Paper Generator

Given $A = {x \in R : |x| < 2}$ and $B = {x \in R : |x – 2| \geq 3}$. We can rewrite the sets as follows: $A = {x \in R : -2 < x < 2} = (-2, 2)$ $B = {x \in R : x – 2 \geq 3 \text{ or } x – 2 \leq -3} = {x \in R : x \geq 5 \text{ or } x \leq -1} = (-\infty, -1] \cup [5, \infty)$ Now, we want to find $B – A$, which is the set of elements in $B$ but not in $A$. $B – A = B \cap A^c$, where $A^c$ is the complement of $A$. $A^c = (-\infty, -2] \cup [2, \infty)$ Then, $B – A = ((-\infty, -1] \cup [5, \infty)) \cap ((-\infty, -2] \cup [2, \infty))$ $B - A = (-\infty, -1] \cup [5, \infty) - (-2,2) = (-\infty, -1] \cup [5, \infty)$. In other words, $B-A$ is the set of all real numbers except those in the interval $(-2, 2)$. So, $B-A = R - (-2, 5)$.

Given $R = {(x, y) : x, y \in Z, x^2 + 3y^2 \le 8}$. So $R = {(0,1), (0,-1), (1,0), (-1,0), (1,1), (1,-1), (-1,1), (-1,-1), (2,0), (-2,0), (2,1), (2,-1), (-2,1), (-2,-1)}$. $\Rightarrow R : {-2, -1, 0, 1, 2} \rightarrow {-1, 0, 1}$. $\therefore R^{-1} : {-1, 0, 1} \rightarrow {-2, -1, 0, 1, 2}$. $\therefore$ Domain of $R^{-1} = {-1, 0, 1}$.

For the roots to be real, the discriminant $D ≥ 0$. $(m+1)^2 - 4(m+4) ≥ 0$ $m^2 + 2m + 1 - 4m - 16 ≥ 0$ $m^2 - 2m - 15 ≥ 0$ $(m-5)(m+3) ≥ 0$ $m ∈ (-∞, -3] ∪ [5, ∞)$ Therefore, $A = (-∞, -3] ∪ [5, ∞)$. Given $B = [-3, 5)$. Analyzing the options: Option A: $A ∩ B = {-3}$. This is true since the intersection contains only $-3$. Option B: $B – A = (-3, 5)$. This is true because removing $A$ from $B$ leaves the interval $(-3, 5)$. Option C: $A ∪ B = R$. This is true, as the union covers all real numbers. Option D: $A - B = (-∞, -3) ∪ (5, ∞)$. This is incorrect. $A - B$ should be $(-∞, -3) ∪ [5, ∞)$. The given interval does not include $5$. Thus, option D is not true.

For $R_1$: Let $a = 1 + \sqrt{2}$, $b = 1 - \sqrt{2}$, $c = \sqrt[4]{8}$. $aR_1b : a^2 + b^2 = 6 \in Q$ $bR_1c : b^2 + c^2 = 3 - 2\sqrt{2} + 2\sqrt{2} = 3 \in Q$ $aR_1c : a^2 + c^2 = 3 + 2\sqrt{2} + 2\sqrt{2} \notin Q$ $\therefore$ $R_1$ is not transitive. For $R_2$: Let $a = 1 + \sqrt{2}$, $b = \sqrt{2}$, $c = 1 - \sqrt{2}$ $aR_2b : a^2 + b^2 = 5 + 2\sqrt{2} \notin Q$ $bR_2c : b^2 + c^2 = 5 - 2\sqrt{2} \notin Q$ $aR_2c : a^2 + c^2 = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} = 6 \in Q$ $\therefore$ $R_2$ is not transitive.

