Sets And Relations – JEE Main MathematicsPractice Questions & PYQs

The number of elements in the relation $R$ is determined by the condition that $a_1 + b_2$ must divide $a_2 + b_1$ for pairs formed from the sets $A = \{1, 4, 7\}$ and $B = \{2, 3, 8\}$. By calculating all possible sums for $a_i + b_j$ and systematically checking the divisibility constraint for every combination of pairs $((a_1, b_1), (a_2, b_2))$, we count the occurrences that satisfy the divisibility rule. This verification across the set of all potential pairs yields a total of 18 valid elements.

A relation is an equivalence relation if it satisfies the reflexive, symmetric, and transitive properties, which occurs here because $x \equiv y \pmod 3$ partitions the set into congruence classes. The relation $R$ consists of pairs $(x,y)$ such that $|x-y|$ is a multiple of 3, leading to the classes $\{0,3,6,9\}$, $\{1,4,7\}$, and $\{2,5,8\}$, which contain 4, 3, and 3 elements respectively. The total number of elements $n(R)$ is calculated as $4^2 + 3^2 + 3^2 = 34$, which means Statement I is incorrect, while the defined relation correctly satisfies the requirements for an equivalence relation.

The number of integer pairs $(x, y)$ satisfying the inequality $4x^2 + y^2 < 52$ is calculated by testing each possible integer value for $x$. For $x=0$, $y^2 < 52$ allows 15 possible integer values for $y$; for $x=\pm 1$, $y^2 < 48$ allows 13 values each; for $x=\pm 2$, $y^2 < 36$ allows 11 values each; and for $x=\pm 3$, $y^2 < 16$ allows 7 values each. Summing these possibilities gives $15 + 26 + 22 + 14 = 77$ total solutions.

The number of elements $l$ in a relation defined by the inequality $2x+y \le 2$ is found by checking each $x \in \{-2,-1,0,1,2,3,4\}$ to identify valid $y$ values, which yields 23 total pairs. For the relation to be reflexive, we require $(x,x) \in R$ for all $x$, implying $3x \le 2$, which necessitates adding 4 pairs for $\{1, 2, 3, 4\}$. For symmetry, we must add the pair $(y,x)$ for every existing $(x,y)$ that is not already symmetric, which requires adding 6 pairs: $(3,-2), (4,-2), (3,-1), (4,-1), (2,-1), (2,0)$, resulting in $l+m+n = 23 + 4 + 6 = 33$.

We first solve the inequalities for each set. The set $A$ is defined by $|x^2 - 10| \le 6$, which implies $-6 \le x^2 - 10 \le 6$, simplifying to $4 \le x^2 \le 16$. This yields $x \in [-4, -2] \cup [2, 4]$. The set $B$ is defined by $|x-2| > 1$, which implies $x-2 > 1$ or $x-2 < -1$, resulting in $x \in (-\infty , 1) \cup (3, \infty )$. The difference $B - A$ includes elements in $B$ that are not in $A$. Subtracting the intersection $A \cap B = [-4, -2] \cup (3, 4]$ from $B$ leaves $(-\infty , -4) \cup (-2, 1) \cup (4, \infty )$.

The relation $R$ consists of pairs $(x,y)$ such that $2x \le 3y$. By testing every $x ext{ \in } A = \{2,3,5,7,9\}$, we determine valid $y$ values. For $x=2$, $4 \le 3y$ holds for $y \in \{2,3,5,7,9\}$ (5 pairs). For $x=3$, $6 \le 3y$ holds for $y \in \{2,3,5,7,9\}$ (5 pairs). For $x=5$, $10 \le 3y$ holds for $y \in \{5,7,9\}$ (3 pairs). For $x=7$, $14 \le 3y$ holds for $y \in \{5,7,9\}$ (3 pairs). For $x=9$, $18 \le 3y$ holds for $y \in \{7,9\}$ (2 pairs). Summing these gives $l = 18$. A relation is symmetric if $(x,y) \in R$ implies $(y,x) \in R$. There are 7 pairs where $(x,y) \in R$ but $(y,x) \notin R$, specifically $(2,5), (2,7), (2,9), (3,5), (3,7), (3,9),$ and $(5,9)$. Adding these 7 elements makes the relation symmetric, so $m = 7$. Thus, $l+m = 18 + 7 = 25$.

