Sets and Relations

Given, $\left|n^2-10 n+19\right|<6$

$\Rightarrow-6 < n^2-10 n+19 < 6$

Take, $-6 < n^2-10 n+19$ and $n^2-10 n+19 < 6$

$ \begin{array}{ll} \Rightarrow n^2-10 n+25 > 0 & \text { and }\quad n^2-10 n+13 < 0 \\\\ \Rightarrow(n-5)^2 > 0 & \text { and } n=\frac{10 \pm \sqrt{100-52}}{2}<0 \end{array} $

$\Rightarrow n \in \mathbb{Z}-\{5\}$

$ \begin{array}{lr} & \therefore n \in[5-2 \sqrt{3}, 5+2 \sqrt{3}] \\\\ & \therefore n \in[13,8.3] \\\\ & \therefore n=2,3,4,5,6,7,8 \end{array} $

Thus, number of element in the set is ' 6 '

We have, $A=\{0,3,4,6,7,8,9,10\}$

Case I : $x-y$ is odd, if one is odd and one is even and $x>y$.

$\therefore$ Possibilites are $\{(3,0),(4,3),(6,3),(7,6),(7,4)$, $(7,0),(8,7),(8,3),(9,8),(9,6),(9,4),(9,0),(10,9),(10$, $7),(10,3)\}$

No. of cases $=15$

Case II : $x-y=2$

$\therefore$ Possibilities are $\{(6,4),(8,6),(9,7),(10,8)\}$

$\therefore$ No. of cases $=4$

So, minimum ordered pair to be added $=15+4=19$

Given sets :

A={1,2,3,4, ............,10}

B={0,1,2,3,4}

We are looking for pairs $(a,b) \in A \times A$ such that :

$ 2(a-b)^2 + 3(a-b) \in B $

Let's break down the relation :

Case 1 : $ a-b = 0 $

$ 2(a-b)^2 + 3(a-b) = 0 $

Pairs : $(1,1), (2,2), (3,3), \ldots, (10,10)$ which gives 10 pairs.

Case 2 : $ a-b = 1 $

$ 2(a-b)^2 + 3(a-b) = 2(1) + 3(1) = 5 $

But 5 is not in B, so no pairs for this case.

Case 3 : $ a-b = -1 $

$ 2(a-b)^2 + 3(a-b) = 2(1) - 3(1) = -1 $

This value is not in B, so no pairs for this case.

Case 4 : $ a-b = 2 $

$ 2(a-b)^2 + 3(a-b) = 2(4) + 3(2) = 8+6 = 14 $

Again, 14 is not in B, so no pairs for this case.

Case 5 : $ a-b = -2 $

$ 2(a-b)^2 + 3(a-b) = 2(4) - 3(2) = 8 - 6 = 2 $

Pairs : $(1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9), (8,10)$ which gives 8 pairs.

For any other $ a-b $ value, the quadratic will grow larger than the maximum value in B, so we don't need to consider them.

In total, we have $ 10 + 8 = 18 $ pairs in the relation $ R $.

Therefore, the number of elements in the relation $ R $ is 18.

Elements of the type $3 \mathrm{k}=3$

Elements of the type $3 \mathrm{k}+1=1,7,9$

Elements of the type $3 \mathrm{k}+2=2,5,11$

Subsets containing one element $S_1=1$

Subsets containing two elements

$ S_2={ }^3 C_1 \times{ }^3 C_1=9 $

Subsets containing three elements

$ \mathrm{S}_3={ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1+1+1=11 $

Subsets containing four elements

$

\mathrm{S}_4={ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2=11 $

Subsets containing five elements

$ \mathrm{S}_5={ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2 \times 1=9 $

Subsets containing six elements $\mathrm{S}_6=1$

Subsets containing seven elements $\mathrm{S}_7=1$

$ \Rightarrow \text { sum }=43 $

$R=\{(a, b)(b, c)(b, d)\}$

$S:\{a, b, c, d\}$

Adding $(a, a),(b, b),(c, c),(d, d)$ make reflexive.

