4 Marks

Question 14 Marks

To confirm that, Is every real number in $R$. $R =\{(a, b): a, b \in R$ and $a-b+\sqrt{3} \in S \}$ where $S$ is set of all irrational numbers, defined, $R$ is reflexive, symmetric and transitive.

Answer

Here relation R , is defined on R such that :
$R =\{(a, b): a, b \in R$ and $a-b+\sqrt{3} \in S\}$
Reflexive : Suppose $a \in R$ (set of real numbers) now $(a, a) \in R$ then $a-a+\sqrt{3}=\sqrt{3} \in S$ so, R is reflexive.
Symmetric : Suppose $a, b \in R$ (set of real numbers)$
\begin{array}{l}
a, b \in R \Rightarrow a-b+\sqrt{3} \in S \\
\Rightarrow b-a+\sqrt{3} \in S \\
\Rightarrow(b, a) \in R
\end{array}
$(set of irrational numbers)
So, R is symmetric relation. ....(ii)
Transitive: Suppose $a, b, c \in R$
now $(a, b) \in R$ and $(b, c) \in R$$
\begin{array}{l}
\Rightarrow a-b+\sqrt{3} \in S \text { and } b-c+\sqrt{3} \in S \\
\Rightarrow a-b+\sqrt{3}+b-c+\sqrt{3} \in S \\
\Rightarrow(a, c) \in R
\end{array}
$
So, R is transitive relation. ....(iii)
It is clear from equation (i), (ii) and (iii) relation R is reflexive, symmetric and transitive.

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Question 24 Marks

In set $I \times I _0$, relation $R$ is defined such that $(a, b)$ $R (c, d) \Leftrightarrow a d=b c$ if $I _{ 0 }$ is set of non-zero integers. Then prove that $R$ is equivalence relation.

Answer

To prove R is equivalence relation in $I \times I _0$
(i) Reflexive : Suppose that $(a, b) \in I \times I _0$, then$
\begin{aligned}
(a, b) \in I \times I_0 & \Rightarrow a \in I, b \in I_0 \\
& \Rightarrow a b=b a
\end{aligned}
$(by commutative law)$
\Rightarrow(a, b) R(a, b)
$$\therefore R$ is reflexive relation.
(ii) Symmetric : Suppose that $(a, b) R (c, d)$ then$
\begin{aligned}
(a, b) R(c, d) & \Rightarrow a d=b c \\
& \Rightarrow b c=a d \\
& \Rightarrow c b=d a
\end{aligned}
$(by commutative law)$
\Rightarrow(c, d) R(a, b)
$$\therefore R$ is symmetric relation.
(iii) Transitive : Suppose that $(a, b) R (c, d)$ and $(c$,
d) $R (e, f)$ then
$(c, d) R (e, f)$ then
and$
(a, b) R(c, d) \Rightarrow a d=b c
$$(c, d) R (e, f) \Rightarrow c f=d e$
$\therefore \quad(a, b) R (c, d)$ and $(c, d) R (e, f)$
on multiplying :$
\begin{array}{lc}
\Rightarrow & (a d)(c f)=(b c)(d e) \\
\Rightarrow & a f=b c \\
\Rightarrow & (a, b) R(c, f)
\end{array}
$
$\therefore R$ is transitive relation.
Hence in set $I \times I _0$, R is equivalence relation.

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Question 34 Marks

In set of real numbers, a relation $R_1$ is defined such that $(a, b) \in R _1 \Leftrightarrow 1+a \cdot b>0 \forall a, b \in R$ prove that $R_1$ is reflexive and symmetric but not transitive.

Answer

Given that :
$R =$ set of real numbers
A relation $R _1$ is defined in R such that :$
(a, b) \in R_1 \Leftrightarrow 1+a b>0 \forall a, b \in R
$
(i) Reflexive : Suppose $a \in R \Leftrightarrow 1+a \cdot a=1+a^2>$$
0\left[\because a^2>0\right]
$
$\Rightarrow a R _1 a, \forall a \in R$
$\therefore R _1$ is reflexive relation.
(ii) Symmetric : Suppose that $a R _1 b$, then
$a R _1 b \Rightarrow 1+a b>0$
$
\begin{array}{l}
\Rightarrow 1+b a>0 \\
\Rightarrow b R_1 a \quad\binom{\because a b=b a,}{\forall a, b \in R}
\end{array}
$
$\therefore R _1$ is symmetric relation.
(iii) Transitive : Suppose that $a R _1 b$ and $b R _1 c$ then $a R _1 b$ and $b R _1 c \Rightarrow 1+a b>0$ and $1+b c>0$
If $a=1, b=1 / 2$ and $c=-1$$
\begin{array}{ll}
& 1+1 \cdot \frac{1}{2}=\frac{3}{2}>0 \Rightarrow 1 R_1 \frac{1}{2} \\
\text { and } & 1+\frac{1}{2}(-1)=\frac{1}{2}>0 \Rightarrow \frac{1}{2} R_1(-1) \\
\text { but } & 1+1(-1)=0 \ngtr 0 \Rightarrow 1 R_1(-1) \\
\therefore & 1 R_1 \frac{1}{2} \text { and } \frac{1}{2} R_1(-1)
\end{array}
$
but $1 R _1^{\prime}(-1)$
hence R is not transitive relation.

