Relations and Functions Test - 68

Given here,

\(A=\left\{x \in R: x^{2}-7 x+12=0\right\}\)

And, \(B=\left\{y \in R: y^{2}-11 y+30=0\right\}\)

\(\because x^{2}-7 x+12=0 \quad\) (Given)

\(\Rightarrow x^{2}-3 x-4 x+12=0\)

\(\Rightarrow(x-3)(x-4)=0\)

\(\Rightarrow x=3\) or 4

\(\Rightarrow A=\{3,4\}\)

Similarly,

\({y}^{2}-11 {y}+30=0 \quad\) (Given)

\(\Rightarrow(y-5)(y-6)=0\)

\(\Rightarrow y=5\) or 6

\(\Rightarrow B=\{5,6\}\)

As we know that,

\(A \times B=\{(a, b) \mid a \in A\) and \(b \in B\}\)

\(\Rightarrow {A} \times {B}=\{(3,5),(3,6),(4,5),(4,6)\}\)

Given:

\(A\) and \(B\) are the domain and range respectively for the relation \(R\) such that

\(R=\{(1\), 3), \((2,4),(3,5),(5,7)\}\)

As we know that,

Domain of \((R)=\{a:(a, b) \in R\}\).

\(\Rightarrow \mathrm{A}=\{1,2,3,5\}\)

We also know that,

Range of \((R)=\{b:(a, b) \in R\}\).

\(\Rightarrow B=\{3,4,5,7\}\)

Given that:

\(A=\{x, y, z\}\)

\(n(A)=3\)

And,

\(B=\{1,2,3,4,5\}\)

\(n(B)=5\)

We know that:

If \(n(A)=x\) and \(n(B)=y\), then:

\(n(A \times B)=n(A) \times n(B)\)

\(=x y\)

\(\therefore n(A \times B)=n(A) \times n(B)\)

\(=3 \times 5\)

\(=15\)

We know that,

\(\Rightarrow  -1 \leq \sin 3 x \leq 1 \text { for all } x \in R \)

\(\Rightarrow 1 \leq 2-\sin 3 x \leq 1 \text { for all } x \in R \) (Adding  2 throughout)

\(\Rightarrow  2-\sin 3 x \neq 0 \text { for any } x \in R \) 

\(\Rightarrow  f(x)=\frac{1}{2-\sin 3 x} \text { is defined for all } x \in R\) 

Hence, domain \((f)=R\).

We know that:

If\({A}\) and \({B}\) be any two sets, then:

\((A \cap B)=\{x, x \in A\) and \(x \in B\}\)

\(A \times B=\{(x, y), x \in A\) and \(y \in B\}\)

Given that,

\(A=\{2,3\}, B=\{4,5\}, C=\{5,6\}\)

Then,

\((B \cap C)=\{5\}\)

\(A \times(B \cap C)=\{(2,5),(3,5)\}\)

Therefore, the number of elements in \(A \times(B \cap C)=2\)

Given:

\(A=\left\{x \in R: x^{2}=2\right\}\)

And, \(B=\left\{y \in R: y^{2}-5 y+6=0\right\}\)

Then,

\({x}^{2}-2=0 \quad\) (Given)

\(\Rightarrow {x}=\pm \sqrt{2}\)

\(\Rightarrow {A}=\{\sqrt{2},-\sqrt{2}\}\)

\(\therefore {n}({A})=2\).

Similarly,

\({y}^{2}-5 {y}+6=0 \quad\) (Given)

\(\Rightarrow y^{2}-3 y-2 y+6=0\)

\(\Rightarrow(y-3)(y-2)=0\)

\(\Rightarrow {y}=2,3\)

\(\Rightarrow {B}=\{2,3\}\)

\(\therefore {n}({B})=2\).

As we know that, if \(n(A)=p, n(B)=q\), then:

\(n(A \times B)=n(A) \times n(B)=p \times q\).

