Sets And Relations – JEE Main MathematicsPractice Questions & PYQs

A relation is reflexive if it contains the identity pairs $(1,1), (2,2),$ and $(3,3)$. Given the inclusion of $(1,2)$ and $(2,3)$, transitivity necessitates the addition of $(1,3)$, forming the base set $\{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}$. This base set is reflexive and transitive but not symmetric, and we can generate additional valid relations by including $(3,2)$ or $(2,1)$ individually to the base, resulting in a total of 3 distinct relations that satisfy all the specified criteria.

The relation $R$ defined by $y = \max\{x, 1\}$ for $x \in \{-2, -1, 0, 1, 2, 3\}$ yields the ordered pairs $(-2,1), (-1,1), (0,1), (1,1), (2,2),$ and $(3,3)$, giving $l=6$. A relation is reflexive if it contains $(x,x)$ for all elements; here, the pairs $(-2,-2), (-1,-1),$ and $(0,0)$ must be added, so $m=3$. For symmetry, if $(x,y) \in R$, then $(y,x)$ must also be in $R$; the pairs $(1,-2), (1,-1),$ and $(1,0)$ are missing, so $n=3$, resulting in $l+m+n = 6+3+3 = 12$.

The relation $R$ consists of ordered pairs $(x,y)$ such that $0 \le x^2 + 2y \le 4$ where $x,y \in \{-3, -2, -1, 0, 1, 2, 3\}$. Calculating these pairs individually yields $R=\{(-3,-3), (-2,-2), (-2,-1), (-2,0), (-1,0), (-1,1), (0,0), (0,1), (0,2), (1,0), (1,1), (2,-2), (2,-1), (2,0), (3,-3)\}$, containing $15$ elements. To make $R$ reflexive, we must include the missing diagonal elements $(-1,-1), (2,2),$ and $(3,3)$, requiring $m=3$ additional elements. Summing the count of existing elements and the required additions gives $15+3=18$.

The number of subsets of the set $\{1, 2, \dots, n\}$ containing no two consecutive integers follows the Fibonacci-based recurrence relation $a_n = a_{n-1} + a_{n-2}$, where $a_n$ represents the count for a set of size $n$. Starting with $a_0=1$ for the empty set and $a_1=2$ for $\{\emptyset, \{1\}\}$, the sequence progresses as $a_2=3, a_3=5, a_4=8,$ and $a_5=13$. Consequently, there are 13 such subsets for the set $\{1, 2, 3, 4, 5\}$.

The relation $R$ is formed by identifying all pairs $(x, y)$ in the set $\{-3, -2, -1, 0, 1, 2, 3\}$ satisfying $2x-y = 0$ or $2x-y = 1$, which results in the 7 pairs: $\{(-1, -2), (-1, -3), (0, 0), (0, -1), (1, 1), (1, 2), (2, 3)\}$. To make $R$ reflexive, we must add 5 pairs of the form $(x, x)$ because $(0, 0)$ and $(1, 1)$ are already present. To make $R$ symmetric, we must add 5 additional pairs to account for the converse of each non-symmetric entry, resulting in $7 + 5 + 5 = 17$.

The relation $\log_e y = x \log_e (2/5)$ is equivalent to $y = (2/5)^x$ for $x \in \mathbb{N} \cup \{0\}$. As $x$ takes values $0, 1, 2, \dots$, the range consists of terms in the sequence $1, 2/5, (2/5)^2, \dots$. This is an infinite geometric series with first term $1$ and common ratio $2/5$, so the sum is given by $\frac{1}{1 - 2/5} = 5/3$.

For each prime $x \in S$, we count elements $y \in A$ such that $x$ divides $y$, where $A$ comprises all products of subsets of $S$. The divisibility condition is satisfied if $x$ is included in the set of prime factors used to form the product $y$. With 9 remaining primes in $S$ after fixing $x$, there are $2^9$ possible subsets forming products that include $x$. Summing over all 10 primes in $S$ gives $10 \times 2^9 = 5120$.

The set $A$ defines a rectangle in the $\alpha\beta$-plane spanning $-3 \le \alpha \le 5$ and $-1 \le \beta \le 11$. The set $B$ represents an ellipse given by the inequality $\frac{(\alpha-2)^2}{9} + \frac{(\beta-6)^2}{16} \le 1$, which implies that $\alpha$ ranges within $[2-3, 2+3] = [-1, 5]$ and $\beta$ ranges within $[6-4, 6+4] = [2, 10]$. Since the closed region of the ellipse is entirely contained within the boundaries of the rectangle defined by set $A$, it follows that $B \subset A$.

The definition $a_i = 2a_{i+1} + 1$ implies $a_1 = 2^k a_{k+1} + (2^k - 1)$ for a sequence of $k$ ordered pairs. Since $a_1 \le 100$, we have $2^k a_{k+1} + 2^k - 1 \le 100$, which simplifies to $2^k(a_{k+1} + 1) \le 101$. Testing integer values for $k$, we find that for $k=5$, $32(a_6+1) \le 101$ is possible when $a_6$ is $1$ or $2$, while $k=6$ would require $64(a_7+1) \le 101$, which is only possible if $a_7+1=1$, contradicting the requirement that $a_7 \ge 1$. Thus, the largest integer $k$ is $5$.

