Sets

$$\begin{gathered} \text{for} m=1\Rightarrow \frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \dots \dots \frac{1}{10}\to 9 \\ \text{for} m=2\Rightarrow \frac{2}{3},\frac{2}{5},\frac{2}{7},\frac{2}{9}\to 4 \\ \text{for} m=3\Rightarrow \frac{3}{4},\frac{3}{5},\frac{3}{7},\frac{3}{8},\frac{3}{10}\to 5 \\ \text{for} m=4\Rightarrow \frac{4}{5},\frac{4}{7},\frac{4}{9}\to 3 \\ \text{for} m=5\Rightarrow \frac{5}{6},\frac{5}{7},\frac{5}{8},\frac{5}{9},\to 4 \\ \text{for} m=6\Rightarrow \frac{6}{7}\to 1 \\ \text{for} m=7\Rightarrow \frac{7}{8},\frac{7}{9},\frac{7}{10},\to 3 \\ \text{for} m=8\Rightarrow \frac{8}{9}\to 1 \\ \text{for} m=9\Rightarrow \frac{9}{10}\to 1 \\ \text{then} n\left(B\right)=31 \end{gathered}$$

Set A:

\(A=\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}:|\alpha-1| \leq 4\) and \(|\beta-5| \leq 6\}\)

This represents a rectangle with:

$-3 \leq \alpha \leq 5$ (from $\alpha - 1 \in [-4, 4]$)

$-1 \leq \beta \leq 11$ (from $\beta - 5 \in [-6, 6]$)

Set B:

\(B=\left\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}: 16(\alpha-2)^2+9(\beta-6)^2 \leq 144\right\}\)

This is an ellipse centered at $(2, 6)$.

Standard form: \(\frac{(\alpha-2)^2}{9}+\frac{(\beta-6)^2}{16} \leq 1\)

So, it has:

Horizontal semi-axis \(= 3\)

Vertical semi-axis \(= 4\)

Hence, the ellipse spans:

$\alpha \in [2 - 3, 2 + 3] = [-1, 5]$

$\beta \in [6 - 4, 6 + 4] = [2, 10]$

Now compare sets:

Rectangle A: $\alpha \in [-3, 5], \beta \in [-1, 11]$

Ellipse B: completely lies within bounds of A

So, B is a subset of A, but A is not a subset of B (e.g., point $(-3, -1)$ is in A but not in B)

\(\begin{aligned}
& 2^{\mathrm{m}}-2^{\mathrm{n}}=56 \\
& 2^{\mathrm{n}}\left(2^{\mathrm{m}-\mathrm{n}}-1\right)=2^3 \times 7 \\
& 2^{\mathrm{n}}=2^3 \text { and } 2^{\mathrm{m}-\mathrm{n}}-1=7 \\
& \Rightarrow \mathrm{n}=3 \text { and } 2^{\mathrm{m}-\mathrm{n}}=8 \\
& \Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}-\mathrm{n}=3 \\
& \Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}=6 \\
& \mathrm{P}(6,3) \text { and } \mathrm{Q}(-2,-3) \\
& \mathrm{PQ}=\sqrt{8^2+6^2}=\sqrt{100}=10
\end{aligned}\)

\(\begin{aligned}
& 2^{\mathrm{m}}-2^{\mathrm{n}}=56 \\
& 2^{\mathrm{n}}\left(2^{\mathrm{m}-\mathrm{n}}-1\right)=2^3 \times 7 \\
& 2^{\mathrm{n}}=2^3 \text { and } 2^{\mathrm{m}-\mathrm{n}}-1=7 \\
& \Rightarrow \mathrm{n}=3 \text { and } 2^{\mathrm{m}-\mathrm{n}}=8 \\
& \Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}-\mathrm{n}=3 \\
& \Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}=6 \\
& \mathrm{P}(6,3) \text { and } \mathrm{Q}(-2,-3) \\
& \mathrm{PQ}=\sqrt{8^2+6^2}=\sqrt{100}=10
\end{aligned}\)

$$\begin{gathered} \text{for} m=1\Rightarrow \frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \dots \dots \frac{1}{10}\to 9 \\ \text{for} m=2\Rightarrow \frac{2}{3},\frac{2}{5},\frac{2}{7},\frac{2}{9}\to 4 \\ \text{for} m=3\Rightarrow \frac{3}{4},\frac{3}{5},\frac{3}{7},\frac{3}{8},\frac{3}{10}\to 5 \\ \text{for} m=4\Rightarrow \frac{4}{5},\frac{4}{7},\frac{4}{9}\to 3 \\ \text{for} m=5\Rightarrow \frac{5}{6},\frac{5}{7},\frac{5}{8},\frac{5}{9}\to 4 \\ \text{for} m=6\Rightarrow \frac{6}{7}\to 1 \\ \text{for} m=7\Rightarrow \frac{7}{8},\frac{7}{9},\frac{7}{10}\to 3 \\ \text{for} m=8\Rightarrow \frac{8}{9}\to 1 \\ \text{for} m=9\Rightarrow \frac{9}{10}\to 1 \\ \text{So,} n\left(B\right)=31 \end{gathered}$$

Let ${(\sqrt{3}+\sqrt{2})}^x=t$

$\Rightarrow {\left(\frac{1}{\sqrt{3}-\sqrt{2}}\right)}^x=t$

$\Rightarrow {(\sqrt{3}-\sqrt{2})}^x=\frac{1}{t}$

Now, ${(\sqrt{3}+\sqrt{2})}^x+{(\sqrt{3}-\sqrt{2})}^x=10$

$\Rightarrow t+\frac{1}{t}=10$

$\Rightarrow t^2-10t+1=0$

$\Rightarrow t=\frac{10\pm \sqrt{100-4}}{2}$

$\Rightarrow t=5\pm 2\sqrt{6}$

$\Rightarrow {(\sqrt{3}+\sqrt{2})}^x=(5\pm 2\sqrt{6})$

So, \(x = 2, x = –2\)

