Sets And Relations – JEE Main MathematicsPractice Questions & PYQs

The set A is defined by the inequality $|x| < 2$, which corresponds to the open interval $(-2, 2)$. The set B, derived from $|x - 2| > 3$, translates to $x - 2 > 3$ or $x - 2 < -3$, resulting in the range $x > 5$ or $x < -1$. To determine the set difference $B - A$, we remove all elements of A from the union of these two intervals, which effectively results in the set of all real numbers excluding the interval $(-2, 5)$.

Transitivity requires that if the relation holds for pairs $(a, b)$ and $(b, c)$, it must also hold for $(a, c)$. For $R_{1}$, one can select values such that $a^{2} + b^{2} \in Q$ and $b^{2} + c^{2} \in Q$, but $a^{2} + c^{2} \notin Q$, and for $R_{2}$, one can select values where $a^{2} + b^{2} \notin Q$ and $b^{2} + c^{2} \notin Q$, but $a^{2} + c^{2} \in Q$. These counterexamples demonstrate that the property of having a rational or irrational sum of squares does not satisfy the transitive property for either relation.

The number of elements in the union of two sets is found using the principle of inclusion-exclusion, represented by the formula $|A \cup B| = |A| + |B| - |A \cap B|$. Within the range of $1$ to $50$, there are $25$ multiples of $2$ and $7$ multiples of $7$. Because the multiples of $14$ (which are $14, 28, 42$) are counted in both sets, we subtract these $3$ overlapping elements, leading to a total of $25 + 7 - 3 = 29$ elements.

The Inclusion-Exclusion Principle determines the number of students who opted for at least one course by calculating $n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$. Given the student counts, we find $n(A)=70, n(B)=46, n(C)=28$ with intersections $n(A \cap B)=23, n(B \cap C)=9, n(A \cap C)=14$ and triple intersection $n(A \cap B \cap C)=4$. Substituting these values gives $70 + 46 + 28 - 23 - 9 - 14 + 4 = 102$ students opted for courses, leaving $140 - 102 = 38$ students who opted for none.

The set A is determined by the equation $2^{(x+2)(x^{2-}5x+6)} = 1$, which requires the exponent to be $0$, yielding the integer roots $\{-2, 2, 3\}$. The set B is formed by integers satisfying $-3 < 2x - 1 < 9$, which simplifies to $-1 < x < 5$, resulting in the set $\{0, 1, 2, 3, 4\}$. The total number of elements in the Cartesian product $A \times B$ is the product of their cardinalities, $3 \times 5 = 15$, and the number of subsets for any set is defined as $2^n$, leading to $2^{15}$ subsets.

A product of integers is even if at least one element in the set is even. The total number of non-empty subsets of a 100-element set is $2^{100}-1$, and the number of subsets containing only odd elements is $2^{50}-1$ since there are 50 odd numbers in the set. Subtracting the number of odd-only subsets from the total number of non-empty subsets yields $(2^{100}-1) - (2^{50}-1)$, which simplifies to $2^{100}-2^{50}$ or $2^{50}(2^{50}-1)$.

The city population can be categorized into three disjoint groups: those who read only newspaper A, only newspaper B, and those who read both. Given that $25\%$ read A, $20\%$ read B, and $8\%$ read both, the proportions for the exclusive groups are $17\%$ (for A only) and $12\%$ (for B only). Applying the specified advertisement viewership rates to these proportions, the total percentage is calculated as $(0.30 \times 0.17) + (0.40 \times 0.12) + (0.50 \times 0.08)$, which equals $13.9\%$.

The definition of set difference $A-C$ indicates that any element belonging to A but not to C must be contained within B. Because the intersection $A \cap B$ is permitted to contain elements also present in C, the condition $(A-C) \subseteq B$ does not enforce that the entire set A must be a subset of B. Since there are scenarios where elements of A that are not in C are in B, while other elements of A remain outside B, the universal implication $A \subseteq B$ is false.

Binary relations on natural numbers define pairs based on specific equations. Solving $2x+y=10$ for $x ext{ \in } N$ leads to $(1,8), (2,6), (3,4), (4,2)$, while solving $x+2y=10$ for $x, y ext{ \in } N$ leads to $(8,1), (6,2), (4,3), (2,4)$. Examining the second relation $R_2$, the possible values for $y$ are $1, 2, 3, 4$, which corresponds exactly to the range of $R_2$ being the set of integers $\{1, 2, 3, 4\}$.