Equivalence relations require reflexivity, symmetry, and transitivity. Reflexivity holds as $x + x = 2x$ is always even. Symmetry holds because $x + y$ is even implies $y + x$ is even. Transitivity is confirmed as $x + y$ even and $y + z$ even imply $(x + y) + (y + z) = x + z + 2y$ is even, which forces $x + z$ to be even.

The number of elements in set $B$ is found by counting pairs $(m,n)$ such that $1 \le m < n \le 10$ and $\gcd(m,n)=1$. Enumerating these pairs for each $n$ from $2$ to $10$ gives $1$ for $n=2$, $2$ for $n=3$, $2$ for $n=4$, $4$ for $n=5$, $2$ for $n=6$, $6$ for $n=7$, $4$ for $n=8$, $6$ for $n=9$, and $4$ for $n=10$. Summing these counts $1+2+2+4+2+6+4+6+4$ results in $31$ elements.

To transform a relation into an equivalence relation, it must be reflexive, symmetric, and transitive. The original set $R$ requires adding $(1,1), (2,2), (3,3),$ and $(4,4)$ for reflexivity, $(2,1), (3,2),$ and $(3,1)$ to satisfy symmetry, and $(1,3)$ to ensure transitivity. Counting these necessary additions, we find that 7 distinct elements must be inserted into the original set $R$.

For a relation to be an equivalence relation, it must satisfy reflexivity, symmetry, and transitivity. Reflexivity is verified since $\sec^2 x - \tan^2 x = 1$ is a standard identity. Symmetry is shown by $\sec^2 x - \tan^2 y = 1 \implies \sec^2 y - \tan^2 x = 1$ using the identity $\sec^2 \theta - \tan^2 \theta = 1$. Transitivity follows from adding $\sec^2 x - \tan^2 y = 1$ and $\sec^2 y - \tan^2 z = 1$ to get $\sec^2 x - \tan^2 z + 1 = 2$, which confirms $x R z$.

The number of equivalence relations on a set corresponds to the number of ways to partition the set, which is given by the Bell number. For the set {1, 2, 3}, the distinct partitions are {{1, 2, 3}}, {{1, 2}, {3}}, {{1, 3}, {2}}, {{2, 3}, {1}}, and {{1}, {2}, {3}}. Counting these results in 5 possible non-empty equivalence relations.

The region $A \cap B$ represents the intersection of the exterior of the strip defined by $|x+y| \ge 3$ and the interior of the square defined by $|x|+|y| \le 3$. Within this intersection, we are restricted to cases where $x=0$ or $y=0$. Solving these conditions simultaneously with the inequalities leads to the boundary points $(-3,0), (3,0), (0,3),$ and $(0,-3)$. Summing the absolute values $|x+y|$ for these specific points yields $3 + 3 + 3 + 3 = 12$.

An equivalence relation partitions a set into disjoint equivalence classes. Since the relation $R$ connects $1$ to $2$, $2$ to $3$, and $1$ to $4$, the transitive and symmetric properties force all these elements into a single equivalence class. This means the equivalence relation $S$ must be the universal relation on the set $A = \{1, 2, 3, 4\}$, containing all $4^2 = 16$ possible ordered pairs.

A relation is symmetric if for every $(x,y) imes R$, the inverse $(y,x)$ is also in $R$. For $R_1$, determined by $2x-3y=2$, there are 6 ordered pairs, none of whose inverses satisfy the equation, requiring 6 additions. Similarly, $R_2$, determined by $-5x+4y=0$, contains 4 pairs whose inverses do not satisfy the equation, requiring 4 additions. Summing these requirements gives $M+N=10$.

The relation R consists of pairs $((a_1, b_1), (a_2, b_2))$ from $A imes B$ that satisfy $a_1+a_2=b_1+b_2$. By setting up the condition $a_1-b_1=b_2-a_2$, we evaluate the combinations of elements from $A$ and $B$ that satisfy this equality. Listing the valid pairs and verifying the sums reveals 25 distinct occurrences that meet the requirement.

An equivalence relation must be reflexive, symmetric, and transitive, so adding $(1, 2)$ and $(1, 3)$ to the set necessitates including their symmetric pairs $(2, 1)$ and $(3, 1)$, and transitivity implies $(2, 3)$ and $(3, 2)$ must also be included. Reflexivity requires the inclusion of the diagonal pairs $(1, 1), (2, 2), (3, 3), (4, 4)$. As element $4$ is not related to the others, it remains in its own singleton class $(4, 4)$, bringing the total number of ordered pairs to $9 + 1 = 10$.

The total number of symmetric relations on a set of $n$ elements is given by $2^{n(n+1)/2}$, while the number of symmetric and reflexive relations is $2^{n(n-1)/2}$. For $n=4$, the total number of symmetric relations is $2^{4(5)/2} = 2^{10} = 1024$. Subtracting the reflexive relations, which must include the $4$ diagonal elements, we calculate $2^{10} - 2^{4(3)/2} = 1024 - 2^6 = 1024 - 64 = 960$.