Maths M.C.Q (1 Marks)

i.e., $f(3)=1$ or $3$

Total function $=1 \times 6 \times 2 \times 6 \times 6 \times 1=432$

$\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1} \Rightarrow[0,1)$

Range of $f(x)=[0,2 \pi)$

Which is not true for every a $\in Z$.

Symmetric:

Take $a =2, b =1 \Rightarrow \operatorname{gcd}(2,1)=1$

Also $2 a=4 \neq b$

Now when $a =1, b =2 \Rightarrow \operatorname{gcd}(1,2)=1$

Also now $2 a =2= b$

Hence $a=2 b$

$\Rightarrow R$ is not Symmetric

Transitive:

Let $a =14, b =19, c =21$

$\operatorname{gcd}( a , b )=1$

$\operatorname{gcd}(b, c)=1$

$\operatorname{gcd}( a , c )=7$

Hence not transitive

$\Rightarrow R$ is neither symmetric nor transitive.

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

For Transitive $(a, b),(b, c) \in R$

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

Now

$1.$ Symmetric

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

$2.$ Transitive

$\therefore(a, b),(b, a) \in R$

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

$\Rightarrow(b, b) \&(c, c) \in R$

$\therefore$ Elements to be added

$\left\{\begin{array}{r}(b, a),(c, b),(a, c),(c, a) \\,(a, a),(b, b),(c, c)\end{array}\right\}$

Number of elements to be added $=7$

Symmetric:

$(c, d) R(a, b) \Rightarrow c b(d-a)=d a(c-b) \Rightarrow$

Symmetric

Reflexive:

(a, b) R $(a, b) \Rightarrow a b(b-a) \neq b a(a-b) \Rightarrow$

Not reflexive

Transitive: $(2,3) R (3,2)$ and $(3,2) R (5,30)$ but $((2,3),(5,30)) \notin R \Rightarrow \quad$ Not transitive

Then,$(b, a)$ on relation $R$

$\Rightarrow b ^2- a ^2=- I$

$\therefore T \text { is symmetric }$

$S =\left\{( a , b ): a , b \in R -\{0\}, 2+\frac{ a }{ b } > 0\right\}$

$2+\frac{ a }{ b } > 0 \Rightarrow \frac{ a }{ b } > -2, \Rightarrow \frac{ b }{ a } < \frac{-1}{2}$

If $(b, a) \in S$ then

$2+\frac{b}{a}$ not necessarily positive

$\therefore S$ is not symmetric

As $3(a-a)+\sqrt{7}=\sqrt{7}$ which belongs to relation so relation is reflexive

Check for symmetric:

Take $a=\frac{\sqrt{7}}{3}, b=0$

Now $(a, b) \in R$ but $(b, a) \notin R$

As $3(b-a)+\sqrt{7}=0$ which is rational so relation is not symmetric.

Check for Transitivity:

Take $(a, b)$ as $\left(\frac{\sqrt{7}}{3}, 1\right)$

$\&(b, c)$ as $\left(1, \frac{2 \sqrt{7}}{3}\right)$

So now $( a , b ) \in R \&( b , c ) \in R$ but $( a , c ) \notin R$ which means relation is not transitive

$P ( S )=$ power set of $S$

$AR , B \Rightarrow( A \cap \overrightarrow{ B }) \cup(\overrightarrow{ A } \cap B )=\phi$

$R 1$ is reflexive, symmetric

For transitive

$( A \cap \overrightarrow{ B }) \cup(\overrightarrow{ A } \cap B )=\phi ;\{ a \}=\phi=\{ b \} A = B$

$( B \cap \overrightarrow{ C }) \cup(\overrightarrow{ B } \cap C )=\phi \therefore B = C \therefore A = C \text { equivalence. }$

$R _2 \equiv A \cup \overrightarrow{ B }=\overrightarrow{ A } \cup B$

$R _2 \rightarrow \text { Reflexive, symmetric }$

for transitive

$\begin{array}{l} A \cup \overrightarrow{ B }=\overrightarrow{ A } \cup B \Rightarrow\{ a , c , d \}=\{ b , c , d \} \\ \{ a \}=\{ b \} \therefore A = B \\ B \cup \overrightarrow{ C }=\overrightarrow{ B } \cup C \Rightarrow B = C \\ \therefore A = C \quad \therefore A \cup \overrightarrow{ C }=\overrightarrow{ A } \cup C \therefore \text { Equivalence }\end{array}$

