2 Marks Questions

Question 12 Marks

Prove that the sets $A$ and $B , A - B = B ^{\prime}- A ^{\prime}$ as true.

Answer

Here we are to prove,
$\begin{array}{l}(A-B) \subseteq\left(B^{\prime}-A^{\prime}\right) \text { and }\left(B^{\prime}-A^{\prime}\right) \subseteq(A-B) \\\text { Let } \quad x \in(A-B) \Rightarrow x \in A \text { and } x \notin B \\\Rightarrow x \notin A^{\prime} \text { and } x \in B^{\prime} \\\Rightarrow x \in B^{\prime} \text { and } x \notin A^{\prime} \\\Rightarrow x \in\left(B^{\prime}-A^{\prime}\right) \\\therefore \quad(A-B) \subseteq\left(B^{\prime}-A^{\prime}\right)\ldots\ldots (1)\end{array}$
$ \begin{aligned}\text {Again}\quad x \in\left(B^{\prime}- A ^{\prime}\right. & \Rightarrow x \in B^{\prime} \text { and } x \notin A^{\prime} \\ & \Rightarrow x \notin B \text { and } x \in A \\ & \Rightarrow x \in A \text { and } x \notin B \\ & \Rightarrow x \in(A- B )\end{aligned}$
$ \therefore \quad\left(B^{\prime}-A^{\prime}\right) \subseteq(A-B)\ldots\ldots (2)$
From equations (i) and (ii), $A - B = B ^{\prime}- A ^{\prime}$ is true.

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Question 22 Marks

Prove $\left( A ^{\prime}\right)^{\prime}= A$ for an empty set $A$.

Answer

Here, $\quad A =\{ \}$
Let $x$ be an universal set, then
$\begin{array}{l}A^{\prime}=\text { The set of all elements of set } U \text { which are not in } A \\=x(\because \text { no element of } U \text { is in } A)\end{array}$
Now, $\quad\left( A ^{\prime}\right)^{\prime}=x^{\prime}$
$=$ Set of those elements of $x$ which are not in $x$
$=\{ \} \quad(\because$ no element of $x$ is in $x)$
$= A$
$\therefore \quad\left( A ^{\prime}\right)^{\prime}= A$

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Question 32 Marks

If $A =\{1,2,3\}, B =\{2,3,4\}$ and $C =\{3,4,5,6\}$ then find $(A \cap B)-C$.

Answer

$A=\{1,2,3\}, B=\{2,3,4\}, C=\{3,4,5,6\}$
$\begin{aligned} \therefore \quad( A \cap B ) & =\{1,2,3\} \cap\{2,3,4\} \\ & =\{2,3\}\end{aligned}$
$\begin{aligned}\text {Now} \ (A \cap B)-C & =\{2,3\}-\{3,4,5,6\} \\ & =\{2\}\end{aligned} $

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Question 42 Marks

If $A =\{x: x=2 n, n$ is a positive integer $\}$ and B $=\{x: x=3 n, n$ is a positive integer $\}$, then find the value of $A \cap B$.

Answer

$\because n$ is an integer $\therefore$ Put $n=1,2,3,4, \ldots \ldots \ldots$
$x=2 n=2,4,6,8,10,12\ldots\ldots\ldots$
then $\quad A =\{2,4,6,8,10,12 \ldots \ldots \ldots \ldots$
and $\quad x=3 n=3,6,9,12,15,18 \ldots \ldots \ldots \ldots$
then $\quad B=\{3,6,9,12,15,18 \ldots \ldots \ldots \ldots\}$
So, $\quad A \cap B =$ Set of common elements of $A$ and $B$
$\begin{array}{l}=\{6,12,18,24 \ldots \ldots \ldots \ldots .\} \\ =\{x: x=6 n, n \text { is a positive integer }\}\end{array}$

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Question 52 Marks

If $A =\{1,2,6,9\}, B=\{2,4,6,8\}$ and $C=\{3,4,9\}$ then prove that $A \cup(B \cap C)=(A$ $\cup$ B) $\cap(A \cup C)$.

Answer

$B \cap C =$ The set of common elements of set $B$ and $C=\{4\}$
$A \cup(B \cap C)=$ Set of all elements of sets $A$ and $(B \cap C)$
$=\{1,2,4,6,9\}$
Again $A \cup B=$ Set of all elements of $A$ and $B$
$=\{1,2,4,6,8,9\}$
$\begin{array}{l}\text { Similarly } A \cup C=\{1,2,3,4,6,9\} \\ \therefore \quad(A \cup B) \cap(A \cup C)=\text { Set of common } \\ \text { elements of }(A \cup B) \text { and }(A \cup C) \\ \therefore \quad\{1,2,4,6,9\} \\ \therefore \quad A \cup(B \cap C)=(A \cup B) \cap(A \cup C)\end{array}$

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Question 62 Marks

If $A =\{x: x \in N, 1 \leq x \leq 3\}$ and $B =\{y: 3 y=9$ or $2 y=2$ or $y-1=1\}$ then is $A=B$ true?

Answer

$\begin{array}{lll}& A=\{x: x \in N, 1 \leq x \leq 3\} \\\text { and } & B=\{y: 3 y=9 \text { or } 2 y=2 \text { or } y-1=1\} \\\text { Now } & 1 \leq x \leq 3 \Rightarrow x=1,2,3 \quad[\because x \in N] \\\therefore & A=\{1,2,3\} \\\text { Again } & 3 y=9 \Rightarrow y=3, \\& 2 y=2 \Rightarrow y=1 \\\text { and } & y-1=1 \Rightarrow y=2 \\\therefore & B=\{3,1,2\}=\{1,2,3\} &\end{array}$
Clearly elements of A and B are equal. Hence statement $A = B$ is true.

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