\(\Rightarrow {n}({A} \times {B})={n}({A}) \times {n}({B})\)

\(=2 \times 2\)

\(=4\)

Since \((1,4),(2,6)\) and \((3,6)\) are elements of \(A \times B\). It follows that \(1,2,3\) are elements of \(A\) and 4, 6 are elements of \(B\). 

It is also given that \(A \times B\) has 6 elements. 

So, \(A=\{1,2,3\}\) and \(B=\{4,6\}\).

Then, \(B \times A=\{4,6\} \times\{1,2,3\}\)

\(=\{(4,1),(4,2),(4,3),(6,1),(6,2),(6,3)\}\)

\(A=\{1,2,3,4,5,6\}\) and \(R\) is a relation on \(A\) such that:

\(R=\{(x, y): y=x+1\), where \(x\) and \(y \in R\}\)

When \(x=1\), then:

\(y=x+1=2 \in A\)

\( \Rightarrow(1,2) \in R\) 

When \(x=2\), then:

\(y=x+1\)

\(=2+1\)

\(=3 \in A\)

\( \Rightarrow (2,3) \in R\) 

When \(x=3\), then: 

\(y=x+1\)

\(=3+1 =4 \in A\)

\( \Rightarrow(3,4) \in R\) 

When \(x=4\), then: 

\(y=x+1=4+1\)

\(=5 \in A\)

\( \Rightarrow(4,5) \in R\) 

When \(x=5\), then: 

\(y=x+1=5+1\)

\(=6 \in A\)

\( \Rightarrow(5,6) \in R\) 

When \(x=6\), then:

\(y=x+1=6+1\)

\(=7 \notin A\)

\( \Rightarrow(6,7) \notin R\) 

So,  \(R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}\)

As we know that, 

Domain \((R)=\{a:(a, b) \in R\}\) 

\(\Rightarrow\) Domain \((R)=\{1,2,3,4,5\}\)

Given:

\(A=\{1,2,3,4,6\}\) and \({R}\) is a relation on \({A}\) such that:

\({R}=\{({a}, {b}): {a}, {b} \in {A}\) and \({b}\) is exactly divisible by \(a\}\)

The given \({R}\) can be re-written in roaster form as:

\({R}=\{(1,1),(1,2),(1,3),(1,4),(1,6),(2, 2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)\}\)

As we know that,

Range \((R)=\{b:(a, b) \in R\}\)

\(\Rightarrow\) Range \((R)=\{1,2,3,4,6\}=A\)

\(\Rightarrow n(A)=5\)

If we express \(f_1, f_2, f_3\) and \(f_4\) as sets of ordered pairs, then we have

\(f_1=\{(1,2),(2,3),(3,4)\} \)

\(f_2=\{(1,4),(4,1),(2,3),(3,2),(2,4),(4,2),(3,4),(4,3)\} \)

\(f_3=\{(2,1),(3,1),(4,1),(3,2),(4,2),(4,3)\} \text { and } f_4=\{(1,4),(2,3),(3,2),(4,1)\}\)

(a) We have, \(f_1=\{(1,2),(2,3),(3,4)\}\).

We observe that an element 4 of the given set has not appeared in first place of any ordered pair of \(f_1\). So, \(f_1\) is not a function from the given set to itself.

(b) We have, \(f_2=\{(1,4),(4,1),(2,3),(3,2),(2,4),(4,2),(3,4),(4,3)\}\).

We observe that 2, 3,4 have appeared more than once as first components of the ordered pairs in \(f_2\). So, \(f_2\) is not a function.

(c) We have, \(f_3=\{(2,1),(3,1),(4,1),(3,2),(4,2),(4,3)\}\).

We observe that 3 and 4 have appeared more than once as first components of the ordered pairs in \(f_3\). So, \(f_3\) is not a function.

(d) We have, \(f_4=\{(1,4),(2,3),(3,2),(4,1)\}\).

For f₄= 1 + 4 = 2 + 3 = 3 + 2 = 4 + 1 = 5

We observe that each element of the given set has appeared as first components in one and only one ordered pair of \(f_4\). So, \(f_4\) is a function in the given set.