Maths M.C.Q (1 Marks)

We have, $a R b: a$ divides $b^2$

For reflexive : $(a, a) \in R$

$\therefore \quad a R a: a$ divides $a^2$.

Hence, $R$ is reflexive.

For symmetric : $(a, b) \in R \Rightarrow(b, a) \in R$

$a$ divides $b^2$ and $b$ not divides $a^2$.

Hence, it is not symmetric.

For transitive : $(a, b) \in R,(b, c) \in R$

$\Rightarrow(a, c) \in R$

$\Rightarrow(8,4): 8$ divides $4^2$

$\Rightarrow(4,2): 4$ divides $2^2$

But $(8 / 2): 8$ not divides $2^2$

$\therefore$ It is not transitive.

Given,

$f(x)=\frac{16 x^2-96 x+153}{x-3}$

Let $\quad f(x)=y$

$\therefore \quad y=\frac{16 x^2-96 x+153}{x-3}$

$\Rightarrow \quad x y-3 y=16 x^2-96 x+153$

$\Rightarrow 16 x^2-x(96+y)+153+3 y=0$

$\therefore \quad x \in R$

$\therefore \quad D \geq 0$

$\because(96+y)^2-4 \times 16(153+3 y) \geq 0$

$(96)^2+192 y+y^2-64(153)-192 y \geq 0$

$\quad y^2 \geq 64(153)-96^2$

$\quad y^2 \geq 9792-9216$

$\quad y^2 \geq 576$

$\therefore \quad y \in(-\infty,-24] \cup[24, \infty)$

$\therefore \text { Least positive value of } f(x)=24$

We have,

$f(x) =\frac{\{x\}}{1+[x]^2}$

$\Rightarrow \quad f(x) =\frac{x-[x]}{1+[x]^2}$

Range of $f(x)=[0,1)$.

$\therefore$ Range of $f$ is semi-closed interval $f$ is discontinuous on integer.

Clearly, $f$ is not one-one function.

$\therefore$ Option $(d)$ is correct.

Given, $f(x y)=f(x)+f(y)$

$f(12)=24 \Rightarrow f(8)=15$

$f(8)=f(2 \cdot 2 \cdot 2)=f(2)+f(2)+f(2)$

$\Rightarrow \quad 15=3 f(2) \Rightarrow f(2)=5$

$\therefore f(48)=f(12 \cdot 2 \cdot 2)=f(12)+f(2)+f(2)$

$=24+5+5=34$

We have,

$P\left(\sin ^2 x\right) =P\left(\cos ^2 x\right), x \in\left[0, \frac{\pi}{2}\right)$

$P\left(\sin ^2 x\right) =P\left(1-\sin ^2 x\right)$

$P(x) =P(1-x), x \in[0,1]$

$P^{\prime}(x) =-P^{\prime}(1-x)$

So, $P^{\prime}(x)$ is symmetric about line $x=\frac{1}{2}$

So, $P^{\prime}(x)$ has highest degree is odd. $\Rightarrow P(x)$ has highest degree is even.

Hence, option $(c)$ is correct.

$f_1(x)=x+\frac{1}{8} \sin (2 \pi x)$

$f_2(x)=x+\frac{1}{8} \sin 2 \pi x+\frac{1}{8} \sin \left(2 \pi\left(x+\frac{1}{8} \sin 2 \pi x\right)\right)$

Similarly, $f _3( x )= x +\frac{1}{8}\left(\sin (2 \pi x)+\sin \left( f _1( x )\right)+\sin \left( f _2( x )\right)\right)$

$\therefore f_n(x)=x+\frac{1}{8}\left(\sin 2 \pi x+\sin \left(f_1(x)\right)+\ldots \cdots+\sin \left(f_{n-1}(x)\right)\right)$

$f_n(x)=0$ at $x=0$ and $f_n(x)=1$ at $x=1$

Therefore $I$ and $III$ are not true

from the figure, $f _{ n }( x )=\frac{1}{2}$ for many value of $x \in[0,1]$

statement $II$ is true

for every $x , \quad 0 \leq f _1( x ) \leq 1$

therefore, for every value of $x, \quad 0 \leq f_n(x) \leq 1$

$\lim _{n \rightarrow 0} f_n(x)$ always exists for $x \in[0,1]$

$IV$ is also not true.