$B=\{0,1,2,3,4\}$

$R=\left\{(a, b) \in A \times A: 2(a-b)^2+3(a-b) \in B\right\}$

$\text { Now } 2(a-b)^2+3(a-b)=(a-b)(2(a-b)+3)$

$\Rightarrow a=b \text { or } a-b=-2$

Now $2(a-b)^2+3(a-b)=(a-b)(2(a-b)+3)$

$\Rightarrow a = b \text { or } a - b =-2$

When $a = b \Rightarrow 10$ order pairs

When $a-b=-2 \Rightarrow 8$ order pairs

Total $=18$

$R =\{ x - y =\text { odd }+ \text { ve or } x - y =2\} 0,4,6,8,10 \rightarrow \text { even }$

${ }^3 C _1 \cdot{ }^5 C _1=15+(6,4),(8,6),(10,8),(9,7)$

$Min ^{ m }$ ordered pairs to be added must be :

$15+4=19$

$a_1=3 \Rightarrow 4 \text { choices of } b_2$

$a_1=4 \Rightarrow 4 \text { choices of } b_2$

$a_1=6 \Rightarrow 2 \text { choices of } b_2$

$a_1=9 \Rightarrow 1 \text { choices of } b_2$

For $\left(a_1, b _2\right) 16$ ways .

Similarly, $b_1=2 \Rightarrow 4$ choices of $a_2$

$b _1=4 \Rightarrow 3 \text { choices of } a _2$

$b _1=5 \Rightarrow 2 \text { choices of } a _2$

$b _1=8 \Rightarrow 1 \text { choices of } a _2$

Required elements in $R =160$

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

$=\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)}$

$=\frac{4^x}{4^x+2}+\frac{2}{2+4^x}$

$=1$

$\Rightarrow f(x)+f(1-x)=1$

$\text { Now } f \left(\frac{1}{2023}\right)+ f \left(\frac{2}{2023}\right)+ f \left(\frac{3}{2023}\right)+\ldots \ldots .+$

$\ldots \ldots \ldots . .+ f \left(1-\frac{3}{2023}\right)+ f \left(1-\frac{2}{2023}\right)+ f \left(1-\frac{1}{2023}\right)$

Now sum of terms equidistant from beginning and end is 1

$\text { Sum }=1+1+1+\ldots \ldots \ldots+1 \text { (1011 times) }$

$=1011$

$f(4)=133$

$f(5)=255$

$133=2 \times 4^{ n }+\lambda......(1)$

$255=2 \times 5^{ n }+\lambda......(2)$

$(2) -(1)$

$122=2\left(5^{ n }-4^{ n }\right)$

$\Rightarrow 5^{ n }-4^{ n }=61$

$\therefore n =3\, and\, \lambda=5$

Now, $f(3)-f(2)=2\left(3^3-2^3\right)=38$

Number of Divisors is $1,2,19,38$; and their sum is $60$.

$f( n )+\frac{1}{ n } f ( n +1)=1, \forall n \in\{1,2,3\}$

$f( n +1)$ must be divisible by $n$

$f(4) \Rightarrow-6,-3,0,3,6$

$f(3) \Rightarrow-8,-6,-4,-2,0,2,4,6,8$

$f(2) \Rightarrow-8, \ldots \ldots \ldots \ldots \ldots, 8$

$f(1) \Rightarrow-8, \ldots \ldots \ldots \ldots \ldots, 8$

$\frac{f(4)}{3}$ must be odd since $f(3)$ should be even therefore $2$ solution possible.

$-\sqrt{2} \leq \sin x-\cos x \leq \sqrt{2}$

$\therefore-2 \leq \sqrt{2}(\sin x-\cos x) \leq 2$

$\quad \text { Assume } \sqrt{2}(\sin x-\cos x)=k)$

$-2 \leq k \leq 2 \quad \ldots( i )$

$f(x)=\log _{\sqrt{m}}( k + m -2)$

$\text { Given, }$

$0  \leq f( x ) \leq 2$

$0 \leq \log _{\sqrt{ m }}( k + m -2) \leq 2$

$\leq k + m -2 \leq m$

$- m +3 \leq k \leq 2 \ldots \text { (ii) }$

From eq. $(i)$ and $(ii)$, we get $- m +3=-2$

$\Rightarrow m=5$

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

$f(2)=\frac{2}{5} \quad f(3)=f(2)+f(1)=\frac{3}{5}$

$f(3)=\frac{3}{5}$

$\therefore \sum \limits_{n=1}^m \frac{f(n)}{n(n+1)(n+2)}$

$=\frac{1}{5} \sum \limits_{n=1}^m\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$

