Question

4 00:06 Let A = { x:x E R. x < 1}; B = {x: XER, |x-1| > 1} and A U B = R - D, then the set D is​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: A = \{x : x \: \in \: R, \: x < 1 \}$

$\rm\implies \:A = x \in \: ( - \infty ,1) - - - (1)$

Also, given that

$\rm \: B = \{x : x \: \in \: R, \: |x - 1| > 1 \}$

Now, Consider

$\rm \: |x - 1| > 1$

We know,

$\boxed{\tt{ |x - z| > y \: \: \rm\implies \:x < z - y \: \: or \: \: x > z + y \: }}$

So, using this result, we get

$\rm \: x - 1 < - 1 \: \: or \: \: x - 1 > 1$

$\rm \: x < - 1 + 1 \: \: or \: \: x > 1 + 1$

$\rm \: x < 0 \: \: or \: \: x > 2$

$\rm\implies \:x \in \: ( - \infty ,0) \:\cup \:(2,\infty)$

$\rm\implies \:B = x \in \: ( - \infty ,0) \:\cup \:(2,\infty) - - - (2)$

From equation (1) and (2), we concluded that

$\rm \: A\cup B = \{x : x \: \in \: R, \: x < 0 or x > 2 \}$

$\rm \: A \cup B = R - [0, 2]$

As it is given that

$\rm \: A \cup B = R - D$

So, on comparing, we get

$\rm \: D\: = \{x : x \: \in \: R, \: 0 \leq x \leq 2 \}$[/tex]

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ADDITIONAL INFORMATION

1. Commutative Law

$\rm \: A \cup B = B \cup A$

$\rm \: A \cap B = B \cap A$

2. Associative Law

$\rm \: (A \cup B) \cup C= A \cup (B \cup C)$

$\rm \: (A \cap B) \cap C= A \cap (B \cap C)$

3. Distributive Law

$\rm \: (A \cup B) \cap C= (A \cap C) \cup (B \cap C)$

$\rm \: (A \cap B) \cup C= (A \cup C) \cap (B \cup C)$

Answer 2

$\sf \underline \red{Solution:−}$

$\: \bullet\rm{ \: A={x:x∈R,x<1}}$

$\bullet\rm{ \: B={x:x∈R,∣x−1∣>1}}$

• Given A U B = R - D then D = ? for A ,

$\rm\rightarrow{ \: for \: A \: = x < 1 = > x ∈( - \infty \: , \: 1)}$

$\rightarrow\rm{for \: B = ∣x−1∣>1}$

So,

$\longmapsto\rm{(x - 1) }$

$\longmapsto\rm{x > 1 \: \cup \: x < 1}$

$\longmapsto\rm{x \: \in \: (2 \: , \: \infty ) \: \cup \: ( - \infty \: , \: 0)}$

$\longmapsto\rm{B=( - \infty \: , \: 0) \: \cup \: (2 \: , \: \infty )}$

Now,

• Given A U B = R - D

$\rm\longmapsto{A \: \cup \: B = ( - \infty \: , \: 0) \: \cup \: (2 \: , \: \infty ) \: \cup \: ( - \infty \: , \: 1)}$

$\longmapsto\rm{( - \infty \: , \: 1) \: \cup \: (2 \: , \: \infty )}$

$\longmapsto\rm{R - D = ( - \infty \: , \: 1) \: \cup \: (2 \: , \: \infty )}$

$\longmapsto\rm{R - ( - \infty \: , \: 1) \: \cup \: (2 \: , \: \infty ) = d}$

$\longmapsto\rm \fbox \red{D = 1,2}$

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More Information :

1. Commutative Law:-

$\begin{gathered}\sf\: A \cup B = B \cup A \\ \end{gathered}$

$\begin{gathered}\sf \: A \cap B = B \cap A \\ \end{gathered}$

2.Associative Law :-

$\begin{gathered}\sf \: (A \cap B) \cap C= A \cap (B \cap C) \\ \end{gathered}$

$\sf{(A∩B)∩C=A∩(B∩C)}$

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@Shivam

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