Adding $(b, a),(c, b),(d, b)$ make Symmetric

And adding $(a, d),(a, c)$ to make transitive

Further $(d, a) \&(c, a)$ to be added to make Symmetricity.

Further $(c, d) \&(d, c)$ also be added.

So total 13 elements to be added to make equivalence.

Here $S = \{ 4,6,9\} $

And $T = \{ 9,10,11,\,\,......,\,\,1000\} $.

We have to find all numbers in the form of $4x + 6y + 9z$, where $x,y,z \in \{ 0,1,2,\,......\} $.

If a and b are coprime number then the least number from which all the number more than or equal to it can be express as $ax + by$ where $x,y \in \{ 0,1,2,\,......\} $ is $(a - 1)\,.\,(b - 1)$.

Then for $6y + 9z = 3(2y + 3z)$

All the number from $(2 - 1)\,.\,(3 - 1) = 2$ and above can be express as $2x + 3z$ (say t).

Now $4x + 6y + 9z = 4x + 3(t + 2)$

$ = 4x + 3t + 6$

again by same rule $4x + 3t$, all the number from $(4 - 1)\,(3 - 1) = 6$ and above can be express from $4x + 3t$.

Then $4x + 6y + 9z$ express all the numbers from 12 and above.

again 9 and 10 can be express in form $4x + 6y + 9z$.

Then set $A = \{ 9,10,12,13,\,....,\,1000\} .$

Then $T - A = \{ 11\} $

Only one element 11 is there.

Sum of elements of $T - A = 11$

As C $\cap$ B $\ne$ $\phi$, c must be not be formed by {1, 2, 4, 5}

$\therefore$ Number of subsets of A = 2⁷ = 128

and number of subsets formed by {1, 2, 4, 5} = 16

$\therefore$ Required no. of subsets = 2⁷ $-$ 2⁴ = 128 $-$ 16 = 112

$\because$ $(B \cup C)' = B'\, \cap C'$

B' is a set containing sub sets of A containing element 1 and not containing 2.

And C' is a set containing subsets of A whose sum of elements is not prime.

So, we need to calculate number of subsets of {3, 4, 5, 6, 7} whose sum of elements plus 1 is composite.

Number of such 5 elements subset = 1

Number of such 4 elements subset = 3 (except selecting 3 or 7)

Number of such 3 elements subset = 6 (except selecting {3, 4, 5}, {3, 6, 7}, {4, 5, 7} or {5, 6, 7})

Number of such 2 elements subset = 7 (except selecting {3, 7}, {4, 6}, {5, 7})

Number of such 1 elements subset = 3 (except selecting {4} or {6})

Number of such 0 elements subset = 1

$n(B'\, \cap C') = 21 \Rightarrow n(B \cup C) = {2^7} - 21 = 107$

Given, ${R}_1=\left\{\left(p, p^n\right): p\right.$ is a Prime and $n \geq 0$ is an integer $\}$

and, set $A=\{1,2,3 \ldots \ldots .50\}$

$p$ is a Prime number which can take 15 values $2,3,5,7,11,13,17,19,23,29,31,37,41,43$ and 47

$\therefore$ We can calculate no. of elements in $\mathrm{R}_1$

$ \mathrm{R}_1=\left(2,2^0\right),\left(2,2^1\right),\left(2,2^2\right)\left(2,2^3\right) \ldots \ldots\left(2,2^5\right)=6$ number of ordered pairs

$\left(3,3^0\right),\left(3,3^1\right),\left(3,3^2\right) \ldots \ldots . .\left(3,3^3\right)=4$ number of order paris

$\left(5,5^0\right),\left(5,5^1\right),\left(5,5^2\right) \ldots \ldots \ldots . .=3$ number of order paris

$\left(7,7^0\right) \ldots \ldots .\left(7,7^2\right) \ldots \ldots \ldots=3$ number of order paris

$\left(11,11^0\right)$ and $\left(11,11^1\right)=2$ number of order paris

$\left(13,13^0\right)$ and $\left(13,13^1\right)=2$ number of order paris

$ \therefore $ For the 11 prime numbers ($11,13,17,19,23,29,31,37,41,43$ and 47), $n$ can only be 0, 1 (two pairs each).