Let $A$ be the set of people who read newspaper A and $B$ be the set of people who read newspaper B. According to the problem, we have: $|A| = 63$ $|B| = 76$ $|A \cup B| = |A| + |B| - |A \cap B|$ Also, $|A \cup B|$ represents the percentage of people who read at least one of the newspapers, and it cannot exceed $100$. We are given that $x$% of the people read both newspapers, so $|A \cap B| = x$. Therefore, $|A \cup B| = 63 + 76 - x = 139 - x$. Since $|A \cup B| \le 100$, we have $139 - x \le 100$, which implies $x \ge 39$. From the Venn diagram, $x$ must be less than or equal to both $63$ and $76$. So, $x \le 63$. Thus, $39 \le x \le 63$. From the given options, the only value that satisfies this inequality is $55$. Therefore, the possible value of $x$ is $55$.

$\bigcup_{i=1}^{50} X_i = X_1, X_2,....., X_{50} = 50$ sets. Given each sets having $10$ elements. So total elements = $50 \times 10$ $\bigcup_{i=1}^{n} Y_i = Y_1, Y_2,....., Y_n = n$ sets. Given each sets having $5$ elements. So total elements = $5 \times n$ Now each element of set $T$ contains exactly $20$ of sets $X_i$. So number of effective elements in set $T = \frac{50 \times 10}{20}$ Also each element of set $T$ contains exactly $6$ of sets $Y_i$. So number of effective elements in set $T = \frac{n \times 5}{6}$ $\therefore \frac{50 \times 10}{20} = \frac{n \times 5}{6}$ $\Rightarrow n = 30$

Let C be the set of persons who like coffee and T be the set of persons who like tea. Given: $n(C) = 73$ and $n(T) = 65$. Also, $n(C \cup T) \le 100$. Using the formula for the union of two sets: $n(C \cup T) = n(C) + n(T) - n(C \cap T)$. Let $x = n(C \cap T)$ be the percentage of people who like both coffee and tea. So, $n(C \cup T) = 73 + 65 - x \le 100$. This implies $138 - x \le 100$, which gives $x \ge 38$. Also, $x$ must be less than or equal to both $73$ and $65$ since $x$ represents the intersection. So, $x \le 73$ and $x \le 65$. Therefore, $38 \le x \le 65$. Since $x$ must be in the range $[38, 65]$, the value $x=36$ is not possible.

Given $R = {(P, Q) | P$ and $Q$ are at the same distance from the origin}. Then the equivalence class of $(1, -1)$ will contain all such points that lie on the circumference of the circle with the center at the origin and passing through the point $(1, -1)$. The radius of the circle $= \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$. Therefore, the equation of the circle is $x^2 + y^2 = (\sqrt{2})^2 = 2$. Required equivalence class of $(S) = {(x, y) | x^2 + y^2 = 2}$.

We need to consider three cases based on the value of $x$: Case 1: $x \le -4$ In this case, $|x| = -x$ and $|x+4| = -(x+4)$. The equation becomes: $(-x - 3)(-x - 4) = 6$ $(x + 3)(x + 4) = 6$ $x^2 + 7x + 12 = 6$ $x^2 + 7x + 6 = 0$ $(x + 1)(x + 6) = 0$ $x = -1$ or $x = -6$ Since $x \le -4$, we only accept $x = -6$. Case 2: $-4 < x < 0$ In this case, $|x| = -x$ and $|x+4| = x+4$. The equation becomes: $(-x - 3)(x + 4) = 6$ $-x^2 - 4x - 3x - 12 = 6$ $-x^2 - 7x - 18 = 0$ $x^2 + 7x + 18 = 0$ The discriminant is $D = 7^2 - 4(1)(18) = 49 - 72 = -23 < 0$. So, there are no real solutions in this case. Case 3: $x \ge 0$ In this case, $|x| = x$ and $|x+4| = x+4$. The equation becomes: $(x - 3)(x + 4) = 6$ $x^2 + 4x - 3x - 12 = 6$ $x^2 + x - 18 = 0$ Using the quadratic formula, $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-18)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 72}}{2} = \frac{-1 \pm \sqrt{73}}{2}$ Since $x \ge 0$, we take the positive root: $x = \frac{-1 + \sqrt{73}}{2} \approx \frac{-1 + 8.54}{2} \approx 3.77$. So, $x = \frac{-1 + \sqrt{73}}{2}$ is a valid solution. Therefore, the solutions are $x = -6$ and $x = \frac{-1 + \sqrt{73}}{2}$. The number of elements in the set is 2.

