Sets And Relations – JEE Main MathematicsPractice Questions & PYQs

The condition $\gcd(n, 45)=1$ restricts elements to those not divisible by 3 or 5, so $A \cap B$ contains even numbers up to 200 that meet this criterion. Summing all even numbers up to 200 gives $10100$, and applying inclusion-exclusion by subtracting the sums of even multiples of 3 and 5, then adding back even multiples of 15, allows us to calculate the target sum $10100 - 3366 - 2100 + 630 = 5264$.

The identity $\min(i,j) + \max(i,j) = i+j$ allows the total sum $A+B$ to be expressed as $\sum_{i=1}^{10} \sum_{j=1}^{10} (i+j)$, which simplifies to $20 \sum_{k=1}^{10} k$. Calculating the sum of the first ten integers as $\frac{10 \times 11}{2} = 55$, we find the final result is $20 \times 55 = 1100$.

To find the number of elements in $B \cup C$, it is efficient to calculate the complement, $(B \cup C)' = B' \cap C'$. The set $B'$ consists of all subsets $T$ of $A$ where $1 \in T$ and $2 \notin T$. We are looking for subsets of $\{3, 4, 5, 6, 7\}$ such that adding $1$ to the sum of elements yields a non-prime number. Counting these cases, we find $1$ way for a $0$-element subset, $3$ ways for a $1$-element subset, $7$ ways for $2$-element subsets, $6$ ways for $3$-element subsets, $3$ ways for $4$-element subsets, and $1$ way for the $5$-element subset. Summing these instances gives $21$ total subsets that fail the condition, leading to $128 - 21 = 107$.

Visualizing sets through Venn diagrams requires that the central region, where all three circles overlap, must be empty because no student plays all three games. If we represent the sets of students playing each game as $A$, $B$, and $C$, the condition that the triple intersection $A \cap B \cap C = \emptyset$ must hold true is paramount. Since the provided diagrams fail to represent this specific configuration where the intersection of all three circles is zero, the valid conclusion is that none of the listed options justify the statement.

Transitivity requires that if $(x,y) \in R$ and $(y,z) \in R$, then $(x,z)$ must also be in $R$. For the relation $0 < |x - y| \leq 1$, we can choose $x=2, y=3$, and $z=4$; here, $|2-3|=1$ and $|3-4|=1$, but $|2-4|=2$, which fails the condition, proving the relation is not transitive.

The absolute value function is defined as $|x|=x$ for x ext{ } oldsymbol{ ext{ge}} ext{ } 0 and $|x|=-x$ for $x < 0$, which allows us to break the equation into three intervals. For $x < -4$, the equation becomes $(-x-3)(-x-4)=6$, simplifying to $x^{2+}7x+6=0$, giving valid solution $x=-6$. In the interval -4 ext{ } oldsymbol{ ext{le}} ext{ } x < 0, the equation simplifies to $x^{2+}7x+18=0$, which has a negative discriminant and no real roots. For x ext{ } oldsymbol{ ext{ge}} ext{ } 0, the equation becomes $(x-3)(x+4)=6$, simplifying to $x^{2+}x-18=0$, yielding one valid solution x = rac{-1+oldsymbol{ ext{sqrt}}oldsymbol{ ext{73}}}{2}. Since there are exactly two valid values, the set contains two elements.

Finding the number of subsets requires identifying the integer range $(A \cap B \cap C)^C$ through inequality analysis. The set intersection results in $A \cap B \cap C = (-\infty , -2) \cup [6, \infty )$, making the complement $[-2, 6)$ which contains the integers $\{-2, -1, 0, 1, 2, 3, 4, 5\}$. As there are 8 elements, the total number of subsets is $2^8 = 256$.

The set $A$ is defined by the inequality $n^{2} - n - 10000 \leq 0$, which evaluates to $n(n-1) \leq 100 \times 100$, confirming that $A = \{1, 2, \ldots, 100\}$. The set $B-C$ contains elements of $B$ that are not multiples of $2$, resulting in the arithmetic sequence $\{7, 13, 19, \ldots\}$ where the terms follow the form $6k+1$ up to $100$. With $16$ terms in this sequence, the sum is calculated using the formula $S_{n} = \frac{n}{2}(a + l)$, which yields $\frac{16}{2}(7 + 97) = 832$.

The number of elements in the intersection of two sets is bounded by the inclusion-exclusion principle, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$, where the union of the sets cannot exceed the total number of patients, $100\%$. Substituting the given values yields $n(A \cap B) = 89 + 98 - n(A \cup B)$, which implies $n(A \cap B) \geq 87$. Since the intersection size cannot exceed the size of the smaller set, $n(A \cap B) \leq 89$, restricting $K$ to the range $87 \leq K \leq 89$. Therefore, $K$ cannot take any value from the set $\{79, 81, 83, 85\}$ as these fall outside the valid range.

The sum of an arithmetic progression is given by S = rac{N}{2}(a+l), where $a$ is the first term, $l$ is the last term, and $N$ is the number of terms. The set $B$ comprises 3-digit numbers $9k+2$, which are oldsymbol{ ext{ extbraceleft}}oldsymbol{ ext{101, 110, ... , 992}}oldsymbol{ ext{ extbraceright}}, resulting in a sum of $50(101+992) = 54650$. The set $C$ consists of 3-digit numbers $9k+l$, and based on the provided total sum of $109600$, the sum of $C$ must be $54950$, which occurs when $l=5$ for the series oldsymbol{ ext{ extbraceleft}}oldsymbol{ ext{104, 113, ... , 995}}oldsymbol{ ext{ extbraceright}}. This confirms $l=5$ as the required value.

