Sets And Relations – JEE Main MathematicsPractice Questions & PYQs

The total number of elements in the relation is the product of the count of valid pairs $(a_1, b_2)$ such that $a_1 rianglelefteq b_2$ and the count of valid pairs $(b_1, a_2)$ such that $b_1 rianglelefteq a_2$. For $a_1 rianglelefteq b_2$, with $a_1 rianglelefteq \{1, 3, 4, 6, 9\}$ and $b_2 rianglelefteq \{2, 4, 5, 8, 10\}$, counting the valid combinations for each $a_1$ gives $5+4+4+2+1 = 16$ pairs. Similarly, for $b_1 rianglelefteq a_2$, with $b_1 rianglelefteq \{2, 4, 5, 8, 10\}$ and $a_2 rianglelefteq \{1, 3, 4, 6, 9\}$, the valid pairs total $4+3+2+1 = 10$. Multiplying these independent counts results in $16 \times 10 = 160$ elements in the relation.

The relation $R_1$ is defined such that the symmetric difference of sets $A$ and $B$ is empty, which implies $A=B$, making it trivially reflexive, symmetric, and transitive. For $R_2$, the condition $A \cup B^c = B \cup A^c$ requires that any element present in $A$ must also be in $B$, and vice versa, because if an element were in $A$ but not $B$, it would create a contradiction between the two sides of the equation. Since both relations effectively force the sets $A$ and $B$ to be identical, they satisfy all properties of an equivalence relation.

To ensure symmetry, we must include the reverse of each existing pair, adding $(b,a)$ and $(c,b)$ to the initial set. Transitivity requires adding $(a,c)$ to connect $a$ to $c$ through $b$, and subsequently adding $(c,a)$ to maintain symmetry. Finally, for the relation to be defined over the set as an equivalence relation, the reflexive pairs $(a,a)$, $(b,b)$, and $(c,c)$ must also be included. Summing these additions results in a total of 7 new elements.

A relation is reflexive if every element relates to itself, symmetric if the existence of $(x, y)$ implies $(y, x)$, and transitive if $(x, y)$ and $(y, z)$ imply $(x, z)$. The relation $R = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$ lacks reflexivity because $(1, 1) \notin R$ as $1+1 \neq 7$. It satisfies symmetry because every ordered pair has its reverse present within the set. However, it fails transitivity because while $(1, 6) \in R$ and $(6, 1) \in R$, the required pair $(1, 1)$ is not included.

The condition $3^n - 3$ is a multiple of $7$ is equivalent to $3^n \equiv 3 \pmod 7$. Evaluating the powers of $3$ modulo $7$ reveals a cyclic sequence of length $6$: $3^1 \equiv 3, 3^2 \equiv 2, 3^3 \equiv 6, 3^4 \equiv 4, 3^5 \equiv 5, 3^6 \equiv 1$, meaning $3^n \equiv 3 \pmod 7$ holds when $n$ is of the form $6k+1$. Solving the inequality $10 \leq 6k+1 \leq 100$ results in $1.5 \leq k \leq 16.5$, where $k$ can be any integer in the range from $2$ to $16$, yielding exactly $15$ values.

The relation $T$ is symmetric because if $a^{2-}b^2$ is an integer $n$, then $b^{2-}a^2$ is $-n$, which is also an integer, ensuring that $(b, a) \in T$ whenever $(a, b) \in T$. Conversely, the relation $S$ is not symmetric because the condition $2+a/b > 0$ is sensitive to the order of division; for instance, if $a=1$ and $b=-3$, the pair satisfies the condition since $2-1/3 > 0$, but reversing the pair yields $2+(-3/1) = -1$, which is not greater than zero. Thus, $T$ is symmetric while $S$ is not.

