5 Marks Questions

Question 15 Marks

Let $R$ be a relation from $N$ to $N$ defined by $R=\left\{(a, b): a, b \in N\right.$ and $\left.a=b^2\right\}$. Are the following true?
(i) $(a, a) \in R$, for all $a \in N$
(ii) $(a, b) \in R \Rightarrow(b, a) \in R$
(iii) $(a, b) \in R,(b, c) \in R \Rightarrow(a, c) \in R$

Answer

Given, $R: N \rightarrow N$ defined by$
R=\left\{(a, b): a, b \in N \text { and } a=b^2\right\}
$
(i) We know that except natural number 1 , none of the natural numbers is equal to its square.
$\therefore(a, a) \notin R$ (It does not satisfy the condition $a=b^2$ )
$\therefore(a, a) \in R$ is false.

(ii) If $a=b^2$, it does not imply that $b=a^2$
i.e., $(a, b) \in R \Rightarrow(b, a) \in R$ is not true.
e.g., $(4,2) \in R$
$\left(\therefore a=4, b=2\right.$ and $\left.a=b^2\right)$ but $(2,4) \notin R\left(\therefore 2 \neq 4^2\right)$

(iii) If $(a, b) \in R \Rightarrow a=b^2 \quad \ldots(i)$
Also, $(b, c) \in R \Rightarrow b=c^2\quad \ldots(ii)$
Putting the value of $b$ from (ii) in (i), we get
$
\begin{array}{l}
a=c^4 \\
\therefore \quad a \neq c^2 \Rightarrow(a, c) \neq R \\
\text { e.g., }(a, b)=(16,4) \in R \text { and }(b, c)=(4,2) \in R
\end{array}
$
(Since, for each pair $a=b^2$ )
But $(a, c)=(16,2) \in R$
$\therefore$ The statement is false.

View full question & answer→

Question 25 Marks

For any two sets $A$ and $B$, prove that $A \cup B=A \cap B$ $\Leftrightarrow A=B$.

Answer

Let $A=B$, then $A \cup B=A$ and $A \cap B=B$
$
\begin{aligned}
A \cup B & =A \cap B \\
Thus,\ A & =B \quad \ldots(i)
\end{aligned}
$
Conversely, let $A \cup B=A \cap B$
Now, let $x \in A$
$
\begin{array}{l}
x \in(A \cup B)(\because A \cup B=A \cap B] \\
x \in(A \cap B) \\
x \in A \text { and } x \in B \\
x \in B \\
A \subseteq B \quad \ldots(ii)
\end{array}
$
Now, let $y \in A$
$\begin{array}{l}y \in A \cup B \\ y \in A \cap B \quad[\because A \cup B=A \cap B] \\ y \in A \text { and } y \in B \\ y \in A\end{array}$
From eqs. (ii) and (iii), we get $A=B$
Thus, $(A \cup B)=(A \cap B)$
$A=B$
From eqs. (iii) and (iv), we get
$A \cup B=A \cap B \Leftrightarrow A=B$
Hence Proved

View full question & answer→

Question 35 Marks

In a group of 500 persons, 300 take tea, 150 take coffee, 250 take cold drink, 90 take tea and coffee, 110 take tea and cold drink, 80 take coffee and cold drink and 50 take all the three drinks.
(i) Find the number of persons who take none of the three drinks.
(ii) Find the number of persons who take only tea.
(iii) Find the number of persons who take coffee and cold drink but not tea.

Answer

Given,
$
\begin{array}{l}
n(U)=500, n(T)=300, n(C o)=150, n(C d)=250, \\
n(T \cap C o)=90, n(T \cap C d)=110 \\
n(C o \cap C d)=80, n(T \cap C d \cap C o)=50
\end{array}
$

Hence,
$\quad \begin{aligned} n(T) & =\text { like tea, } \\ n(C o) & =\text { like coffee, } \\ n(C d) & =\text { like cold drink }\end{aligned}$

(i) Number of persons who take none of three drinks$
\begin{aligned}
n(U)-n(T \cup C o \cup C d) & \\
n(T \cup C o \cup C d)= & n(T)+n(C o)+n(C d) \\
& -n(T \cap C o)-n(T \cap C d) \\
- & n(C o \cap C d)+n(T \cap C d \cap C o) \\
& =300+150+250-90-110-80+50 \\
& =750-280 \\
& =470 \\
n(U)-n(T \cup C o \cup C d) & =500-470=30
\end{aligned}
$
$\therefore$ Number of persons who take none of three drinks $=30$.

(ii) Number of persons who take only tea
$
\begin{array}{l}
=n(T)-n(T \cap C o)-n(T \cap C d)+n(T \cap C d \cap C o) \\
=300-90-110+50 \\
=350-200=150 \\
\therefore \text { Number of persons who take only tea }=150
\end{array}
$

(iii) Number of persons who take coffee and cold drink but not tea
$
\begin{array}{l}
=n(C o \cap C d)-n(T \cap C o \cap C d) \\
=80-50=30
\end{array}
$
$\therefore$ Number of persons who take coffee and cold drink but not tea $=30$.

View full question & answer→

Applied Maths 5 Marks Questions

Test yourself on this topic