Maths M.C.Q (1 Marks)

$f ( x )=( x +1)( ax + b )$

$f (-2)+ f (3)=-1(-2 a + b )+4(3 a + b )=0$

$2 a - b +12 a +4 b =0$

$14 a +3 b =0$

$\frac{- b }{ a }=\frac{14}{3}$

Sum of roots $=\left(-1+\frac{-b}{a}\right)=-1+\frac{14}{3}=\frac{11}{3}$

$(4,4)-$ all have one choice for image.

$(2,1),(1,2),(2,2)$ - all have three choices for image

$(3,2),(2,3),(3,1),(1,3),(3,3)-$ all have two choices for image.

So the total functions $=3 \times 3 \times 2 \times 2 \times 2=72$

Case $1$ : None of the pre-images have $3$ as image

Total functions $=2 \times 2 \times 1 \times 1 \times 1=4$

Case $2$ : None of the pre-images have $2$ as image

Total functions $=2 \times 2 \times 2 \times 2 \times 2=32$

Case $3$ : None of the pre-images have either $3$ or $2$

as image

Total functions $=1 \times 1 \times 1 \times 1 \times 1=1$

$\therefore$ Total onto functions $=72-4-32+1=37$

$-\pi / 2 \leq \sin ^{-1} \frac{1}{4 x^{2}-1} \leq \pi / 2$

Always $-1 \leq \frac{1}{4 x^{2}-1} \leq 1$

$x \in\left(\infty, \frac{1}{\sqrt{2}}\right) \cup\left[\frac{1}{\sqrt{2}}, \infty\right)$

$f(3)>f(9)>f(15) \ldots . .>f(99)$

So number of ways $={ }^{50} C_{17} \cdot 1 \cdot 33$ !

$={ }^{50} P_{33}$

$B =\{1,2,3,4,5,6\}$

Here $f(3)$ can be $2, 3, 4, 5, 6$

$f (3)=2,( f (1), f (2)) \rightarrow(1,1) \rightarrow 6 \text { cases }$

$f (3)=3,( f (1), f (2)) \rightarrow(1,2),(2,1)$

$\rightarrow 2 \times 6=12 \text { cases }$

$f (3)=4,( f (1), f (2)) \rightarrow(1,3),(3,1),(2,2)$

$\rightarrow 3 \times 6=18 \text { cases }$

$f (3)=5,( f (1), f (2)) \rightarrow(1,4),(4,1),(2,3),(3,2)$

$\rightarrow 4 \times 6=24 \text { cases }$

$f (3)=6,( f (1), f (2)) \rightarrow(1,5),(5,1),(2,4),(4,2),(3,3)$

$\rightarrow 5 \times 6=30 \text { cases }$

Total number of cases $=6+12+18+24+30=90$

$f(x / 3)-f\left(x / 3^{2}\right)=x / 3^{2}$

$f(x)-\lim _{n \rightarrow \infty} f\left(\frac{x}{3^{n}}\right)=x\left(\frac{1}{3}+\frac{1}{3^{2}} \ldots \infty\right)$

$f(x)-f(0)=\frac{x}{2}$

$f(8)=7 ; f(0)=3$

$f(x)=x / 2+3$

$f(14)=10$

Where $\alpha$ is max of powers of prime $P$ such that $p ^{\alpha}$ divides a. Also $g ( a )= a +1$

$f(2)=1$ $g(2)=3$

$f(3)=1$ $g(3)=4$

$f(4)=2$ $g(4)=5$

$f(5)=1$ $g(5)=6$

$f(2)+g(2)=4$

$(f(3)+g(3))=5$

$f(4)+g(4)=7$

$f(5)+g(5)=7$

$\therefore$ Many one $f(x)+g(x)$ does not cotain 1

into function

Ans.$(D)$ [neither one-one nor onto]

$\therefore f (0)=7-\lambda= c$

$f (1)= a + b + c =3$

$f (3)=9 a +3 b + c =4 \quad$...(ii)

$f (-2)=4 a -2 b + c =\lambda \quad$...(iii)