The equivalence relation $(a, b) \sim (c, d)$ defined by $ad = bc$ implies that the ratio $\frac{a}{b} = \frac{c}{d}$ must hold. For the given pair $(4, 3)$, the condition simplifies to $\frac{a}{b} = \frac{4}{3}$, which means $a = 4k$ and $b = 3k$ for some positive integer $k$. Given that both $a$ and $b$ must be within the set $\{2, 3, \ldots, 30\}$, we require $4k \leq 30$ and $3k \leq 30$, leading to $k \leq 7.5$. Thus, there are $7$ possible values for $k$ where $k \in \{1, 2, 3, 4, 5, 6, 7\}$, resulting in $7$ ordered pairs.

A relation is an equivalence relation if it is reflexive, symmetric, and transitive. For any matrix $A$, $A = IAI^{-1}$ holds with the identity matrix, confirming reflexivity; if $A = PAP^{-1}$, then $A = P^{-1}BP$ shows symmetry; and if $B = PAP^{-1}$ and $C = Q B Q^{-1}$, substituting $B$ reveals $C = (QP)A(QP)^{-1}$, which confirms transitivity. Since all three properties are satisfied, the relation is an equivalence relation.

An equivalence class under the relation of equal distance from the origin consists of all points $(x,y)$ such that their Euclidean distance from the origin equals the distance of the given point $(1,-1)$. The distance of $(1,-1)$ from the origin is oldsymbol{ ext{sqrt}}oldsymbol{ ext{1}^2 + (-1)^2} = oldsymbol{ ext{sqrt}}oldsymbol{ ext{2}}. Therefore, the set of all such points $(x,y)$ satisfies the equation oldsymbol{ ext{sqrt}}oldsymbol{ ext{x}^{2+}y^2} = oldsymbol{ ext{sqrt}}oldsymbol{ ext{2}}, which simplifies to $x^{2+}y^2=2$.

The given equation factors into $(x - y)(x + y)(x - 3y) = 0$, meaning a pair $(x, y)$ belongs to the relation if $x = y$, $x = -y$, or $x = 3y$. Since the condition $x = y$ is satisfied for all $x \in \mathbb{N}$, the relation is reflexive. However, it fails to be symmetric because $(3, 1)$ satisfies $x = 3y$ while $(1, 3)$ does not satisfy any of the conditions, and it is not transitive since $(9, 3)$ and $(3, 1)$ belong to the relation but $(9, 1)$ does not.

For the quadratic equation $x^{2} - (m + 1)x + m + 4 = 0$ to have real roots, the discriminant $D = (m + 1)^{2} - 4(m + 4)$ must be greater than or equal to $0$. Simplifying this leads to $m^{2} - 2m - 15 \ge 0$, which factors as $(m - 5)(m + 3) \ge 0$, yielding set $A = (-\infty , -3] \cup [5, \infty )$. Comparing this with $B = [-3, 5)$, the set difference $A - B$ results in $(-\infty , -3) \cup [5, \infty )$, making the statement in option A incorrect as it fails to include the value 5.

Using the Principle of Inclusion-Exclusion, the percentage of people liking both coffee and tea, $x$, satisfies n(C oldsymbol{ ext{\cup}} T) = n(C) + n(T) - n(C oldsymbol{ ext{\cap}} T), where n(C oldsymbol{ ext{\cup}} T) oldsymbol{ ext{le}} ext{ } 100. Substituting the values, 73 + 65 - x oldsymbol{ ext{le}} ext{ } 100 leads to x oldsymbol{ ext{ge}} ext{ } 38. Additionally, $x$ cannot exceed the percentage of people who like tea or coffee individually, meaning x oldsymbol{ ext{le}} ext{ } 65. Thus, $x$ must be in the range $[38, 65]$, making 36 an impossible value.

The number of subsets of a set with $k$ elements is $2^k$. Given the difference in subsets is 112, we solve the equation $2^m - 2^n = 112$. Factoring this as $2^n(2^{m-n} - 1) = 2^4 imes 7$, we equate powers to find $n=4$ and $2^{m-4} = 8$, which implies $m-4=3$ or $m=7$. The product $m imes n$ is therefore $7 imes 4 = 28$.

The number of distinct elements in the set $T$ is determined by the total count of elements across all sets $X_{i}$ divided by the frequency of each element's repetition, which is $50 \times 10 / 20 = 25$. Applying this same logic to the sets $Y_{i}$ with $n$ total sets, where each contains 5 elements and each element appears 6 times, the total count of elements $5n$ must satisfy $5n / 6 = 25$, resulting in $n = 30$.

The domain of the inverse relation $R^{-1}$ consists of all integer values $y$ for which there exists an integer $x$ satisfying $x^{2} + 3y^{2} \le 8$. Evaluating this for integer values of $y$, we find that for $y = 0$, $x^{2} \le 8$ permits $x \in \{0, \pm 1, \pm 2\}$, and for $y = \pm 1$, $x^{2} \le 5$ also allows integer solutions for $x$, while for $|y| \ge 2$, $3y^{2} \ge 12$ makes the inequality impossible, confirming the domain of $R^{-1}$ is $\{-1, 0, 1\}$.

The percentage of people reading at least one newspaper is given by $n(A \cup B) = n(A) + n(B) - n(A \cap B)$, which equals $63 + 76 - x = 139 - x$. Since this union cannot exceed $100\%$, it follows that $139 - x \le 100$, or $x \ge 39$. Because the intersection $x$ cannot be greater than the percentage reading either newspaper, $x \le 63$, identifying the possible range as $[39, 63]$, where 55 is a valid value.