$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots .+\frac{1}{m+1}-\frac{1}{m+2}\right)$

$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12}$

$\therefore m=10$

$f(1)+2 f(2)+\ldots \ldots+x f(x)=x(x+1) f(x)$

$\text { replace } x \text { by } x +1$

$\Rightarrow \quad x(x+1) f(x)+(x+1) f(x+1)$

$=(x+1)(x+2) f(x+1)$

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

$\Rightarrow \quad x f(x)=(x+1) f(x+1)=\frac{1}{2}, x \geq 2$

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

$\text { Now } f (2022)=\frac{1}{4044}$

$f(2028)=\frac{1}{4056}$

So, $\frac{1}{f(2022)}+\frac{1}{f(2028)}=4044+4056=8100$

$5 f (0)= f (0)^2 \Rightarrow f (0)=5$

$5 f ( x +1)= f ( x ) \cdot f (1)$

$\frac{ f ( x +1)}{ f ( x )}=\frac{ f (1)}{5}$

$\frac{ f (1)}{ f (0)} \cdot \frac{ f (2)}{ f (1)} \cdot \frac{ f (3)}{ f (2)}=\left(\frac{ f (1)}{5}\right)^3$

$\frac{320}{5}=\frac{( f (1))^3}{5^3} \Rightarrow f (1)=20$

$5 f ( x +1)=20 \cdot f ( x ) \Rightarrow f ( x +1)=4 f ( x )$

$\sum \limits_{ n =0}^5 f ( n )=5+5.4+5.4^2+5.4^3+5.4^4+5.4^5$

$=\frac{5\left[4^6-1\right]}{3}=6825$

$n(S)=6$

$P(S)=\left\{\begin{array}{c}\phi,\{1\}, \ldots\{6\},\{1,2\}, \ldots, \\\{5,6\}, \ldots,\{1,2,3,4,5,6\}\end{array}\right\}$