$(ii) - (iii)$

$a+b=\frac{4-\lambda}{5}$ put in equation (i)

$\frac{4-\lambda}{5}+7-\lambda=3$

$6 \lambda=24 ; \quad \lambda=4$

$f(x)=\left[\left(2-x^{25}\right)\left(2+x^{25}\right)\right]^{\frac{1}{50}}$

$\quad=\left(4-x^{50}\right)^{1 / 50}$

$\begin{aligned} f(f(x)) =\left(4-\left(\left(4-x^{50}\right)^{1 / 50}\right)^{50}\right)^{1 / 50}=x \\ g(x)=f(f(f(x)))+f(f(x)) \\ \quad=f(x)+x \\ g(1)=f(1)+1=3^{1 / 50}+1 \\g(1)]=\left[3^{1 / 50}+1\right]=2 \end{aligned}$

$\therefore \quad f ( g ( n ))=\left\{\begin{array}{lll} n +1  ;& n \in \text { odd } \\ n -1 ;  &n \in \text { even } \end{array}\right.$

$\Rightarrow \quad g ( n ) =\left\{\begin{array}{ll} f ^{-1}( n +1) ;& n \in \text { odd } \\ f ^{-1}( n -1)  ; &n \in \text { even } \end{array}\right.$

$\therefore \quad g ( n ) =\left\{\begin{array}{ll} \frac{ n +1}{2} ;  &n \in \text { odd } \\ \frac{ n +10}{2} ; & n \in \text { even } \end{array}\right.$

$g (10) \cdot[ g (1)+ g (2)+ g (3)+ g (4)+ g (5)]$

$=10 \cdot[1+6+2+7+3]=190$

$\frac{x^{2}-5 x+6}{x^{2}-9}-1 \leq 0$

$\frac{1}{x+3} \geq 0$

$x \in(-3, \infty) \ldots \ldots(1)$

$\frac{x^{2}-5 x+6}{x^{2}-9}+1 \geq 0$

$\frac{2 x+1}{x+3} \geq 0$

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

after taking intersection

$x \in\left[-\frac{1}{2}, \infty\right)$

$x^{2}-3 x+2>0$

$x \in(-\infty, 1) \cup(2, \infty)$

$x^{2}-3 x+2 \neq 1$

$x \neq \frac{3 \pm \sqrt{5}}{2}$

after taking intersection of each solution

$\left[-\frac{1}{2}, 1\right) \cup(2, \infty)-\left\{\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\right\}$

$f(x+y)=f(x)+f(y)-x y \quad \forall x y \in R$

$\lim \limits_{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}=\lim \limits_{y \rightarrow 0} \frac{f(y)-x y}{y}\Rightarrow f^{\prime}(x)=f^{\prime}(0)-x$

$f(x)=-\frac{1}{2} x^{2}+f^{\prime}(0) \cdot x+\lambda \quad \text { but } f(0)=0 \Rightarrow \lambda=0$

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

$\therefore \quad$ as $f ^{\prime}(0)=1- c ^{2}$

Comparing equation $(1)$ and $(2)$

We obtain, $c =-\frac{3}{2}$

$\therefore \quad f ( x )=-\frac{1}{2} x ^{2}-\frac{5}{4} x$

Now $\left|2 \sum \limits_{x=1}^{20} f(x)\right|=\sum \limits_{x=1}^{20} x^{2}+\frac{5}{2} \cdot \sum \limits_{x=1}^{20} x$

$=2870+525$

$=3395$

It is given that $f(0)=p \Rightarrow \alpha \beta=p$

and $f(1)=\frac{1}{3} \Rightarrow(1-\alpha)(1-\beta)=\frac{1}{3}$

Now, let us assume that $\alpha$ is the common root of $f(x)=0$ and $f \circ f \circ f o f(x)=0$

$fofofof(x)=0$

$fofof(0) =0$

$f o f(p)=0$

So, $f(p)$ is either $\alpha$ or $\beta$.