We have,

$P(x)=4 x^3-3 x x \in\left(-\frac{1}{2}, \frac{1}{2}\right)$

$P^{\prime}(x)=12 x^2-3=3\left(4 x^2-1\right)$

$P^{\prime}(x)<0 x \in\left(-\frac{1}{2}, \frac{1}{2}\right)$

Given,

$\Rightarrow f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1$

$\text { Put } x=1-x \text {, we get }$

$f(1-x)+\left(1-x+\frac{1}{2}\right) f(1-(1-x)=1$

$\Rightarrow f(1-x)+\left(\frac{3}{2}-x\right) f(x)=1$

Eq.$(ii)$ multiply by $\left(x+\frac{1}{2}\right)$ we get

$\left(\frac{3}{2}-x\right)\left(x+\frac{1}{2}\right) f(x)+\left(x+\frac{1}{2}\right)$

$f(1-x)=x+\frac{1}{2}$

On subtracting Eq. $(iii)$ from Eq.$(i)$, we get

$f(x)\left[1-\left(\frac{3}{4}+x-x^2\right)\right]=1-x-\frac{1}{2}$

$\Rightarrow f(x)=\frac{\frac{1}{2}-x}{x^2-x+\frac{1}{4}} \Rightarrow f(0)=2 \text { and } f(1)=-2$

$\because 2 f(0)+3 f(1)-2(2)+3(-2)=4-6=-2$

We have,

$\sum \limits_{n=0}^{1947} \frac{1}{2^n+\sqrt{2^{1947}}}$

Let $\quad f(n)=\frac{1}{2^n+\sqrt{2^{1947}}}$

$f(0)+f(1947)=\frac{1}{1+2^{\frac{1947}{2}}}+\frac{1}{2^{1947}+2^{\frac{1947}{2}}}$

$=\frac{1}{1+2^{\frac{1947}{2}}}+\frac{1}{2^{\frac{1947}{2}}\left(2^{\frac{1947}{2}}+1\right)}$

$=\frac{2^{\frac{1947}{2}}+1}{2^{\frac{1947}{2}}\left(2^{\frac{1947}{2}+1}\right)}=\frac{1}{2^{\frac{1947}{2}}}$

Similarly, $f(1)+f(1946)=\frac{1}{2^{\frac{1947}{2}}}$

$\because \sum \limits_{n=0}^{1947} f(x)=f(0)+f(1)+f(2)+f(3)+\ldots .+f(1947)$

$=(f(0)+f(1947)+(f(1)+f(1946))+\ldots +(f(973)+f(974))$

$=974 \times \frac{1}{2^{\frac{1917}{2}}}$

$\sum \limits_{n=0}^{1947} f(n)=\frac{2 \times 487}{2 \times 2^{\frac{1945}{2}}}=\frac{487}{\sqrt{2^{1945}}}$

We have, $f\left(\frac{1}{2}\right)=100$

$f(x) \leq 100, \forall x \in R$

$\therefore \quad f(x)=a\left(x-\frac{1}{2}\right)$

$\left[a_0 x^{n-1}+a_1 x^{n-2}+\ldots+a_{n-1}\right]+100$

If $f(x) \leq 100, \forall x \in R$

$\therefore a < 0$ and $f(x)$ must be even degree polynomial.

Since, there may be more value of $x$ at which $f(x)$ attains maximum.

$\therefore$ If $x \neq \frac{1}{2}$, then $f(x) < 100$ may be false.