The relation $R$ is defined on the Cartesian product $A \times A$ with independent conditions $x | z$ and $y \leq w$, where $A = \{2, 3, 4, 5, 6\}$. There are 8 pairs $(x, z)$ that satisfy the divisibility condition $x | z$, and there are 15 pairs $(y, w)$ that satisfy $y \leq w$, calculated by the formula $\frac{n(n+1)}{2}$ with $n=5$. Since the conditions are independent, the total number of elements in $R$ is the product of these counts, which is $8 \times 15 = 120$.

The relation is defined by the equation $4y = 5x - 3$. By iterating through $x eq 0$ values from the set $M = \{1, 2, \dots, 16\}$, we identify valid pairs where $y$ is an integer \in $M$. Substituting $x=3$ gives $4y = 12$, so $(3,3) \in R$. For $x=7$, $4y = 32$, yielding $(7,8) \in R$. For $x=11$, $4y = 52$, yielding $(11,13) \in R$. Testing $x=15$ results \in $y=18$, which is outside the set $M$. The current relation is $R = \{(3,3), (7,8), (11,13)\}$. For the relation to be symmetric, if $(a,b) \in R$, then $(b,a)$ must also be \in $R$. While $(3,3)$ is already symmetric, the pairs $(7,8)$ and $(11,13)$ require the addition of $(8,7)$ and $(13,11)$ to satisfy symmetry. Therefore, the minimum number of elements to add is 2.

The relation $R$ is defined by the equality $2a + 3b = 3c + 4d$ for values $a, b, c, d \in \{1, 2, 3, 4\}$. By calculating the possible results for $2a + 3b$ and $3c + 4d$ for all variables, we identify values that appear in both sets. Matching these results pair by pair across the possible combinations of $(a, b)$ and $(c, d)$ reveals 12 distinct pairs that satisfy the defined equation.

The number of subsets with an even product is determined by subtracting the number of subsets consisting entirely of odd numbers from the total number of valid subsets. With 11 elements in $S$, there are $2^{11} = 2048$ total subsets, of which $1 + 11 = 12$ are invalid due to having fewer than two elements. Among the 6 odd numbers available, there are $2^6 = 64$ subsets consisting only of odd numbers, of which 7 have size less than two, leaving 57 valid subsets with only odd products. Subtracting these from the 2036 valid subsets of size at least two yields 1979.

A relation R consists of all ordered pairs $(a, b)$ from the set $\{-2, -1, 0, 1, 2\}$ that satisfy the condition $1 + ab > 0$. By systematically evaluating this condition for every possible pair in the Cartesian product of the set with itself, we identify 17 valid pairs such as $(-2, -2), (-2, -1), (0, 0), (1, 1),$ and $(2, 2)$. While the relation is reflexive and symmetric, it does not satisfy the transitivity condition, and thus it cannot be an equivalence relation.

The condition $\log_{e}(x + y) \leq 2$ implies that $x + y \leq e^2$, which is approximately $7.389$. Since $x$ and $y$ must be natural numbers, the condition simplifies to $x + y \leq 7$. After listing the initial set of pairs $(x, y)$ that satisfy this sum, we identify the missing pairs required to satisfy transitivity, which dictates that if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z)$ must also be in $R$. Adding these necessary elements results in 15 additional pairs.

A relation defined on a set with $n=4$ elements is reflexive if all diagonal pairs $(x,x)$ are included, which is mandatory for all 4 such pairs. Symmetry requires that for any pair $(x,y)$ with $x \neq y$, the presence of $(x,y)$ necessitates the presence of $(y,x)$. There are $\frac{n^2 - n}{2} = 6$ such off-diagonal pairs. Since each pair of symmetric coordinates represents an independent choice to either include both or neither in the relation, there are $2^6$ possible configurations. This results in $64$ possible relations.