case $-1$

$f(6)=S$ i.e. $1$ option,

$f(5)=$ any $5$ element subset $A$ of $S$ i.e. $6$ options,

$f(4)=$ any $4$ element subset $B$ of $A$ i.e. $5$ options,

$f (3)=$ any $3$ element subset $C$ of $B$ i.e.$4$ options,

$f (2)=$ any $2$ element subset $D$ of $C$ i.e. $3$ options,

$f (1)=$ any $1$ element subset $E$ of $D$ or empty subset i.e. $3$

options,Total functions $=1080$

Case $-2$

$f (6)=$ any $5$ element subset $A$ of $S$ i.e. $6$ options,

$f(5)=$ any $4$ element subset $B$ of $A$ i.e. $5$ options,

$f^{\prime}(4)=$ any $3$ element subset $C$ of $B$ i.e. $4$ options,

$f (3)=$ any $2$ element subset $D$ of $C$ i.e. $3$ options,

$f' (2) =$ any $1$ element subset $E$ of $D$ i.e. $2$ options,

$f(1)=$ empty subset i.e.$1$ option Total functions $=720$

Case $-3$

$f(6)=S$

$f(5)=$ any $4$ element subset $A$ of' $S$ i.e. $15$ options,

$f(4)=$ any $3$ element subset $B$ of $A$ i.e. $4$ options,

$f(3)=$ any $2$ element subset $C$ of B i.e. $3$ options,

$f(2)=$ any $1$ element subset $D$ of $C$ i.e. $2$ options,

$f (1)=$ empty subset i.e. $1$ option

Total functions $=360$

Case $-4$

$f(6)=S$

$f(5)=$ any $5$ element subset $A$ of $S$ i.e. $6$ options,

$f(4)=$ any $3$ element subset $B$ of $A$ i.e. $10$ options,

$f(3)=$ any $2$ element subset $C$ of $B$ i.e. $3$ options,

$f(2)=$ any $1$ element subset $D$ of $C$ i.e. $2$ options,

$f(1)=$ empty subset i.e. $1$ option

Total functions $=360$

Case $-6$

$f (6)= S$

$f (5)=$ any $5$ element subset $A$ of $S$ i.e. $6$ options,

$f (4)=$ any $4$ element subset $B$ of $A$ i.e. $5$ options,

$f (3)=$ any $3$ element subset $C$ of $B$ i.e. $4$ options,

$f (2)=$ any $1$ element subset $D$ of $C$ i.e. $3$ options,

$f (1)=$ empty subset i.e. 1 option

Total functions $=360$

$\therefore$ Number of such functions $=3240$

$\Rightarrow f^2(x)=\frac{13 x+12}{12 x+13}$

$\Rightarrow f^3(x)=\frac{63 x+62}{62 x+63}$

$\therefore f^5(x)=\frac{1563 x+1562}{1562 x+1563}$

$a+b=3125$

$x =2 \Rightarrow f (2)+ f (-1)=3$

$x =-1 \Rightarrow f (-1)+ f \left(\frac{1}{2}\right)=0$

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

$(1)+(3)-(2) \Rightarrow 2 f (2)=\frac{9}{2}$

$\therefore f (2)=\frac{9}{4}$

If $x \in I \lceil x \rceil=[ x ]$ (greatest integer function)

If $x \notin I \lceil x \rceil=[ x ]+1$

$\Rightarrow f(x)=\left\{\begin{array}{l}\frac{1}{\sqrt{[x]-x}}, x \in I \frac{1}{\sqrt{[x]+1-x}}, x \notin I\end{array}\right.$

$\begin{aligned} & \Rightarrow f(x)=\left\{\begin{array}{l}\frac{1}{\sqrt{-\{x\}}}, x \in I, \text { (does not exist) } \\ \frac{1}{\sqrt{1-\{x\}}}, x \notin I\end{array}\right. \\ & \Rightarrow \text { domain of } f(x)=R-I\end{aligned}$

$\text { Now, } f(x)=\frac{1}{\sqrt{1-\{x\}}}, x \notin I$

$\Rightarrow 0 < \{x\} < 1$

$\Rightarrow 0 < \sqrt{1-\{x\}} < 1$

$\Rightarrow \frac{1}{\sqrt{1-\{x\}}} > 1$

$\Rightarrow \text { Range }(1, \infty)$

$\Rightarrow A=R-I$

$B=(1, \infty)$

$\text { So, } A \cap B=(1, \infty)-N$

$A \cup B \neq(1, \infty)$

$\Rightarrow S 1 \text { is only correct }$

$\frac{(3 x+1)(2 x+1)}{2 x-1} > 0$

$-\frac{1}{2}-\frac{1}{3} \quad \frac{1}{2}$

$x \in\left(\frac{-1}{2}, \frac{-1}{3}\right) \cup\left(\frac{1}{2}, \infty\right).....(A)$

$x \in\left[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \cup\left(\frac{5}{3}, \infty\right)..........(B)$

$x < \frac{5}{3}........(C)$

$A \cap B \cap C \equiv\left(\frac{-1}{2}, \frac{-1}{3}\right) \cup\left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right]$

$\text { So } 18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right)=18\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{4}+\frac{1}{2}\right)$

$=18+2=20$

$\frac{[5}{\left\lfloor{3}\lfloor 2\right.} \times\lfloor 4=240$

Now when $f(a)=1$

$\lfloor 4+\frac{\lfloor 4}{\lfloor 2\lfloor 2} \times\lfloor 3=24+36=60$

so required $f ^{ n }=240-60=180$

$f(f(x))=\frac{A\left(\frac{A x+B}{C x-A}\right)+B}{C\left(\frac{A x+B}{C x-A}\right)-A}=x$

$f\left(f\left(\frac{4}{x}\right)\right)=\frac{4}{x}f(f(x))+f\left(f\left(\frac{4}{x}\right)\right)=x+\frac{4}{x} \geq 4(b y A M . \geq G . M .)$

${[x] < -2 \text { or }[x] > 5}$

$\Rightarrow t^4-t^3-3 t^2-t+1=0$

$\Rightarrow t^2+\frac{1}{t^2}-\left(t+\frac{1}{t}\right)-3=0$

$\Rightarrow\left(t+\frac{1}{t}\right)^2-\left(t+\frac{1}{t}\right)-5=0$

$\Rightarrow t+\frac{1}{t}=\frac{1+\sqrt{21}}{2}$

Two real values of $t$.