So, number of elements is \(2\)

$$\begin{gathered} \text{Given, total no. of families =1003} \\ \text{Let TV be set A and radio be set B} \\ \mathrm{n}( \mathrm{A})=187, \mathrm{n}( \mathrm{B})=794 \\ \mathrm{n}(\overline{A}\cap \overline{B})=63 \\ \therefore \mathrm{n}( \mathrm{A}\cup \mathrm{B})=1003-63=940 \end{gathered}$$

$$\begin{gathered} \mathrm{n}( \mathrm{A}\cap \mathrm{B})=\mathrm{n}( \mathrm{A})+\mathrm{n}( \mathrm{B})-\mathrm{n}( \mathrm{A}\cap \mathrm{B}) \\ 940=187+794-n(A\cap B) \\ n(A\cap B)=41 \end{gathered}$$

Set A is an AP with:

first term = 1, common difference = 5

General term of A:

a = 1 + 5k

Set B is an AP with:

first term = 9, common difference = 7

General term of B:

B = 9+7m

Number of elements in A = 2025

Number of elements in B = 2025

To find common elements, solve:

1+5k = 9+7m

5k – 7m = 8

One solution is k = 3, m = 1 giving common element 16 .

LCM of 5 and 7 = 35, so common elements form an AP:

16, 51, 86, …

Largest common element possible within first 2025 terms:

Last term of A = 1+2024 × 5 = 10121

Common terms are $16+35r\le 10121$

$35r\le 10105$

$\text{r}\le 288$

Hence number of common elements = 289.

Therefore,

$n\left(A\cup B\right)=n\left(A\right)+n\left(B\right)-n\left(A\cap B\right)$

= 2025 + 2025 –289 = 3761

Answer: 3761

Ａ＝｛1,６,１1,１6,２1,２6,３1,３6,４1,４6,５1,５6,６1,６6,７1,７6,８1,８6,９1,．．．．．｝

Ｂ＝｛9,１6,２3,３0,３7,４４,５1,５8,６5,７２,７9,８6,９3,１00,．．．．．．

Ａ ∩ Ｂ ＝ ｛１6,５１,８６,．．．．｝

For set ′Ａ′ ⇒ ${\mathrm{T}}_{2025}$ ＝１＋(2025 －１)(５) ＝ 10121

For set $B^{\text{'}}\Rightarrow T_{2025}=9+(2025-1)\left(7\right)=14177$

So, for $(A \cap B) \Rightarrow T_n = 16 + (n - 1)(35) \leq 10121$

$$(n - 1) \leq \frac{10121 - 16}{35} = 288.71$$

$$n \leq 289.71 \Rightarrow n = 289$$

$\therefore n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$$=2025+2025-289=3761$

The total number of integers is \(700-100+1=601\)

The number of multiples of 3 between 100 and 700 \(=200\)

The number of multiples of 4 between 100 and 700 \(=151\)

The number of multiples of 12 between 100 and 700 \(=50\)

Required Count is \(601-200-151+50=300\)

The total number of integers is 700-100+1=601

The number of multiples of 3 between 100 and 700 is =200

The number of multiples of 4 between 100 and 700 is =151

The number of multiples of 12 between 100 and 700 is =50

The number of integers is 601-200-151+50=300.

$$\begin{gathered} \text{for} m=1\Rightarrow \frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \dots \dots \frac{1}{10}\to 9 \\ \text{for} m=2\Rightarrow \frac{2}{3},\frac{2}{5},\frac{2}{7},\frac{2}{9}\to 4 \\ \text{for} m=3\Rightarrow \frac{3}{4},\frac{3}{5},\frac{3}{7},\frac{3}{8},\frac{3}{10}\to 5 \\ \text{for} m=4\Rightarrow \frac{4}{5},\frac{4}{7},\frac{4}{9}\to 3 \\ \text{for} m=5\Rightarrow \frac{5}{6},\frac{5}{7},\frac{5}{8},\frac{5}{9},\to 4 \\ \text{for} m=6\Rightarrow \frac{6}{7}\to 1 \\ \text{for} m=7\Rightarrow \frac{7}{8},\frac{7}{9},\frac{7}{10},\to 3 \\ \text{for} m=8\Rightarrow \frac{8}{9}\to 1 \\ \text{for} m=9\Rightarrow \frac{9}{10}\to 1 \\ \text{then} n\left(B\right)=31 \end{gathered}$$

$$\begin{gathered} \text{for} m=1\Rightarrow \frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \dots \dots \frac{1}{10}\to 9 \\ \text{for} m=2\Rightarrow \frac{2}{3},\frac{2}{5},\frac{2}{7},\frac{2}{9}\to 4 \\ \text{for} m=3\Rightarrow \frac{3}{4},\frac{3}{5},\frac{3}{7},\frac{3}{8},\frac{3}{10}\to 5 \\ \text{for} m=4\Rightarrow \frac{4}{5},\frac{4}{7},\frac{4}{9}\to 3 \\ \text{for} m=5\Rightarrow \frac{5}{6},\frac{5}{7},\frac{5}{8},\frac{5}{9},\to 4 \\ \text{for} m=6\Rightarrow \frac{6}{7}\to 1 \\ \text{for} m=7\Rightarrow \frac{7}{8},\frac{7}{9},\frac{7}{10},\to 3 \\ \text{for} m=8\Rightarrow \frac{8}{9}\to 1 \\ \text{for} m=9\Rightarrow \frac{9}{10}\to 1 \\ \text{then} n\left(B\right)=31 \end{gathered}$$