A binary relation $R$ on a set $A$ is symmetric if $(x, y) \in R \implies (y, x) \in R$, and it is transitive if $(x, y) \in R$ and $(y, z) \in R \implies (x, z) \in R$. For $R_2$, every pair $(x, y) \in R_2$ has its counterpart $(y, x) \in R_2$, confirming that it is symmetric, but the relation is not transitive because $(b, a) \in R_2$ and $(a, c) \in R_2$ exist without the required element $(b, c) \in R_2$. Meanwhile, $R_1$ is neither symmetric, as $(b, c) \in R_1$ while $(c, b) \notin R_1$, nor transitive, because $(b, c) \in R_1$ and $(c, a) \in R_1$ appear in the set while $(b, a) \notin R_1$.

The constraints for set $A$ require both $a$ and $b$ to fall within the open interval $4 < a < 6$ and $4 < b < 6$, which defines a square region centered at $(5, 5)$. The boundary for set $B$ is given by the inequality $4(a-6)^2 + 9(b-5)^2 \leq 36$, which can be rearranged as $\frac{(a-6)^2}{9} + \frac{(b-5)^2}{4} \leq 1$, representing the interior of an ellipse centered at $(6, 5)$. Within the region defined by $A$, the values are constrained such that $(a-6)^2 < 4$ and $(b-5)^2 < 1$, which makes the expression $4(a-6)^2 + 9(b-5)^2$ strictly less than 25. Since this upper bound of 25 is less than the required value of 36, every point in $A$ necessarily satisfies the condition for $B$, proving $A \subset B$.

The definitions for sets $P$ and $Q$ can be simplified by isolating the trigonometric terms. For set $P$, the equation $\sin\theta - \cos\theta = \sqrt{2} \cos\theta$ rearranges to $\sin\theta = (\sqrt{2}+1)\cos\theta$. For set $Q$, the equation $\sin\theta + \cos\theta = \sqrt{2} \sin\theta$ rearranges to $\cos\theta = (\sqrt{2}-1)\sin\theta$, which is equivalent to $\sin\theta = \frac{1}{\sqrt{2}-1}\cos\theta = (\sqrt{2}+1)\cos\theta$. Since both conditions represent the same relationship between $\sin\theta$ and $\cos\theta$, the two sets are identical.

The Cartesian product $A \times B$ contains $4 \times 2 = 8$ distinct elements. To find the number of subsets with at least three elements, we subtract the subsets with zero, one, or two elements from the total number of possible subsets, which is $2^8 = 256$. Using the combination formula $\binom{n}{r}$, we subtract $\binom{8}{0} + \binom{8}{1} + \binom{8}{2} = 1 + 8 + 28 = 37$ from $256$, resulting in $219$.

The binomial expansion of $4^n$ can be written as $(1+3)^n = 1 + n(3) + \binom{n}{2}3^2 + \dots + 3^n$, which simplifies to $1 + 3n + 9(\sum_{k=2}^{n} \binom{n}{k}3^{k-2})$. By subtracting $3n+1$ from this expression, it is evident that every element in set $X$ is a multiple of $9$, meaning $X$ is a subset of the set of all non-negative multiples of $9$. Since set $Y$ is defined as $9(n-1)$ for $n \in N$, which represents the entire set of non-negative multiples of $9$, the union of $X$ and $Y$ is simply $Y$.

To determine the number of ordered pairs of disjoint subsets $Y$ and $Z$ from a set $X$ with $5$ elements, consider the placement of each individual element. For any element $x \in X$, it must either be in $Y$, in $Z$, or in neither set, but it cannot belong to both sets simultaneously since $Y \cap Z = \emptyset$. This yields $3$ mutually exclusive options for each of the $5$ elements. Consequently, the total number of ways to form these subsets is calculated by multiplying the $3$ choices for each of the $5$ elements, resulting in $3^5$.

For a relation to be an equivalence relation, it must satisfy the properties of being reflexive, symmetric, and transitive. Relation $S$ fails the reflexive property because for any $x \in (0, 2)$, the element $(x, x)$ cannot belong to $S$ as $x$ never equals $x + 1$. Relation $T$ satisfies all three properties because $x - x = 0$ is an integer for all $x \in R$, $x - y \in \mathbb{Z}$ implies $y - x \in \mathbb{Z}$, and the sum of two integers $x - y$ and $y - z$ is also an integer $x - z$. Therefore, $T$ functions as an equivalence relation while $S$ does not.