We have, $f(x)=\sqrt{2-x-x^2}$ and

$g(x) =\cos x$

$(g(x))^2 =\cos ^2 x$

$f\left((g(x))^2\right) =\sqrt{2-(g(x))^2-(g(x))^4}$

$=\sqrt{2-\cos ^2 x-\cos ^4 x}$

$f\left((g(x))^2\right)$ is defined if

$2-\cos ^2 x-\cos ^4 x \geq 0$

$\Rightarrow \quad \cos ^4 x+\cos ^2 x-2 \leq 0$

$\Rightarrow \cos ^4 x+2 \cos ^2 x-\cos ^2 x-2 \leq 0$

$\Rightarrow \quad\left(\cos ^2 x+2\right)\left(\cos ^2 x-1\right) \leq 0$

$\Rightarrow \quad \cos ^2 x \in[-2,1]$

$\therefore \quad f(g(x))=\sqrt{2-\cos x-\cos ^2 x}$

Similarly, for domain of $f(g(x))$ is $x \in R$

Hence, domain of $f\left(\left(g(x)^2\right)\right)=$ Domain of $f(g(x))$

We have, $f(x)=\sqrt{4-\sqrt{2 x+5}}$

$f(x)$ is defined if

$4-\sqrt{2 x+5} \geq 0$ and $2 x+5 \geq 0$

$\Rightarrow 4 \geq \sqrt{2 x+5}$ and $x \geq-5 / 2$

$\Rightarrow 16 \geq 2 x+5$ and $x \geq-5 / 2$

$\Rightarrow x \leq \frac{11}{2}$ and $x \geq-5 / 2$

$\therefore$ Domain of $f(x) x \in\left[\frac{-5}{2}, \frac{11}{2}\right]$

Mid-point of domain $\frac{\frac{-5}{2}+\frac{11}{2}}{2}=\frac{3}{2}$

We have, $f^1(x)=\frac{x+1}{x-1}$

$f^3(x) =f(x), f^4(x)=x$

$P =f^1(2) \cdot f^2(3) \cdot f^3(4) \cdot f^4(5)$

$P =3 \times 3 \times \frac{5}{3} \times 5=75$

$\therefore$ Multiple of $P$ is $375$

We have,

$\begin{array}{c}f: X \times P(X) \rightarrow R \\f(x, A)=\left\{\begin{array}{ll} 1, & \text { if } x \in A \\0, & \text { if } x \notin A\end{array}\right. \\f(x, A \cup B)=\left\{\begin{array}{ll}1, & \text { if } x \in A \cup B \\0, & \text { if } x \notin A \cup B\end{array}\right. \\\text { If } x \in A, x \in B \Rightarrow f(x, A \cup B)=1 \\\text { If } x \in A, x \notin B \Rightarrow f(x, A \cup B)=1 \\\text { If } x \notin A, x \in B \Rightarrow f(x, A \cup B)=1 \\ \text { If } x \notin A, x \notin B \Rightarrow f(x, A \cup B)=0\end{array}$

Set $A$ have $m$-elements,

Set $B$ have $n$-elements

$\left(a, b_1\right) \in K,\left(a, b_2\right) \in R \Rightarrow\left(b_1=b_2\right)$

By condition relation is a function.

$\therefore$ Total number of function (Relation) $=n^m$

Therefore reflexive

Let $(a, b) R(c, d) \Rightarrow a d-b c$ is divisible by $5$

$\Rightarrow \mathrm{bc}-\mathrm{ad}$ is divisible by $5 \Rightarrow(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b})$

Therefore symmetric

Relation not transitive as $(3,1) \mathrm{R}(10,5)$ and $(10,5) \mathrm{R}(1,1)$ but $(3,1)$ is not related to $(1,1)$

Minimum order pairs are

$ (1,1),(2,2),(3,3),(4,4),(3,1),(2,1),(2,3),(3,2),  (1,3),(1,2)$

Thus no. of elements $=10$

Total number of symmetric relation $=2^{\left(\frac{\mathrm{n}^2+\mathrm{n}}{2}\right)}$

$\Rightarrow$ Then number of symmetric relation which are not reflexive

$ \Rightarrow $ $ 2^{\frac{n(n+1)}{2}}-2^{\frac{n(n-1)}{2}}$

$ \Rightarrow $ $2^{10}-2^6 $

$\Rightarrow $  $1024-64 $

$ =960$

$ +2+\underbrace{1+\ldots+1}_{10 \text { times }}$

$\mathrm{n}\left(\mathrm{R}_1\right)=66$

$\mathrm{R}_1 \cap \mathrm{R}_2=\{(1,1),(2,2), \ldots(20,20)\}$

$\mathrm{n}\left(\mathrm{R}_1 \cap \mathrm{R}_2\right)=20$

$\mathrm{n}\left(\mathrm{R}_1-\mathrm{R}_2\right)=\mathrm{n}\left(\mathrm{R}_1\right)-\mathrm{n}\left(\mathrm{R}_1 \cap \mathrm{R}_2\right)$

$=\mathrm{n}\left(\mathrm{R}_1\right)-20$

$=66-20$

$\mathrm{R}_1-\mathrm{R}_2=46 \text { Pair }$

(a, b) $R_2(c, d) \Leftrightarrow a+d=b+c ;(a, b),(c, d) \in N$

for $R_1$ : Not reflexive symmetric not transitive

for $R_2: R_2$ is reflexive, symmetric and transitive

Hence only $R_2$ is equivalence relation.