Given that $(a,b) \simeq (4,3)$, we have $3a=4b$, which means $a = \frac{4}{3}b$. Since $a$ must be an integer, $b$ must be a multiple of $3$. The possible values of $b$ are ${3, 6, 9, 12, 15, 18, 21, 24, 27, 30}$. For each value of $b$, we can find the corresponding value of $a$: If $b=3$, $a=\frac{4}{3}(3) = 4$. If $b=6$, $a=\frac{4}{3}(6) = 8$. If $b=9$, $a=\frac{4}{3}(9) = 12$. If $b=12$, $a=\frac{4}{3}(12) = 16$. If $b=15$, $a=\frac{4}{3}(15) = 20$. If $b=18$, $a=\frac{4}{3}(18) = 24$. If $b=21$, $a=\frac{4}{3}(21) = 28$. If $b=24$, $a=\frac{4}{3}(24) = 32$. But $a$ must be less than or equal to $30$, so we stop here. If $b=27$, $a=\frac{4}{3}(27) = 36$, which is also greater than $30$. If $b=30$, $a=\frac{4}{3}(30) = 40$, which is also greater than $30$. The pairs $(a, b)$ that satisfy the condition are $(4, 3), (8, 6), (12, 9), (16, 12), (20, 15), (24, 18), (28, 21)$. There are $7$ such ordered pairs.

The question states that none of the students play all three games. This means that the intersection of all three sets (circles) in the Venn diagram must be empty. Observing the provided Venn diagrams (P, Q, and R), we see that in each case, there is an intersection between all three circles. Thus, none of the diagrams satisfy the condition that no student plays all three games. Therefore, the correct answer is 'None of these'.

For reflexive relation, $\forall (A, A) \in R$ for matrix $P$. $\Rightarrow A = PAP^{-1}$ is true for $P = I$. So, $R$ is reflexive relation. For symmetric relation, Let $(A, B) \in R$ for matrix $P$. $\Rightarrow A = PBP^{-1}$. After pre-multiply by $P^{-1}$ and post-multiply by $P$, we get $P^{-1}AP = B$. So, $(B, A) \in R$ for matrix $P^{-1}$. So, $R$ is a symmetric relation. For transitive relation, Let $ARB$ and $BRC$. So, $A = PBP^{-1}$ and $B = PCP^{-1}$. Now, $A = P(PCP^{-1})P^{-1} \Rightarrow A = (P)^2C(P^{-1})^2 \Rightarrow A = (P)^2 \cdot C \cdot (P^2)^{-1}$. $\therefore (A, C) \in R$ for matrix $P^2$. $\therefore R$ is transitive relation. Hence, $R$ is an equivalence relation.

Given the relation $x^3 - 3x^2y - xy^2 + 3y^3 = 0$. We can factor this as follows: $x(x^2 - y^2) - 3y(x^2 - y^2) = 0$ $(x - 3y)(x^2 - y^2) = 0$ $(x - 3y)(x - y)(x + y) = 0$ Thus, the relation holds if $x = 3y$, $x = y$, or $x = -y$. Reflexivity: For $R$ to be reflexive, $(x, x) \in R$ for all $x \in N$. If $x = y$, then $(x, x) \in R$. So, the relation is reflexive. Symmetry: For $R$ to be symmetric, if $(x, y) \in R$, then $(y, x) \in R$. Let's consider $x = 3y$. If $(3, 1) \in R$ since $3 = 3(1)$, but $(1, 3) \notin R$ since $1 \neq 3(3)$. So, the relation is not symmetric. Transitivity: For $R$ to be transitive, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$. Let $x = 3y$ and $y = z$. Then $(x, y) = (3y, y)$ and $(y, z) = (y, y)$. For transitivity, we need $(3y, y) \in R$ and $(y, y) \in R$ implying $(3y, y) \in R$. However, if we choose $x=3$, $y=1$ and $z=1$, then $x = 3y$. But if $x = 3$ and $y = 1$, we don't have $x=z$. So the relation is not transitive. Since $R$ is reflexive but neither symmetric nor transitive, the correct option is B.