An equivalence relation on the set containing four elements must connect every element to every other element, requiring all $4^2 = 16$ possible ordered pairs to be present. Given that the initial set contains three pairs, we must calculate the difference between the complete set of 16 required pairs and the three existing ones. This process requires adding the four reflexive pairs, the symmetric inverses for the given pairs, and all necessary transitive links, resulting in the addition of 13 specific elements.

For a relation to be reflexive, the condition must hold when $a=b$, which results in $3(0) + \sqrt{7} = \sqrt{7}$, an irrational number, confirming reflexivity. Symmetry fails because if $3a-3b+\sqrt{7}$ is irrational, the reversed expression $3b-3a+\sqrt{7}$ can result in a rational number if the irrational components cancel out, such as when $a-b = \sqrt{7}/3$. Transitivity also fails because one can find values for $a, b,$ and $c$ such that $(a, b) \in R$ and $(b, c) \in R$, but the resulting $3a-3c+\sqrt{7}$ simplifies to a rational number, thereby excluding $(a, c)$ from the relation.

A relation $R$ is reflexive if $(x,x) \in R$ for all $x \in A$ and symmetric if $(x,y) \in R \implies (y,x) \in R$. The current relation $R = \{(-4,4), (-3,3), (0,0), (0,1), (1,1), (3,3), (3,-2), (4,4)\}$ is missing $(-4,-4), (-3,-3), (-2,-2)$ to be reflexive, which constitutes 3 additions. To satisfy symmetry, we must add the inverses of existing pairs that do not already have them, specifically $(4,-4), (3,-3), (-2,3), (1,0)$, adding 4 more elements, which brings the minimum total additions to 7.

To determine the properties of the relation, evaluate the defining condition $ad(b-c) = bc(a-d)$ against the axioms of an equivalence relation. Symmetry holds because swapping $(a,b)$ and $(c,d)$ yields the same equality after algebraic manipulation. Reflexivity fails because the condition becomes $ab(b-a) = ba(a-b)$, which simplifies to $ab(b-a) = -ab(b-a)$, a statement that is not true for all natural numbers. Transitivity also fails, as specific values can demonstrate that while the relation holds for intermediate pairs, it does not hold between the first and last elements.

Solving the inequality $-6 < n^{2-}10n+19 < 6$ involves treating the two boundaries separately. The left inequality $n^{2-}10n+19 > -6$ simplifies to $(n-5)^2 > 0$, which holds for all integers $n \neq 5$. The right inequality $n^{2-}10n+13 < 0$ has roots at $n = 5 \pm \sqrt{12}$, approximately $1.54$ and $8.46$, meaning $n$ must be in the range $\{2, 3, 4, 5, 6, 7, 8\}$. Removing $n=5$ from this set leaves $\{2, 3, 4, 6, 7, 8\}$, resulting in 6 valid integer values.

A relation $R$ is reflexive if every element relates to itself, symmetric if $aRb$ implies $bRa$, and transitive if $aRb$ and $bRc$ imply $aRc$. Given $R = \{(x, y) : x \in A_i \iff y \in A_i\}$, the definition of logical equivalence ensures reflexivity because $x \in A_i$ is always equivalent to itself. Symmetry follows because the biconditional operator is commutative, meaning $x \in A_i \iff y \in A_i$ is identical to $y \in A_i \iff x \in A_i$. Transitivity holds because if $x \in A_i \iff y \in A_i$ and $y \in A_i \iff z \in A_i$, then by the transitivity of logical equivalence, $x \in A_i \iff z \in A_i$, which defines an equivalence relation.

An equivalence relation must satisfy the transitive property for all elements. For $R_1$, consider $(2, 11) \in R_1$ and $(11, 19) \in R_1$ because the absolute differences are $9$ and $8$, respectively, both being $\leq 13$, yet the absolute difference between $2$ and $19$ is $17$, which is not $\leq 13$, failing transitivity. For $R_2$, consider $(13, 3) \in R_2$ and $(3, 26) \in R_2$ because their differences are not $13$, but the absolute difference $|13 - 26| = 13$, meaning $(13, 26) \notin R_2$, also failing transitivity. Since neither relation is transitive, neither qualifies as an equivalence relation.