$\mathrm{x}_1, \mathrm{y}_1 \in \mathrm{N} \therefore \mathrm{R}$ is reflexive

$((1,1),(2,3)) \in \mathrm{R}$ but $((2,3),(1,1)) \notin \mathrm{R}$

$\therefore \mathrm{R}$ is not symmetric

$((2,4),(3,3)) \in \mathrm{R}$ and $((3,3),(1,3)) \in \mathrm{R}$ but $((2,4)$,

$(1,3)) \notin \mathrm{R}$

$\therefore \mathrm{R}$ is not transitive

$\mathrm{f}\left(\frac{1}{2}\right)<0$

$\mathrm{f}(1)>0 \Rightarrow(\mathrm{A})$ is correct.

$f(x)=\sqrt{2}\left(4 x^3-3 x\right)-1=0$

Let $\cos \alpha=\mathrm{x}$,

$\cos 3 \alpha=\cos \frac{\pi}{4} \Rightarrow \alpha=\frac{\pi}{12}$

$\mathrm{x}=\cos \frac{\pi}{12}$

$(4)$ is correct.

$x=1 $  $ y=7 $

$x=4 $  $ y=5 $

$x=7 $  $ y=3 $

$x=10 $  $ y=1 $

$A $  $ B $

$(1,7) $  $ 1 $

$(4,5) $  $ 4 $

$(7,3) $  $ 7 $

$(10,1) $  $ 10$

The number of one-one functions from $\mathrm{A}$ to $\mathrm{B}$ is equal to $4$ !

$f(n)=$ The highest prime factor of $n$.

$f(2)=2$

$f(4)=2$

$\Rightarrow$ many one

$4$ is not image of any element

$\Rightarrow$ into

Hence many one and into

Neither one-one nor onto.

Let $g(x)=x^2-4 x+9$

$ D < 0 $

$ g(x) > 0$ for $x \in R$

$\therefore\left[\begin{array}{l}\mathrm{f}(-5)=0 \\ \mathrm{f}(3)=0\end{array}\right.$

So, $\mathrm{f}(\mathrm{x})$ is many-one.

again,

$ y x^2-4 x y+9 y=x^2+2 x-15 $

$ x^2(y-1)-2 x(2 y+1)+(9 y+15)=0 $

$ \text { for } \forall x \in R \Rightarrow D \geq 0 $

$ D=4(2 y+1)^2-4(y-1)(9 y+15) \geq 0 $

$ 5 y^2+2 y+16 \leq 0 $

$ (5 y-8)(y+2) \leq 0$

$Image$

$\mathrm{y} \in\left[-2, \frac{8}{5}\right]$ range

Note : If function is defined from $f: R \rightarrow R$ then only correct answer is option ($3$)

$\Rightarrow$ Bonus

$ g(x)=\frac{4\left(\frac{4 x+3}{6 x-4}\right)+3}{6\left(\frac{4 x+3}{6 x-4}\right)-4}=\frac{34 x}{34}=x $

$ g(x)=x \therefore g(g(g(4)))=4$

$\mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}e^{-x},(-\infty, 0] \\ e^{\ln x},(0,1) \\ \ln x,[1, \infty)\end{array}\right.$

Graph of $g(f(x))$

$\mathrm{g}(\mathrm{f}(\mathrm{x})) \Rightarrow$ Many one into

$\mathrm{n}(\mathrm{R})=33 $

$ \therefore 66$

$R=\{(A, B): A \cap B \neq \phi ; A, B \in M\}$

For Reflexive,

$M$ is subset of ' $S$ '

So $\phi \in \mathrm{M}$

for $\phi \cap \phi=\phi$

$\Rightarrow$ but relation is $\mathrm{A} \cap \mathrm{B} \neq \phi$

So it is not reflexive.

For symmetric,

$\mathrm{ARB}$

$\mathrm{A} \cap \mathrm{B} \neq \phi,$

$\Rightarrow \mathrm{BRA} \quad \Rightarrow \mathrm{B} \cap \mathrm{A} \neq \phi$,

So it is symmetric.

