Paper Generator

First, we determine the pairs $(x, y)$ that satisfy the condition $0 \le x^2 + 2y \le 4$ for the given set $A = \{-3, -2, -1, 0, 1, 2, 3\}$. For $y = -3$, $x^2 + 2(-3) \le 4 \implies x^2 \le 10$, so $x$ can be $-3$ or $3$. For $y = -2$, $x^2 + 2(-2) \le 4 \implies x^2 \le 8$, so $x$ can be $-2$ or $2$. For $y = -1$, $x^2 + 2(-1) \le 4 \implies x^2 \le 6$, so $x$ can be $-2$ or $2$. For $y = 0$, $x^2 + 2(0) \le 4 \implies x^2 \le 4$, so $x$ can be $-2, -1, 0, 1, 2$. For $y = 1$, $x^2 + 2(1) \le 4 \implies x^2 \le 2$, so $x$ can be $-1, 0, 1$. For $y = 2$, $x^2 + 2(2) \le 4 \implies x^2 \le 0$, so $x$ can only be $0$. Thus, the relation R consists of the following pairs: $R = \{(3, -3), (-3, -3), (-2, -2), (2, -2), (-2, -1), (2, -1), (-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0), (-1, 1), (0, 1), (1, 1), (0, 2)\}$. The number of elements in R is $l = 15$. To make R reflexive, it must include all pairs $(x, x)$ for every $x \in A$. The set A has 7 elements, so the reflexive relation must contain $(-3, -3), (-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2), (3, 3)$. Currently, R has $(-3,-3), (-2, -2), (0,0), (1,1)$. So the missing pairs are $(-1, -1), (2, 2), (3, 3)$, meaning $m = 3$. Therefore, $l + m = 15 + 3 = 18$.

The given sets are $A = \{(x, y) \in R \times R : x^2 + y^2 = 25\}$ and $B = \{(x, y) \in R \times R : x^2 + 9y^2 = 144\}$. To find the intersection of A and B, we solve the system of equations. Subtracting the equation for A from the equation for B gives: $x^2 + 9y^2 - (x^2 + y^2) = 144 - 25$, which simplifies to $8y^2 = 119$, so $y^2 = \frac{119}{8}$. Substituting this into the equation for A, we get $x^2 = 25 - \frac{119}{8} = \frac{200 - 119}{8} = \frac{81}{8}$. Thus, $x = \pm \frac{9}{\sqrt{8}}$ and $y = \pm \frac{\sqrt{119}}{\sqrt{8}}$. Since $D = A \cap B$, the set D contains 4 points. Now, consider the set $C = \{(x, y) \in Z \times Z : x^2 + y^2 \le 4\}$. The integer pairs (x, y) that satisfy this condition are: $(\pm 2, 0)$, $(0, \pm 2)$, $(\pm 1, 0)$, $(0, \pm 1)$, $(\pm 1, \pm 1)$, and $(0, 0)$. Counting these points, we find that there are 13 such pairs. The number of one-to-one functions from D to C is the number of ways to choose 4 elements from C and arrange them, which is $P(13, 4) = \frac{13!}{(13-4)!} = \frac{13!}{9!} = 13 \times 12 \times 11 \times 10 = 17160$.

For set A, we have $|x - 1| \leq 4$ and $|y - 5| \leq 6$. This implies $-4 \leq x - 1 \leq 4$ and $-6 \leq y - 5 \leq 6$. Therefore, $-3 \leq x \leq 5$ and $-1 \leq y \leq 11$. For set B, we have $16(x - 2)^2 + 9(y - 6)^2 \leq 144$, which simplifies to $\frac{(x - 2)^2}{9} + \frac{(y - 6)^2}{16} \leq 1$. This represents an ellipse centered at $(2, 6)$ with semi-major axis 4 and semi-minor axis 3. From the diagram, it's clear that $B \subset A$.

To determine the elements in the relation $R$, we need to find all pairs $(x, y)$ such that $\max(x, y) \in {3, 4}$. The set $A = {0, 1, 2, 3, 4, 5}$. If $\max(x, y) = 3$, the pairs are $(0, 3), (1, 3), (2, 3), (3, 3), (3, 0), (3, 1), (3, 2)$. If $\max(x, y) = 4$, the pairs are $(0, 4), (1, 4), (2, 4), (3, 4), (4, 4), (4, 0), (4, 1), (4, 2), (4, 3)$. Thus, the total number of elements in $R$ is $7 + 9 = 16$. Therefore, statement S1 is false. For statement S2, we check the properties of $R$: Reflexivity: $(0, 0)$ is not in $R$, so $R$ is not reflexive. Symmetry: If $(x, y) \in R$, then $\max(x, y) \in {3, 4}$, which implies $\max(y, x) \in {3, 4}$, so $(y, x) \in R$. Thus, $R$ is symmetric. Transitivity: $(0, 3) \in R$ and $(3, 1) \in R$, but $(0, 1)$ is not in $R$. Therefore, $R$ is not transitive. Statement S2 is true since $R$ is symmetric but neither reflexive nor transitive.