$ \therefore n\left(\mathrm{R}_1\right)=6+4+3+3+(2 \times 11)=38 $

$\mathrm{R}_2=\left(p, p^n\right) $, where n = 0 or 1

$ \left(2,2^0\right),\left(2,2^1\right)\left(3,3^0\right)\left(3,3^1\right) \ldots . .\left(47,47^0\right)\left(47,47^1\right) $

Two ordered pairs of each element $n\left({R}_2\right)=2 \times 15=30$ elements

Hence $ R_1-R_2=38-30=8$

Sum of all elements of A $\cap$ B = 2 [Sum of natural numbers upto 100 which are neither divisible by 3 nor by 5]

$ = 2\left[ {{{100 \times 101} \over 2} - 3\left( {{{33 \times 34} \over 2}} \right) - 5\left( {{{20 \times 21} \over 2}} \right) + 15\left( {{{6 \times 7} \over 2}} \right)} \right]$

$ = 10100 - 3366 - 2100 + 630$

$ = 5264$

$\sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\{ i,\,j\} } } $

= {1, 1} {1, 2} {1, 3} ..... {1, 10}

{2, 1} {2, 2} {2, 3} ..... {2, 10}

{3, 1} {3, 2} {3, 3} ..... {3, 10}

$ \vdots $

{10, 1} {10, 2} {10, 3} ..... {10, 10}

Now, $A = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {min\{ i,\,j\} } } $

= minimum between i and j in all sets and summation of all those values.

and $B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \{ i,\,j\} } } $

= maximum between i and j in all sets and summation of all those values.

For 1 :

1 is minimum in sets =

{1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {1, 8}, {1, 9}, {1, 10}, {2, 1}, {3, 1}, {4, 1}, {5, 1}, {6, 1}, {7, 1}, {8, 1}, {9, 1}, {10, 1}

$\therefore$ 1 is minimum in 19 sets

1 is maximum in {1, 1} sets.

$\therefore$ 1 is maximum and minimum in total 20 sets.

$\therefore$ Sum of 1 in all those sets = 1 $\times$ 20 = 20

For 2 :

2 is minimum in sets =

{2, 2}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {2, 7}, {2, 8}, {2, 9}, {2, 10}, {3, 2}, {4, 2}, {5, 2}, {6, 2}, {7, 2}, {8, 2}, {9, 2}, {10, 2}

$\therefore$ 2 is minimum in 17 sets

2 is maximum in sets = {1, 2}, {2, 1}, {2, 2}

$\therefore$ 2 is maximum and minimum in 20 sets.

$\therefore$ Sum of 2 in all those sets = 2 $\times$ 20 = 40

Similarly 3 is maximum and minimum in 20 sets.

$\therefore$ Sum of 3 in all those sets = 20 $\times$ 3 = 60

$ \vdots $

Similarly, 10 is maximum and minimum in 20 sets.

$\therefore$ Sum of 10 in all those sets = 20 $\times$ 10 = 200

$\therefore$ A + B = 20 + 20 $\times$ 2 + 20 $\times$ 3 + ....... + 20 $\times$ 10

= 20(1 + 2 + 3 + ...... + 10)

= 20 $\times$ ${{10 \times 11} \over 2}$

= 1100

The numbers upto 24 which gives g.c.d. with 24 equals to 1 are 1, 5, 7, 11, 13, 17, 19 and 23.