For transitive,

If $\mathrm{A}=\{(1,2),(2,3)\}$

$ B=\{(2,3),(3,4)\} $

$C=\{(3,4),(5,6)\}$

$ARB$ $BRC$ but $A$ does not relate to $C$

So it not transitive

$ \mathrm{R}_1=\{(\mathrm{x}, \mathrm{y}): 2 \mathrm{x}-3 \mathrm{y}=2\} $

$ \mathrm{R}_2=\{(\mathrm{x}, \mathrm{y}):-5 \mathrm{x}+4 \mathrm{y}=0\}$

$ \mathrm{R}_1=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\} $

$ \mathrm{R}_2=\{(4,5),(8,10),(12,15),(16,20)\}$

in $\mathrm{R}_1 6$ element needed

in $\mathrm{R}_2 4$ element needed

So, total $6+4=10$ element

then

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

$ (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\}$

i.e. $16$ elements.

i.e. $\mathrm{m}=16$

Now to make $\mathrm{R}$ a symmetric relation add

$\{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)\}$

i.e. $\mathrm{n}=9$

So $\mathrm{m}+\mathrm{n}=25$

$ \mathrm{B}=\{1,4,5,10,15\} \quad 3 \mathrm{ad}-7 \mathrm{bc} $

$ \text { Reflexive : }(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b})$

$\Rightarrow 3 \mathrm{ab}-7 \mathrm{ba}=-4 \mathrm{ab}$ always even so it is reflexive.

Symmetric : If $3 \mathrm{ad}-7 \mathrm{bc}=$ Even

Case $-I$ : odd odd

Case $-II$ : even even

$(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \Rightarrow 3 \mathrm{bc}-3 \mathrm{ab}$

Case $-I$ : odd odd

Case $-II$ : even even

so symmetric relation

Transitive :

Set $(3,4) R(6,4)$ Satisfy relation

Set $(6,4) R(3,1)$ Satisfy relation

but $(3,4) R(3,1)$ does not satisfy relation so not transitive.

$ B=\{2,5,6,8\} $

$ \left(a_1, b_1\right) R\left(a_2, b_2\right) $

$ a_1+a_2=b_1+b_2$

$1$. $(2,4) \mathrm{R}(6,4) \quad$ 2. $(2,4) \mathrm{R}(7,5)$

$3$. $(2,5) \mathrm{R}(7,4) \quad$ 4. $(3,4) \mathrm{R}(6,5)$

$5$. $(3,5) \mathrm{R}(6,4) \quad$ 6. $(3,5) \mathrm{R}(7,5)$

$7$. $(3,6) \mathrm{R}(7,4) \quad$ 8. $(3,4) \mathrm{R}(7,6)$ $\times 2$

$9$. $(6,5) \mathrm{R}(7,8) \quad$ 10. $(6,8) \mathrm{R}(7,5)$

$11$. $(7,8) \mathrm{R}(7,6) \quad$ 12. $(6,8) \mathrm{R}(6,4)$

$13$. $(6,6) \mathrm{R}(6,6)$

Total $24+1=25$

$\frac{2 \times+3}{(4 x-3)(x+1)}>0 \quad \frac{3 x+1}{x+2} \geq 0 \quad \& \quad \frac{x-3}{x+2} \leq 0 $

$(-\infty,-2) \cup\left[\frac{-1}{3}, \infty\right) \quad \ldots . .(1) $

$(-2,3] \quad \ldots . .(2)$

${\left[\frac{-1}{3}, 3\right] \ldots \ldots(3) \quad(1) \cap(2) \cap(3)} $

$ \left(\frac{3}{4}, 3\right] $

$ \alpha=\frac{3}{4} \beta=3 $

$ 5 \beta-4 \alpha=15-3=12$

$ \mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x})))=\frac{2^3 \mathrm{x} / \sqrt{1+9 \mathrm{x}^2}}{\sqrt{1+9\left(1+2^2\right) \frac{2^2 \mathrm{x}^2}{1+9 \mathrm{x}^2}}}=\frac{2^3 \mathrm{x}}{\sqrt{1+9 \mathrm{x}^2\left(1+2^2+2^4\right)}} $

$ \therefore \text { By observation } $

$ \alpha=1+2^2+2^4+\ldots+2^{18}=1\left(\frac{\left(2^2\right)^{10}-1}{2^2-1}\right)=\frac{2^{20}-1}{3} $