$f(1)+f(2) \leq 5$

Case $(i)$ $f(1)=1 \Rightarrow f(2)=1,2,3,4 \Rightarrow 4$ mappings

Case $(ii)$ $f(1)=2 \Rightarrow f(2)=1,2,3 \Rightarrow 3$ mappings

Case $(iii)$ $f(1)=3 \Rightarrow f(2)=1,2 \Rightarrow 2$ mappings

Case $(iv)$ $f(1)=4 \Rightarrow f(2)=1 \Rightarrow 1$ mapping

$f(5)$ and $f(6)$ both have $6$ mappings each

Number of functions $=(4+3+2+1) \times 6 \times 6=360$

$+\cos ^{-1}\left(\frac{10 x+6}{3}\right)$

$\text { (i) } 4 x^2+11 x+6 > 0$

$4 x^2+8 x+3 x+6 > 0$

$(4 x+3)(x+2) > 0$

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

$\text { (ii) } 4 x+3 \in[-1,1]$

$x \in[-1,-1 / 2]$

$\text { (iii) } \frac{10 x+6}{3} \in[-1,1]$

$x \in\left[-\frac{9}{10},-\frac{3}{10}\right]$

$x \in\left(-\frac{3}{4},-\frac{1}{2}\right]$

$\alpha+\beta=-\frac{5}{4}$

$36|\alpha+\beta|=45$

$\Rightarrow h ^{-1}( x )=\left(\frac{ x -7}{2}\right)^{\frac{1}{3}}$

$\Rightarrow h ( x )= fog ( x )=2 x ^3+7$

$\text { fog }(x)=a\left(x^b+c\right)-3$

$\Rightarrow a =2, b =3, c =5$

$\Rightarrow \operatorname{fog}( ac )= fog (10)=2007$

$g \left( f ( x )=(2 x -3)^3+5\right.$

$\Rightarrow \operatorname{gof}( b )=\operatorname{gof}(3)=32$

$\Rightarrow \operatorname{sum}=2039$

$=\lim _{ k \rightarrow 0} g (-2+1) \because f ( x )=-1 \forall x < 0$

$= g (-1)=1$

$RHL =\lim _{ k \rightarrow 0} g ( h ( k )) , k > 0$

$=\lim _{ k \rightarrow 0} g (-1) , \because f ( x )=1, \forall x > 0$

$=1$

$f \circ g(x)=x+3-\sqrt{x}$

$=(\sqrt{x}+1)^2-3(\sqrt{x}+1)+5$

$=g^2(x)-3 g(x)+5$

$\Rightarrow f(x)=x^2-3 x+5$

$\therefore f(0)=5$

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

$2 a +3 b =4 c +5 d =\alpha \text { let }$

$2 a =\{2,4,6,8\} \quad 4 c =\{4,8,12,16\}$

$3 b =\{3,6,9,12\} \quad 5 d =\{5,10,15,20\}$

$2 a +3 b =\left\{\begin{array}{l}5,8,11,14 \\ 7,10,13,16 \\ 9,12,15,18 \\ 11,14,17,20\end{array}\right\} \quad 4 c +5 d \left\{\begin{array}{l}9,14,19,24 \\ 13,18 \ldots . \\ 17,22 \ldots \\ 21,26 \ldots\end{array}\right\}$

Possible value of $\alpha=9,13,14,14,17,18$

Pairs of $\{( a , b ),( c , d )\}=6$

$\left|\frac{2 x}{5 x+3}\right| \geq 1 \Rightarrow|2 x| \geq|5 x+3|$

$(2 x)^2-(5 x+3)^2 \geq|5 x+3|$

$(7 x+3)(-3 x-3) \geq 0$

$\frac{-\quad-\quad-}{-1} \quad-\frac{3}{7}$

$\text { domain }\left[-1, \frac{-3}{5}\right) \cup\left(\frac{-3}{5}, \frac{-3}{7}\right]$

$\alpha=-1, \beta=\frac{-3}{5}, \gamma=\frac{-3}{5}, \delta=\frac{-3}{7}$

$3 \alpha+10(\beta+\gamma)+21 \delta=-3$

$-3+10\left(\frac{-6}{5}\right)+\left(\frac{-3}{7}\right) 21=-24$

$\text { this gives } x \in\left(0, \frac{1}{27}\right) \ldots \text { (1) }$

$-1 \leq \log _{3 x}\left(\frac{6+2 \log _3 x}{-5 x}\right) \leq 1$

$3 x \leq \frac{6+2 \log _3 x}{-5 x} \leq \frac{1}{3 x}$

$15 x^2+6+2 \log _3 x \geq 0 \quad 6+2 \log _3 x+\frac{5}{3} \geq 0$

$x \in\left(0, \frac{1}{27}\right) \quad \ldots(2) \quad x \geq 3^{-\frac{23}{6}}$

from (1), (2) and (3)