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Question 44 Marks

Suppose $N$ is set of natural numbers and $R$, is defined in $N \times N$ such that :
$(a, b) R (c, d) \Leftrightarrow a d(b+c)=b c(a+d)$. Prove that $R$ in equivalence in $N \times N$.

Answer

Reflexive : Suppose $(a, b)$ is a arbitrary element in $N \times N$.
then $(a, b) \in N \times N$
$\Rightarrow a, b \in N$
$\Rightarrow a b(b+a)=b a(a+b)$
[from commutativity]$
\Rightarrow(a, b) R(a, b) \forall(a, b) \in N \times N
$$\therefore R$ is reflexive in $N \times N$.
Symmetric : Suppose $(a, b),(c, d) \in N \times N$ such that $(a, b) R (c, d)$ then$
\begin{array}{l}
\Rightarrow a d(b+c)=b c(a+d) \\
\Rightarrow b c(a+d)=a d(b+c) \\
\Rightarrow c b(d+a)=d a(c+b)
\end{array}
$[from commutativity]$
\Rightarrow(c, d) R(a, b)
$thus $(a, b) R (c, d) \Rightarrow(c, d) R (a, b) \forall(a, b),(c, d)$ $\in N \times N$.
So, R is symmetric in $N \times N$.
Transitive : Suppose $(a, b),(c, d),(e, f) \in N \times N$ such that :
$(a, b) R (c, d)$ and $(c, d) R (e, f)$ then $(a, b) R (c, d)$$
\begin{array}{l}
\Rightarrow a d(b+c)=b(a+d) \\
\Rightarrow \frac{b+c}{b c}=\frac{a+d}{a d} \Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}
\end{array}
$$(c, d) R (e, f)$$
\begin{array}{l}
\Rightarrow c f(d+e)=d e(c+f) \\
\Rightarrow \frac{d+e}{d e}=\frac{c+f}{c f} \Rightarrow \frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f} .
\end{array}
$on adding (i) and (ii)$
\begin{array}{l}
\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{d}+\frac{1}{e}\right)=\left(\frac{1}{a}+\frac{1}{d}\right)+\left(\frac{1}{c}+\frac{1}{f}\right) \\
\Rightarrow \frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f} \Rightarrow \frac{b+e}{b e}=\frac{a+f}{a f}
\end{array}
$$
\begin{array}{l}
\Rightarrow a f(b+e)=b e(a+f) \\
\Rightarrow(a, b) R(e, f) \\
\text { Thus }(a, b) R(c, d) \text { and }(c, d) R(e, f) \\
\Rightarrow(a, b) R(e, f) \forall(a, b),(c, d),(e, f) \in N \times N
\end{array}
$so, $R$ is transitive.
Hence R is equivalence relation in $N \times N$.

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Question 54 Marks

If function $f: R \rightarrow R , f(x)=x^2+2$ and $g: R \rightarrow R$ $g(x)=\frac{x}{x-1}, x \neq 1$ then find $f o g$ and $g o f$ and also find $( fog )(2)$ and $( gof )(-3)$ ?

Answer

$
\begin{aligned}
(f o g)(x) & =f[g(x)]=f\left[\frac{x}{x-1}\right] \\
& =\left(\frac{x}{x-1}\right)^2+2
\end{aligned}
$
and$
\begin{aligned}
(g \circ f)(x) & =g[f(x)]=g\left[x^2+2\right] \\
& =\frac{x^2+2}{x^2+2-1}=\frac{x^2+2}{x^2+1}
\end{aligned}
$
now $\quad(f \circ g)(2)=\left(\frac{2}{2-1}\right)^2+2=4+2=6$ Ans.
$( gof )(-3)=\frac{(-3)^2+2}{(-3)^2+1}=\frac{11}{10}$

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Maths 4 Marks

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