$ 3 \alpha+1=2^{20} \rightarrow \sqrt{3 \alpha+1}=2^{10}=1024$

$ B=\{2,4,5,7,8,10,12\} $

$ f(1)+f(3)=14 $

$ \text { (i) } 2+12 $

$ \text { (ii) } 4+10 $

$ 2 \times(2 \times 5 \times 4 \times 3)=240$

$= 3 \times 11 \times 7 \times 2 \times 5$

$ A=\{2,3,5,7,11\} $

$ f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right]$

$ f(2)=\left[\log _2(5)\right]=2 $

$ f(3)=\left[\log _2(14)\right]=3 $

$ f(5)=\left[\log _2(25+25)\right]=5 $

$ f(7)=\left[\log _2(117)\right]=6 $

$ f(11)=\left[\log _2 387\right]=8$

Range of $f: B=\{2,3,5,6,8\}$

No. of one-one functions $=5 !=120$

$ 2 x+3 \neq 0 \& x \neq \frac{-3}{2} \text { and }\left|\frac{(x-1)}{2 x+3}\right| \leq 1 $

$|x-1| \leq|2 x+3|$

$Image$

$ \text { For }|2 x+3| \geq|x-1| $

$ x \in(-\infty,-4] \cup\left(-\frac{2}{3}, \infty\right) $

$ \alpha=-4 \& \beta=-\frac{2}{3}: 12 \alpha \beta=32$

fog $\left[\begin{array}{cc}\left|e^x-1\right| & x \geq 0 \\ |x+1-1| & x \leq 0\end{array}\right.$

$f o g\left[\begin{array}{cc}e^x-1 & x \geq 0 \\ -x & x \leq 0\end{array}\right.$

$Image$

$y=f|x|$

$Image$

$y=|f(x)|$

$Image$

$g(x)=\frac{f(|x|)-|f(x)|}{2}$

$Image$

$ \mathrm{~g}(\mathrm{x})=\frac{|\mathrm{x}|+1}{2 \mathrm{x}+5}, \mathrm{x} \neq-\frac{5}{2}$

Domain of $f(g(x))$

$f(g(x))=\frac{2 g(x)+3}{2 g(x)+1}$

$x \neq-\frac{5}{2}$ and $\frac{|x|+1}{2 x+5} \neq-\frac{1}{2}$

$x \in R-\left\{-\frac{5}{2}\right\}$ and $x \in R$

$\therefore$ Domain will be $R-\left\{-\frac{5}{2}\right\}$

By $(1)$ $x \in \phi$

And by $(2)$ $x \in[-3,0]$ and $x \in[0,1]$

Range of $f(g(X))$ is $[0,1]$

In order to make it equivalence relation as per given set, $R$ must be

$\{(a, a),(b, b),(c, c),(d, d),(a, b),(b, a),(b, c),(c, b)$, $(b, d),(d, b),(a, c),(a, d),(c, d),(d, c),(c, a),(d, a)\}$

There already given so $13$ more to be added.

So reflexive

$a R b \Rightarrow 2 a +3 b =5 \alpha$,

Now b R a

$2 b+3 a=2 b+\left(\frac{5 \alpha-3 b}{2}\right) \cdot 3$

$=\frac{15}{2} \alpha-\frac{5}{2} b=\frac{5}{2}(3 \alpha-b)$

$=\frac{5}{2}(2 a+2 b-2 \alpha)$

$=5(a+b-\alpha)$

Hence symmetric

$\text { a R b } \quad \Rightarrow 2 a+3 b=5 \alpha \text {. }$

$\text { b R c } \quad \Rightarrow 2 b+3 c=5 \beta$

$\text { Now } \quad 2 a+5 b+3 c=5(\alpha+\beta)$

$\Rightarrow 2 a +5 b +3 c =5(\alpha+\beta)$

$\Rightarrow 2 a +3 c =5(\alpha+\beta- b )$

$\Rightarrow a R c$

Hence relation is equivalence relation.

Each element has 2 choices

$\Rightarrow 3 \times 2=6$

$a_2$ divides $b_1$

Each element has $2$ choices

$\Rightarrow 3 \times 2=6$

$\text { Total }=6 \times 6=36$

For Reflexive $(1,1)(2,2),(3,3) \in R$

For transitive : $(1,2)$ and $(2,3) \in R \Rightarrow(1,3) \in R$

Not symmetric : $(2,1)$ and $(3,2) \notin R$

$R _1=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$

$R _2=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)(2,1)\}$

$R _3=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)(3,2)\}$

For reflexive, add $\Rightarrow(-2,-2),(-4,-4),(-3,-3)$

For symmetric, add $\Rightarrow(4,-4),(3,-3),(-2,3),(1,0)$