Consider the relation defined by $0 < |x - y| \le 1$. We need to show that this relation is symmetric but not transitive. A relation is symmetric if whenever $(a, b)$ is in the relation, $(b, a)$ is also in the relation. Since $|x-y| = |y-x|$, if $0 < |x-y| \le 1$, then $0 < |y-x| \le 1$. Therefore, if $(a, b)$ is in the relation, so is $(b, a)$, so the relation is symmetric. A relation is transitive if whenever $(a, b)$ and $(b, c)$ are in the relation, $(a, c)$ is also in the relation. Let $x = 0.2, y = 0.9, z = 1.5$. Then $|x - y| = |0.2 - 0.9| = |-0.7| = 0.7$, and $|y - z| = |0.9 - 1.5| = |-0.6| = 0.6$. So $0 < |x - y| \le 1$ and $0 < |y - z| \le 1$. However, $|x - z| = |0.2 - 1.5| = |-1.3| = 1.3$, which is not less than or equal to $1$. So $(x, z)$ is not in the relation. Therefore, the relation is not transitive. So, option B is not correct.

To check if $R_1$ and $R_2$ are equivalence relations, we need to verify if they are reflexive, symmetric, and transitive. For $R_1 = \{(a, b) \in N \times N : |a - b| \le 13\}$: Reflexive: For any $a \in N$, $|a - a| = 0 \le 13$. So, $(a, a) \in R_1$. Thus, $R_1$ is reflexive. Symmetric: If $(a, b) \in R_1$, then $|a - b| \le 13$. This implies $|b - a| \le 13$, so $(b, a) \in R_1$. Thus, $R_1$ is symmetric. Transitive: Consider $a = 2, b = 11, c = 19$. Then $|2 - 11| = 9 \le 13$, so $(2, 11) \in R_1$. Also, $|11 - 19| = 8 \le 13$, so $(11, 19) \in R_1$. However, $|2 - 19| = 17 > 13$, so $(2, 19) \notin R_1$. Thus, $R_1$ is not transitive. Since $R_1$ is not transitive, $R_1$ is not an equivalence relation. For $R_2 = \{(a, b) \in N \times N : |a - b| \ne 13\}$: Reflexive: For any $a \in N$, $|a - a| = 0 \ne 13$. So, $(a, a) \in R_2$. Thus, $R_2$ is reflexive. Symmetric: If $(a, b) \in R_2$, then $|a - b| \ne 13$. This implies $|b - a| \ne 13$, so $(b, a) \in R_2$. Thus, $R_2$ is symmetric. Transitive: Consider $a = 13, b = 3, c = 26$. Then $|13 - 3| = 10 \ne 13$, so $(13, 3) \in R_2$. Also, $|3 - 26| = 23 \ne 13$, so $(3, 26) \in R_2$. However, $|13 - 26| = 13$, so $(13, 26) \notin R_2$. Thus, $R_2$ is not transitive. Since $R_2$ is not transitive, $R_2$ is not an equivalence relation. Therefore, neither $R_1$ nor $R_2$ is an equivalence relation.

