmetric, and transitive properties. For reflexivity, consider $a=2$, where $\gcd(2, 2)=2 \neq 1$, failing the condition. For symmetry, $(2, 1) \in R$ because $\gcd(2, 1)=1$ and $4 \neq 1$, but $(1, 2) \notin R$ as $2(1)=2$ violates $2a \neq b$. Transitivity also fails because we can find triples such as $(14, 19)$ and $(19, 21)$ where the pairs $(14, 19)$ and $(19, 21)$ belong to $R$, yet $(14, 21) \notin R$ since $\gcd(14, 21)=7$.

A relation is defined by the condition $2(a-b)^2 + 3(a-b) \in \{0, 1, 2, 3, 4\}$. Let $k = a-b$. If $k=0$, the expression evaluates to $0$, which is in the set $B$, yielding 10 pairs where $a=b$. If $k=-2$, the expression evaluates to $2(4) + 3(-2) = 2$, which is also in set $B$, yielding 8 pairs where $a-b=-2$, specifically the pairs $(1,3), (2,4), \dots, (8,10)$. Any other integer value for $k$ results in an expression value outside the set $B$, meaning the total number of ordered pairs is $10 + 8 = 18$.

A relation is an equivalence relation if it is reflexive, symmetric, and transitive. For reflexivity, $2a+3a = 5a$ is clearly divisible by 5 for all $a$. Symmetry holds because $2a+3b$ being a multiple of 5 implies $2b+3a = 5(a+b) - (2a+3b)$, which is also a multiple of 5 since both terms are. Transitivity follows because if $2a+3b$ and $2b+3c$ are multiples of 5, their combination allows us to isolate $2a+3c$ as a multiple of 5 by eliminating the intermediate $b$ term, confirming the relation satisfies all criteria for an equivalence relation.

The number of elements in this relation is the product of the number of valid pairs for each independent condition defined for the ordered pairs. For $a_1$ dividing $b_2$, where $a_1 \in \{2,3,4\}$ and $b_2 \in \{8,9,12\}$, the valid pairs are $(2,8), (2,12), (3,9), (3,12), (4,8), (4,12)$, totaling 6 pairs. Similarly, for the condition $a_2$ dividing $b_1$, there are 6 possible pairs. Multiplying these independent counts yields $6 \times 6 = 36$ total elements in the relation.