Sum of these numbers = 96

There are four such blocks and a number 97 is there upto 100.

$\therefore$ Complete sum

= 96 + (24 $\times$ 8 + 96) + (48 $\times$ 8 + 96) + (72 $\times$ 8 + 96) + 97

= 1633

A = ($-$$\infty$, 1) $\cup$ (3, $\infty$)

B = ($-$$\infty$, $-$2) $\cup$ (2, $\infty$)

C = ($-$$\infty$, 2] $\cup$ [6, $\infty$)

So, A $\cap$ B $\cap$ C = ($-$$\infty$, $-$2) $\cup$ [6, $\infty$)

z $\cap$ (A $\cap$ B $\cap$ C)' = {$-$2, $-$1, 0, $-$1, 2, 3, 4, 5}

Hence, no. of its subsets = 2⁸ = 256.

B $-$ C $ \equiv $ {7, 13, 19, ......, 97, .......}

Now, n² $-$ n $\le$ 100 $\times$ 100

$\Rightarrow$ n(n $-$ 1) $\le$ 100 $\times$ 100

$\Rightarrow$ A = {1, 2, ......., 100}.

So, A$\cap$(B $-$ C) = {7, 13, 19, ......., 97}

Hence, sum = ${{16} \over 2}(7 + 97) = 832$

In this problem, we're dealing with 3-digit numbers in set $A$, and subsets $B$ and $C$ which represent numbers of specific forms.

1. First, we consider the numbers of the form $9k + 2$ (Set $B$) within the 3-digit range, which starts at 101 and ends at 992.

2. We calculate the sum of these numbers, denoted as $s_1$. To calculate $s_1$, you use the formula for the sum of an arithmetic series :

$(n/2) \times (\text{{first term}} + \text{{last term}})$

Here, $n$ is the total count of such numbers. These are 3-digit numbers of the form $9k + 2$, and we can find the total count by subtracting the smallest such number (101) from the largest (992), dividing the result by 9 (because we're considering numbers with a difference of 9), and then adding 1.

The sum $s_1$ is calculated as follows :

$(100/2) \times (101 + 992) = 54650$

3. According to the problem, the sum of all elements of the set $A \cap (B \cup C)$ is $274 \times 400 = 109600$.

Since the set $A \cap (B \cup C)$ is the union of two disjoint sets (the set of all three-digit numbers of form $9k + 2$ and the set of all three-digit numbers of form $9k + l$), we can write this sum as :

$s_1 $(for numbers of the form 9k + 2) + $s_2$ (for numbers of the form 9k + l) = 109600

4. Solving this equation for $s_2$ (the sum of numbers of the form $9k + l$), we get :

$s_2 = 109600 - s_1 = 109600 - 54650 = 54950$

5. The sum $s_2$ can be expressed as $(n/2) \times (\text{{first term}} + \text{{last term}})$, where $n$ is the count of numbers of the form $9k + l$. The first term here is the smallest 3-digit number of this form, which is $99 + l$, and the last term is the largest such number, which is $990 + l$.

We equate this to $s_2$ to solve for $l$ :

$54950 = (100/2)[(99 + l) + (990 + l)]$

6. Simplifying this equation, we get :

$2l + 1089 = 1099$

Solving for $l$, we find :

$l = 5$

So, the correct answer is 5.

Number of subsets of A = 2^m

Number of subsets of B = 2ⁿ

Given = 2^m – 2ⁿ = 112

$ \therefore $ m = 7, n = 4 (2⁷ – 2⁴ = 112)

$ \therefore $ m $ \times $ n = 7 $ \times $ 4 = 28

X = {1, 2, 3, 4, …, 50}

A = {2, 4, 6, 8, …, 50} = 25 elements

B = {7, 14, 21, 28, 35, 42, 49} = 7 elements

Here n(A$ \cup $B) = n(A) + n(B) – n(A$ \cap $B)

= 25 + 7 – 3 = 29