Question

25. Taking the set of natural numbers as the universal set if A={X/XE N.2x+1>10)
and B={ x/x € 1,3x-1>8} find a) B-A,b) A' and B'​

Answer 1

B - A  = { 4 }  ,  A' = { 1 , 2 , 3 , 4} ,  B' = { 1 , 2 , 3 } if A={X/XE N.2x+1>10) and B={ x/x € 1,3x-1>8}

Step-by-step explanation:

A={X/XE N.2x+1>10)

=> 2 x + 1 > 10

=> 2x > 9

=> x  > 4.5

=> x  > 4

=> A' = { 1 , 2 , 3 , 4}

B={ x/x € 1,3x-1>8}

=> 3x  - 1 > 8

=> 3x > 9

=> x > 3

=> B' = { 1 , 2 , 3 }

=> B has 4 as one value which A does not have

=> B - A  = { 4 }

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

Considering the set of natural numbers as the universal set

$\sf{A = \{ \: x : x \in \mathbb{ N} \: , \: 2x + 1 > 10 \} }$

$\sf{B = \{ \: x : x \in \mathbb{ N} \: , \: 3x - 1 > 8 \: \} }$

TO DETERMINE

a ) B-A

b ) A'

c ) B'

CALCULATION

Here the set of natural numbers as the universal set

$\sf{So \: \: U = \mathbb{ N} = \{ \: 1,2,3,4,....... \} \: \: }$

$\sf{A = \{ \: x : x \in \mathbb{ N} \: , \: 2x + 1 > 11 \} }$

Now

$\sf{ \: 2x + 1 > 10 }$

$\implies \sf{ \: 2x > 10 - 1 }$

$\implies \sf{ \: 2x > 9 }$

$\displaystyle \: \implies \sf{ \: x > \frac{9}{2} }$

$\sf{ \therefore \: \: \: A = \{ \: 5,6,7,8,9,....... \}}$

Again

$\sf{B = \{ \: x : x \in \mathbb{ N} \: , \: 3x - 1 > 8 \: \} }$

Now

$\sf{ 3x - 1 > 8 \: \: \: } \: \: gives$

$\implies \: \sf{ 3x > 9 \: \: \: }$

$\implies \: \sf{ x > 3\: \: \: }$

$\sf{ \therefore \: B = \{ \: 4,5,6,7,8..... \} }$

So

$\sf{B-A = \{ \: x : x \in B, x \notin A \: \}}$

$= \sf{ \{ \: 4 \: \}\: }$

Now

$\sf{A' = \{ \: x \in U : \: x \notin A \: \}}$

$= \sf{ \{ \: {1,2,3,4} \: \}}$

Again

$\sf{B' = \{ \: x \in \: U : x \notin \: B \: \}}$

$= \sf{ \{ \: {1,2,3} \: \}}$

RESULT

$\sf{B-A = \{ \: 4 \: \}}$

$\sf{A' = \sf{ \{ \: {1,2,3,4} \: \}}}$

$\sf{B' = \sf{ \{ \: {1,2,3} \: \}}}$

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