Paper Generator

Given, $4(a-6)^2 + 9(b-5)^2 ≤ 36$. Let $a-6 = x$ and $b-5 = y$. Therefore, $4x^2 + 9y^2 ≤ 36$, which implies $\frac{x^2}{9} + \frac{y^2}{4} ≤ 1$. This is the equation of an ellipse. According to set A, $|a-5| < 1$, which implies, since $a-6=x$, then $a-5 = x+1$. Thus $|x+1| < 1$, implying $-1 < x+1 < 1$, which leads to $-2 < x < 0$. Similarly, $|b-5| < 1$, implying $|y| < 1$, which leads to $-1 < y < 1$. To check if the entire set A is inside of B, we can test a point of set A in the inequality $\frac{x^2}{9} + \frac{y^2}{4} ≤ 1$. Consider the point $(-2, 1)$. Substituting into the inequality gives LHS = $\frac{(-2)^2}{9} + \frac{1^2}{4} = \frac{4}{9} + \frac{1}{4} = \frac{25}{36} < 1$. Since the inequality holds, $(-2, 1)$ is inside the ellipse. Similarly, $(-2, -1)$ is also inside the ellipse. Hence, we can say that entire set A is inside of set B, therefore $A ⊂ B$.

For $R_1$, $2x + y = 10$ and $x, y \in N$, the possible values for $x$ and $y$ are: $x = 1, y = 8 \implies (1, 8)$ $x = 2, y = 6 \implies (2, 6)$ $x = 3, y = 4 \implies (3, 4)$ $x = 4, y = 2 \implies (4, 2)$ Therefore, $R_1 = \{(1, 8), (2, 6), (3, 4), (4, 2)\}$. Hence, the range of $R_1$ is $\{2, 4, 6, 8\}$. $R_1$ is not symmetric. $R_1$ is not transitive as $(3, 4), (4, 2) \in R_1$, but $(3, 2) \notin R_1$. For $R_2$, $x + 2y = 10$ and $x, y \in N$, the possible values for $x$ and $y$ are: $x = 8, y = 1 \implies (8, 1)$ $x = 6, y = 2 \implies (6, 2)$ $x = 4, y = 3 \implies (4, 3)$ $x = 2, y = 4 \implies (2, 4)$ Therefore, $R_2 = \{(8, 1), (6, 2), (4, 3), (2, 4)\}$. Hence, the range of $R_2 = \{1, 2, 3, 4\}$. $R_2$ is not symmetric and transitive.

Let $A$ be the set of students who opted for Mathematics, $B$ be the set of students who opted for Physics, and $C$ be the set of students who opted for Chemistry. We are given: Total number of students $= 140$ $n(A) = \lfloor\frac{140}{2}\rfloor = 70$ $n(B) = \lfloor\frac{140}{3}\rfloor = 46$ $n(C) = \lfloor\frac{140}{5}\rfloor = 28$ $n(A \cap B) = \lfloor\frac{140}{6}\rfloor = 23$ $n(B \cap C) = \lfloor\frac{140}{15}\rfloor = 9$ $n(C \cap A) = \lfloor\frac{140}{10}\rfloor = 14$ $n(A \cap B \cap C) = \lfloor\frac{140}{30}\rfloor = 4$ Using the principle of inclusion-exclusion: $n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$ $n(A \cup B \cup C) = 70 + 46 + 28 - 23 - 9 - 14 + 4 = 102$ The number of students who did not opt for any of the three courses is: $140 - n(A \cup B \cup C) = 140 - 102 = 38$

Let $S = {1, 2, 3, ..., 100}$. The total number of non-empty subsets is $2^{100} - 1$. The number of subsets with an odd product consists only of odd numbers. There are 50 odd numbers in the set $S$. So, the number of such subsets is $2^{50} - 1$. Therefore, the number of subsets with an even product is $(2^{100} - 1) - (2^{50} - 1) = 2^{100} - 2^{50} = 2^{50}(2^{50} - 1)$.

Given $A = {x \in Z : 2(x+2)(x^2 - 5x + 6) = 1}$. Since $2(x+2)(x^2 - 5x + 6) = 1$, we can rewrite it as $2(x+2)(x^2 - 5x + 6) = 2^0$. This implies that $x = -2, 2, 3$, so $A = {-2, 2, 3}$. Also, $B = {x \in Z : -3 < 2x - 1 < 9}$. Adding 1 to all sides, we get $-2 < 2x < 10$. Dividing by 2, we get $-1 < x < 5$. Thus, $B = {0, 1, 2, 3, 4}$. Now, $A \times B$ has $3 \times 5 = 15$ elements. The number of subsets of $A \times B$ is $2^{15}$.

Let the total population be $100$. Then: People who read A only: $25 - 8 = 17$ People who read B only: $20 - 8 = 12$ People who read A and look into advertisements: $0.30 \times 17 = 5.1$ People who read B and look into advertisements: $0.40 \times 12 = 4.8$ People who read both A and B and look into advertisements: $0.50 \times 8 = 4$ Total percentage of people who look into advertisements: $5.1 + 4.8 + 4 = 13.9$ Therefore, the percentage of the population who look into advertisement is $13.9$%.

The question provides that $A \cap B \subseteq C$ and $A \cap B \neq \phi$. Analyzing each statement: Statement A: If $(A - B) \subseteq C$, then $A \subseteq C$. This statement can be verified using Venn diagrams. If the part of A that is not in B is contained in C, then A is a subset of C. Statement B: $B \cap C \neq \phi$. This means that the intersection of B and C is not empty. Given that $A\cap B \subseteq C$, this is true. Statement C: $(C \cup A) \cap (C \cup B) = C$. Using the distributive property of sets, $(C \cup A) \cap (C \cup B) = C \cup (A \cap B)$. Since $A \cap B \subseteq C$, then $C \cup (A \cap B) = C$. So, this statement is true. Statement D: If $(A - C) \subseteq B$, then $A \subseteq B$. This statement is NOT necessarily true. It is possible to have $A - C \subseteq B$, $A \cap B \subseteq C$ and $A \cap B \neq \phi$, but it does NOT require that $A \subseteq B$. This is because A can contain elements outside of B that are also outside of C. Therefore, the statement that is not true is D.

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.