$x \in\left[3^{-\frac{23}{6}}, \frac{1}{27}\right)$

$\therefore \alpha$ is small positive quantity

$\beta=\frac{1}{27}$

$\therefore \alpha^2+\frac{5}{\beta}$ is just greater than $135$

Bonus

Fav. relation $=\phi,\{( x , x )\},\{( y , y )\},\{( x , x )( y , y )\}$

                            $\{(x, x),(y, y),(x, y)(y, x)\}$

Prob. $=\frac{5}{16}$

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

For reflexive $x \times x \geq 0$ which is true.

For symmetric

If $x y \geq 0 \Rightarrow y x \geq 0$

If $x =2, y =0$ and $z =-2$

Then $x . y \geq 0 \& y . z \geq 0$ but $x . z \geq 0$ is not true $\Rightarrow$ not transitive relation.

$R_{ I }$ is not equivalence

$R _{2}$ if $a \geq b$ it does not implies $b \geq a$

$R_{2}$ is not equivalence relation

$D$

$\Rightarrow 3 x +\alpha x =7 x \Rightarrow(3+\alpha) x =7 K$

$\Rightarrow 3+\alpha=7 \lambda \Rightarrow \alpha=7 \lambda-3=7 N +4, K , \lambda, N \in I$

$\therefore$ when $\alpha$ divided by $7$, remainder is $4$.

$R$ to be symmetric $x R y \Rightarrow y R x$

$3 x +\alpha y =7 N _{1}, 3 y +\alpha x =7 N _{2}$

$\Rightarrow(3+\alpha)( x + y )=7\left( N _{1}+ N _{2}\right)=7 N _{3}$

Which holds when $3+\alpha$ is multiple of $7$

$\therefore \alpha=7 N +4 \text { (as did earlier) }$

$R$ to be transitive

$xRy \& yRz \Rightarrow xRz$.

$3 x +\alpha y =7 N _{1} \& 3 y +\alpha z =7 N _{2}$

$3 x +\alpha z =7 N _{3}$

$\therefore 3 x +7 N _{2}-3 y =7 N _{3}$

$\therefore 7 N _{1}-\alpha y +7 N _{2}-3 y =7 N _{3}$

$\therefore 7\left( N _{1}+ N _{2}\right)-(3+\alpha) y =7 N _{3}$

$\therefore(3+\alpha) y =7\,N$

Which is true again when $3+\alpha$ divisible by $7$ when $\alpha$ divided by $7$, remainder is $4$ .

$(R \in N)$

$Note$ that for any element, it will fall into exactly. one of these sets.

$\{y: y=4 R ; y \in N\}$

$\{y: y=4 R-2 ; y \in N\}$

$\{ y : y =2 R -1 ; y \in N \}$

Corresponding to that $y$, we will get exactly one value of $n$.

Thus, $f$ is one - one and onto.

$f ( g ( x ))= g ( x )-1$

$\quad=\frac{ x ^{2}}{ x ^{2}-1}-1=\frac{ x ^{2}- x ^{2}+1}{ x ^{2}-1}$

$f ( g ( x ))=\frac{1}{x^{2}-1} ; x \neq \pm 1 \text {, even function }$

$\rightarrow$ Hence $f ( g ( x ))$ is many one function

$y=\frac{1}{x^{2}-1}$

$y \cdot x^{2}-y=1$

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

$\left(\frac{1+y}{y}\right) \geq 0$

Range:- $y \in(-\infty,-1) \cup(0, \infty)$

Hence, Range $\neq$ Co-domain $\Rightarrow f ( g ( x ))$ is into function

$g\left(f(x)=4 x^{2}+6 x+1\right.$

So, $g ( x )=2 x -1$       $g(2)=3$

$f ( x )=2 x ^{2}+3 x +1$

$f(2)=8+6+1=15$

Ans. $18$

For continuity $a =4$ and $b =-15$

$g(f(2))+f(g(-2))$

$=g(2)+f(-1)=-8$

$R_{2}=\{(a, b) \in N \times N:|a-b| \neq 13\}$.