$xRy \Leftrightarrow 2x - y \in \{0, 1\}$ $\Rightarrow y = 2x$ or $y = 2x - 1$ $A=\{-3,-2,-1, 0, 1, 2, 3\}$ $R = \{(-1,-2), (0,0), (1, 2), (-1, -3), (0, -1), (1, 1), (2,3)\}$ $\Rightarrow I=7$ For $R$ to be reflexive $(0,0), (1,1) \in R$ But other $(a, a)$ such that $2a - a \in \{0, 1\}$ $\Rightarrow a \in \{0, 1\}$ 5 other pairs needs to be added $\Rightarrow m=5$ $xRy \Rightarrow yRx$ to be symmetric $(-1, -2) \Rightarrow (-2, -1)$ $(1, 2) \Rightarrow (2, 1)$ $(-1, -3) \Rightarrow (-3, -1)$ $(0, -1) \Rightarrow (-1, 0)$ $(2, 3) \Rightarrow (3, 2)$ $\Rightarrow 5$ needs to be added, $n=5$ $\Rightarrow l+m+n=17$

$A: |x-1| \leq 4$ and $|y-5| \leq 6$ $\Rightarrow -4 \leq x-1 \leq 4 \Rightarrow -6 \leq y-5 \leq 6$ $\Rightarrow -3 \leq x \leq 5 \Rightarrow -1 \leq y \leq 11$ $B : 16(x-2)^{2} + 9(y-6)^{2} \leq 144$ $B : \frac{(x-2)^{2}}{9} + \frac{(y-6)^{2}}{16} \leq 1$ From Diagram $B \subset A$

Probability distribution $x = 0$, $p = \frac{^7C_2}{^{10}C_2} = \frac{42}{90}$ $x = 1$, $p = \frac{^7C_1 \times ^3C_1}{^{10}C_2} = \frac{42}{90}$ $x = 2$, $p = \frac{^3C_2}{^{10}C_2} = \frac{6}{90}$ Now, $\mu = \sum x_i p_i = \frac{42}{90} + \frac{12}{90} = \frac{54}{90}$ $\sigma^2 = \sum p_i x_i^2 - \mu^2 = \frac{42}{90} + \frac{24}{90} - (\frac{54}{90})^2$ $\Rightarrow \frac{66}{90} - (\frac{54}{90})^2$ $\sigma^2 \Rightarrow \frac{28}{75}$

$i^{k_1} + i^{k_2} \ne 0 \Rightarrow i^{k_1} \rightarrow 4$ option for $i, -1, -i, 1$ Total cases $ \Rightarrow 4 \times 4 = 16$ Unfovourble cases $ \Rightarrow i^{k_1} + i^{k_2} = 0$ $ \{\begin{array}{c}1, -1 \\ -1, 1 \\ i, -i \\ -i, i\end{array}\}$ 4 Cases $\Rightarrow$ Probability $ = \frac{16-4}{16} = \frac{3}{4}$

A, E,G R D N Probabllity $(P) = \frac{\text{favourable case}}{\text{Total case}}$ (when A & E are in order) Total case = $6!$ Favourable case = ${6}C_2 . 4! = (15)4! = (30)4!$ $P = \frac{(15)4!}{(30)4!} = \frac{1}{2}$ Probability when not in order = $1 - \frac{1}{2} = \frac{1}{2}$

$E_1$: Bag $B_1$ is selected $B_1$: 6 W 4 B $B_2$: 4 W 6 B $B_3$: 5 W 5 B $E_2$: bag $B_2$ is selected We have to find $P(\frac{E_2}{A})$ $P(\frac{E_2}{A}) = \frac{P(E_2)P(\frac{A}{E_2})}{P(E_1)P(\frac{A}{E_1}) + P(E_2)P(\frac{A}{E_2}) + P(E_3)P(\frac{A}{E_3})} = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}} = \frac{4}{15}$

Bag $1 = \{4 W, 5 B\}$ Bag $2 = \{nW, 3B\}$ $P(\frac{W}{Bag2}) = \frac{29}{45}$ $\Rightarrow P(\frac{W}{B_1}) \times P(\frac{W}{B_2}) + P(\frac{B}{B_1}) \times P(\frac{W}{B_2}) = \frac{29}{45}$ $\frac{4}{9} \times \frac{n+1}{n+4} + \frac{5}{9} \times \frac{n}{n+4} = \frac{29}{45}$ $n = 6$

Probability that a Red ball comes from Bag I ($p$): $p(B_1/R) = \frac{p(B_1) \cdot p(R/B_1)}{p(R)}$ Substituting the values, $p(B_1/R) = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}}$ Simplify to find $p$: $=\frac{\frac{1}{10}}{\frac{1}{3}(1.2)} = \frac{1}{4}$ So, $p = \frac{1}{4}$. Probability that a Green ball comes from Bag III ($q$): $p(B_3/G) = \frac{p(B_3) \cdot p(G/B_3)}{p(G)}$ Substitute the values, $p(B_3/G) = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10}}$ Simplify to find $q$: $=\frac{\frac{2}{15}}{\frac{1}{3} \times 1.2} = \frac{1}{3}$ So, $q = \frac{1}{3}$. Calculation of $(\frac{1}{p} + \frac{1}{q})$: $\frac{1}{p} = 4$, $\frac{1}{q} = 3$ Therefore, $(\frac{1}{p} + \frac{1}{q}) = 4 + 3 = 7$

Since |x| is always non-negative, x^2 + 5|x| + 6 is the sum of positive terms plus 6, which can never be zero. Thus, no real solution.

For the block to remain stationary relative to the wedge, the pseudo force ma must balance the component of gravity along the incline. ma cosθ = mg sinθ => a = g tanθ.

Haematite is Fe2O3 and Magnetite is Fe3O4. Bauxite is Al, Malachite is Cu.

Let z = e^(iθ). Then z/(1+z^2) = e^(iθ) / (1 + e^(2iθ)). Simplifying this using Euler formula results in a real number.

Hydrogen bonding occurs when H is bonded to highly electronegative atoms like F, O, or N.

Velocity v = dx/dt = 3t^2 - 12t + 3. Acceleration a = dv/dt = 6t - 12. For a=0, 6t=12 => t=2s. At t=2, v = 3(4) - 12(2) + 3 = -9.