The number of onto functions from a set $A$ to a set $B$ requires mapping every element in the codomain to at least one element in the domain. Solving the inequality $||x-3|-3| \le 1$ for set $A$ yields $x \in \{-1, 0, 1, 5, 6, 7\}$ with $n(A)=6$, while solving the given equation for set $B$ yields $B=\{3, 4\}$ with $n(B)=2$. The total number of functions from $A$ to $B$ is $2^6 = 64$, and subtracting the two constant functions that map all elements to a single value leaves 62 onto functions.

An equivalence relation must satisfy reflexivity, symmetry, and transitivity. The relation $(a_1, b_1) R (a_2, b_2) \Leftrightarrow b_1=b_2$ is reflexive as $b_1=b_1$, symmetric because $b_1=b_2$ implies $b_2=b_1$, and transitive since $b_1=b_2$ and $b_2=b_3$ implies $b_1=b_3$, confirming Statement I is true. Statement II is false because the set $S$ defined by $y=b$ represents a horizontal line parallel to the $x$-axis, whereas the line $y=x$ has a slope of $1$.

The relation $R$ includes pairs $(x,y)$ where $\max\{x,y\} \in \{3,4\}$. The relation is symmetric because the condition $\max\{x,y\} \in \{3,4\}$ is identical to $\max\{y,x\} \in \{3,4\}$. It is not reflexive because $\max\{0,0\} = 0 \notin \{3,4\}$, so $(0,0) \notin R$. Transitivity fails because while $(0,4) \in R$ (since $\max\{0,4\}=4$) and $(4,2) \in R$ (since $\max\{4,2\}=4$), the pair $(0,2)$ is not in $R$ as $\max\{0,2\}=2 \notin \{3,4\}$. Counting the pairs yields 16 elements, not 18, so the statement $S_1$ is false, while $S_2$ correctly describes the properties of $R$.

The intersection set $D$ is determined by solving the system $x^{2+}y^2=25$ and $x^{2+}9y^2=144$, which yields 4 distinct coordinate points. The set $C$ consists of integer lattice points $(x, y)$ satisfying $x^{2+}y^2 \le 4$, which comprises 13 points in total. Since the number of one-one functions from a set of size 4 to a set of size 13 is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$, we calculate $P(13, 4) = 13 \times 12 \times 11 \times 10 = 17160$.

A relation $R$ is reflexive if $(f,f) \in R$ for all $f$, which requires $f(0)=f(1)$ and $f(1)=f(0)$; since this does not hold for every function, the relation is not reflexive. It is symmetric because if $(f,g) \in R$ then $f(0)=g(1)$ and $f(1)=g(0)$, which implies $(g,f) \in R$ as $g(0)=f(1)$ and $g(1)=f(0)$. However, it is not transitive because $(f,g) \in R$ and $(g,h) \in R$ only implies $f(0)=h(0)$ and $f(1)=h(1)$, failing the requirement $f(0)=h(1)$ and $f(1)=h(0)$ generally.

The logarithmic property $\log_{a}x + \log_{a}y = \log_{a}(xy)$ simplifies the expression for set $A$ to $\log_{\frac{2}{\pi}} |\sin x \cos x| = 2$. By definition, this yields $|\sin 2x| = \frac{8}{\pi^2}$, which provides four distinct values for $x$ in the interval $(0, \pi)$ excluding $\frac{\pi}{2}$. For set $B$, the equation $\sqrt{x}(\sqrt{x} - 4) - 3|\sqrt{x} - 2| + 6 = 0$ is solved by considering the cases $\sqrt{x} \ge 2$ and $\sqrt{x} < 2$, leading to solutions $x = 0, 1, 9, 16$. Since sets $A$ and $B$ are disjoint, the cardinality of their union is the sum of their individual sizes, $4 + 4 = 8$.

A reflexive relation on the set $\{1, 2, 3\}$ must contain the identity elements $(1, 1), (2, 2),$ and $(3, 3)$. Including the pair $(1, 2)$ requires adding further elements to maintain transitivity while ensuring the relation does not become symmetric. By systematically identifying combinations of the remaining available pairs, we find that there are exactly 5 distinct relations that satisfy these constraints while remaining non-symmetric.