For $R_{1}$ :

$(i)\,Reflexive \,\,relation$

$(a, a) \in N \times N:|a-a| \leq 13$

$(ii)\, Symmetric\,\, relation$

$( a , b ) \in R _{1},( b , a ) \in R _{1}:| b - a | \leq 13$

$(iii) \,Transitive\, \,relation$

$( a , b ) \in R _{1},( b , c ) \in R _{1},( a , c ) \in R _{1}:$

$(1,3) \in R _{1,}(3,16) \in R _{1,} \text { but }(1,16) \notin R _{1}$

For $R _{2}$ :

$(i) \,Reflexive\,\, relation$

$(a, a) \in N \times N:|a-a| \neq 13$

$(ii)\, Symmetric\,\, relation$

$(b, a) \in N \times N:|b-a| \neq 13$

$(iii)\, Transitive \,\,relation$

$( a , b ) \in R _{2},( b , c ) \in R _{2},( a , c ) \in R _{2}$

$(1,3) \in R _{2,}(3,14) \in R _{2} \text { but }(1,14) \notin R _{2}$

Now p can take values

$2,3,5,7,11,13,17,23,29,31,37,41,43$ and $47 .$

we can calculate no. of elements in $R$, as

$\left(2,2^{\circ}\right),\left(2,2^{1}\right) \ldots\left(2,2^{5}\right)$

$\left(3,3^{\circ}\right), \ldots\left(3,3^{3}\right)$

$\left(5,5^{\circ}\right), \ldots\left(5,5^{2}\right)$

$\left(7,7^{\circ}\right), \ldots\left(7,7^{2}\right)$

$\left(11,11^{\circ}\right), \ldots\left(11,11^{1}\right)$

And rest for all other two elements each

$n \left( R _{1}\right)=6+4+3+3+(2 \times 10)=36$

Similarly for $R _{2}$

$\left(2,2^{\circ}\right),\left(2,2^{1}\right)$

$\left(47,47^{\circ}\right),\left(47,47^{1}\right)$

$n \left( R _{2}\right)=2 \times 14=28$

$n \left( R _{1}\right)- n \left( R _{2}\right)=36-28=8$

Using fundamental principle of counting

Number of one-one function is $31$

$f (1)=2$,

$\sum_{ k =1}^{10} f (\alpha+ k )=2 f (\alpha) \sum_{ k =1}^{10} f ( k )$

$=2 f (\alpha)( f (1)+ f (2)+\ldots .+ f (10))$

From $(1)$

$f (2)=2 f ^{2}(1)=2^{3}$

$\left.f (3)=2 f (2) f (1)=2^{5}\right)$

$\vdots$

$f (10)=2^{9} f ^{10}(1)=2^{19}$

$f (\alpha)=2^{2 \alpha-1} ; \alpha \in N$

from $(2)$

$\sum_{ k =1}^{10} f (\alpha+ k )=2\left(2^{2 \alpha-1}\right)\left(2+2^{3}+2^{5}+\ldots .+2^{19}\right)$

$\frac{512}{3}\left(2^{20}-1\right)=2^{2 \alpha}\left(2 \frac{\left(2^{20}-1\right)}{3}\right)$

Hence $\alpha=4$

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

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

$\Rightarrow f^{4}(x)=f\left(\frac{x+1}{1-x}\right)=-\frac{1}{x}$

$\Rightarrow f^{6}(x)=-\frac{1}{x} \Rightarrow f^{6}(6)=-\frac{1}{8}$

$f^{7}(x)=\left(-\frac{1}{x}\right)=\frac{x+1}{1-x}$

$\Rightarrow f^{7}(7)=\frac{8}{-6}=-\frac{4}{3}$

$\therefore-\frac{1}{6}+-\frac{4}{3}=-\frac{3}{2}$

$=2\left[\frac{e^{2 x-1}}{e^{2 x-1}+1}+\frac{1}{1+e^{2 x-1}}\right]=2$

$f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+f\left(\frac{3}{100}\right)+\ldots .+f\left(\frac{99}{100}\right)$

$=\left\{f\left(\frac{1}{100}\right)+f\left(\frac{99}{100}\right)\right\}+\left\{f\left(\frac{2}{100}\right)+f\left(\frac{98}{100}\right)\right\}+\ldots .+f\left\{\left(\frac{49}{100}\right)+f\left(\frac{51}{100}\right)\right\}+f\left(\frac{1}{2}\right)$

$=(2+2+2+----49$ times $)+\frac{2 e}{e+e}$

$=98+1=99$