An equivalence relation requires reflexivity, symmetry, and transitivity. $R_1$ is reflexive as $a^2 \geq 0$ and symmetric as $ab \geq 0 \iff ba \geq 0$, but it fails transitivity since $2 R_1 0$ and $0 R_1 -2$ are true, yet $2 R_1 -2$ is false because $2 \times -2 < 0$. $R_2$ is reflexive ($a \geq a$) and transitive ($a \geq b, b \geq c \implies a \geq c$), but it fails symmetry because $5 \geq 2$ does not imply $2 \geq 5$. Since neither relation satisfies all three properties, neither is an equivalence relation.

The relation $R_1$ contains pairs $(p, p^n)$ for any prime $p$ where $p^n \leq 50$ and $n \geq 0$, while $R_2$ restricts these to $n = 0$ or $1$. The set $R_1 \setminus R_2$ consists of pairs where $n \geq 2$ and $p^n \leq 50$. Listing these for each prime $p$, we find $(2, 2^2=4), (2, 2^3=8), (2, 2^4=16), (2, 2^5=32)$ for $p=2$, $(3, 3^2=9), (3, 3^3=27)$ for $p=3$, $(5, 5^2=25)$ for $p=5$, and $(7, 7^2=49)$ for $p=7$. Counting these distinct pairs results in a total of $8$ elements.

Any number representable as $4x+6y+9z$ for non-negative integers $x, y, z$ forms the set $A$. Specifically, 9 is in $S$, 10 is $4+6$, 11 cannot be formed by any combination of these numbers, and 12 is $4+4+4$. For any number $n \geq 12$, combinations exist because 4, 6, and 9 can combine to cover all subsequent integers. Thus, $T-A$ contains only the value 11, making the sum of elements equal to 11.

For a relation to be reflexive, the condition $(a, a) \in R$ must hold for all $a \in \mathbb{N}$, implying $3a + \alpha a = a(3+\alpha)$ must be a multiple of 7. This requires $3+\alpha = 7k$, or $\alpha = 7k-3$, which is equivalent to $\alpha \equiv 4 \pmod 7$. Given this condition, symmetry holds because $3a + \alpha b \equiv 0 \implies 3b + \alpha a \equiv 0 \pmod 7$ is satisfied through modular arithmetic, and transitivity follows from the additive property of linear congruences. Thus, the condition for an equivalence relation is that 4 is the remainder when $\alpha$ is divided by 7.

A set $C$ that does not intersect $B$ must consist entirely of elements from $A$ that are not in $B$. The total number of subsets of $A$ is $2^{7} = 128$, while the number of subsets that avoid $B$ is determined by the elements in $A \setminus B$, which are $\{1, 2, 4, 5\}$. Since there are $4$ such elements, there are $2^{4} = 16$ subsets that satisfy $C \cap B = \phi$. Subtracting these disjoint cases from the total yields $128 - 16 = 112$.

Integers $\alpha$ coprime to 24 must not be divisible by 2 or 3, meaning they follow a periodic pattern every 24 units. In the first 24 integers, the sum of such numbers is 96, and because there are four full blocks of 24 up to 96, we compute the sum as $96 + 288 + 480 + 672 + 97 = 1633$.

The relation depends entirely on the product of two primes $p, q \geq 3$. Listing these products $b \leq 60$ yields $3 \times 3=9, 3 \times 5=15, 3 \times 7=21, 3 \times 11=33, 3 \times 13=39, 3 \times 17=51, 3 \times 19=57, 5 \times 5=25, 5 \times 7=35, 5 \times 11=55, 7 \times 7=49$. There are exactly 11 such values for $b$. Since $a$ can be any of the 60 numbers in the set, each $b$ pairs with 60 values of $a$, resulting in $60 \times 11 = 660$ total elements.