$R = \{(x, y) : y \in A_i, iff x \in A_i, 1 \le i \le k\}$ (1) Reflexive: $(a, a) \Rightarrow a \in A_i$ iff $a \in A_i$ (2) Symmetric: $(a, b) \Rightarrow a \in A_i$ iff $b \in A_i$. $(b, a) \in R$ as $b \in A_i$ iff $a \in A_i$ (3) Transitive: $(a, b) \in R$ & $(b, c) \in R$. $\Rightarrow a \in A_i$ iff $b \in A_i$ & $b \in A_i$ iff $c \in A_i$. $\Rightarrow a \in A_i$ iff $c \in A_i$. $\Rightarrow (a, c) \in R$. Therefore, the relation is an equivalence relation.

For $R_1$: $aR_1b \Leftrightarrow ab \ge 0$. Reflexive: $(a,a) \in R_1$ since $a^2 \ge 0$ for all $a \in R$. Symmetric: If $(a,b) \in R_1$, then $ab \ge 0$, so $ba \ge 0$, and $(b,a) \in R_1$. Transitive: If $(a,b) \in R_1$ and $(b,c) \in R_1$, then $ab \ge 0$ and $bc \ge 0$. This does not necessarily imply $ac \ge 0$. For example, $a = -5$, $b = 0$, $c = 5$. Then $ab = 0 \ge 0$ and $bc = 0 \ge 0$, but $ac = -25 < 0$. Thus $R_1$ is not transitive, and hence not an equivalence relation. For $R_2$: $aR_2b \Leftrightarrow a \ge b$. Reflexive: $(a,a) \in R_2$ since $a \ge a$ for all $a \in R$. Symmetric: If $(a,b) \in R_2$, then $a \ge b$. This does not necessarily imply $b \ge a$. For example, $a = 2$, $b = 1$. Then $2 \ge 1$, but $1 \ngeq 2$. Thus $R_2$ is not symmetric, and hence not an equivalence relation. Therefore, neither $R_1$ nor $R_2$ is an equivalence relation.

For $R$ to be an equivalence relation, it needs to be reflexive, symmetric and transitive. Reflexive: $(a, a) \in R$ for all $a \in N$. So, $3a + \alpha a = 7k$, where $k$ is an integer. This implies $(3 + \alpha)a = 7k$. Thus, $3 + \alpha$ must be a multiple of $7$, so $3 + \alpha = 7k_1$. Therefore, $\alpha = 7k_1 - 3 = 7k_1 + 4$ for some integer $k_1$. This indicates that when $\alpha$ is divided by $7$, the remainder is $4$. Symmetric: If $(a, b) \in R$, then $(b, a) \in R$. Given $3a + (7k - 3)b = 7m$, then $3(a - b) + 7kb = 7m$. If $3(b - a) + 7ka = 7m$, then $aRb \implies bRa$. Thus, $R$ will be symmetric for $\alpha = 7k_1 - 3$. Transitive: If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. So, $3a + (7k - 3)b = 7k_1$ and $3b + (7k_2 - 3)c = 7k_3$. Adding $3a + 7kb + (7k_2 - 3)c = 7(k_1 + k_3)$. Then $3a + (7k_2 - 3)c = 7m$. Thus, $(a, c) \in R$. Therefore, $R$ is transitive. Since $\alpha = 7k - 3 = 7k + 4$, the remainder when $\alpha$ is divided by $7$ is $4$.

Let $S = \{1, 2, 3, \dots, 60\}$. Relation $R = \{(a,b) : b=pq, p,q \geq 3, p,q \text{ are primes}\}$. Prime numbers greater than or equal to $3$ and less than or equal to $60$ are $3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$. We need to find values of $b = p \times q \leq 60$. If $p=3$, then $q$ can be $3, 5, 7, 11, 13, 17, 19$. That gives us $7$ possibilities. If $p=5$, then $q$ can be $5, 7, 11$. That gives us $3$ possibilities. If $p=7$, then $q$ can be $7$. That gives us $1$ possibility. Total possibilities are $7+3+1 = 11$. Since $a$ can be any number in the set $S$, there are $60$ possible values for $a$ for each of the $11$ values of $b$. Therefore, total number of elements in the relation $R$ is $60 \times 11 = 660$.