$(p-\alpha)(p-\beta)=\alpha$

$(\alpha \beta-\alpha)(\alpha \beta-\beta)=\alpha \Rightarrow(\beta-1)(\alpha-1) \beta=1$

$(\because \alpha \neq 0)$

So, $\beta=3$

$(1-\alpha)(1-3)=\frac{1}{3}$

$\alpha=\frac{7}{6}$

$f(x)=\left(x-\frac{7}{6}\right)(x-3)$

$f(-3)=\left(-3-\frac{7}{6}\right)(3-3)=25$

$x \in[1,9] \Rightarrow A =\{1,2,3,4,5,6,7,8,9\}$

$f ( x ) \leq( x -3)^{2}+1$

$x =1: f (1) \leq 5 \Rightarrow 1^{2}, 2^{2}$

$x =2: f (2) \leq 2 \Rightarrow 1^{2}$

$x =3: f (3) \leq 1 \Rightarrow 1^{2}$

$x =4: f (4) \leq 2 \Rightarrow 1^{2}$

$x =5: f (5) \leq 5 \Rightarrow 1^{2}, 2^{2}$

$x =6: f (6) \leq 10 \Rightarrow 1^{2}, 2^{2}, 3^{2}$

$x =7: f (7) \leq 17 \Rightarrow 1^{2}, 2^{2}, 3^{2}, 4^{2}$

$x =8: f (8) \leq 26 \Rightarrow 1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}$

$x =9: f (9) \leq 37 \Rightarrow 1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2}$

Total number of such function

$=2(6 !)=2(720)=1440$

$f ( x )=2 x ^{2}- x -1$

$| f ( x )| \leq 800$

$2 n ^{2}- n -801 \leq 0$

$n^{2}-\frac{1}{2} n-\frac{801}{2} \leq 0$

$\left(n-\frac{1}{4}\right)^{2}-\frac{801}{2}-\frac{1}{16} \leq 0$

$\left(n-\frac{1}{4}\right)^{2}-\frac{6409}{16} \leq 0$

$\left(n-\frac{1}{4}-\frac{\sqrt{6409}}{4}\right)\left(n-\frac{1}{4}+\frac{\sqrt{6409}}{16}\right) \leq 0$

$\frac{1-\sqrt{6409}}{4} \leq n \leq \frac{1+\sqrt{6409}}{4}$

$n=\{-19,-18-17, \ldots \ldots 0,1,2, \ldots \ldots, 20\}$

$\sum_{n \in s} f(x)=\sum\left(2 x^{2}-x-1\right)$

$=2\left[19^{2}+18^{2}+\ldots . .+1^{2}+1^{2}+2^{2}+\ldots .+19^{2}+20^{2}\right]$

$=4\left[1^{2}+2^{2}+\ldots . .+19^{2}\right]+2\left[20^{2}\right]-20-40$

$=\frac{4 \times 19 \times 20 \times(2 \times 19+1)}{6}+2 \times 400-60$

$=\frac{4 \times 19 \times 20 \times 39}{6}+800-60-9880+800-60$

$=10620$

at $(0,0)$ and radius $=\sqrt{2}$

$\Rightarrow x^{2}+y^{2}=2$

$S=\left\{(x, y) \mid x^{2}+y^{2}=2\right\}$

but $(1,3)$ does not satisfy it so

$0 \leq|\mathrm{x}-\mathrm{y}| \leq 1$ is symmetric but not transitive

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

$\Rightarrow f\left(\frac{1}{20}\right)+f\left(\frac{2}{20}\right)+\ldots+f\left(\frac{39}{20}\right)$

$=\left(f\left(\frac{1}{20}\right)+f\left(\frac{39}{20}\right)\right)+\ldots+\left(f\left(\frac{19}{20}\right)+f\left(\frac{21}{20}\right)+f\left(\frac{20}{20}\right)\right)$

$=19+\frac{1}{2}=\frac{39}{2}$

$y={ }^{15} C_{3} \times 3 !=15 \times 14 \times 13=30 \times 91$

$\therefore 2 y=91 x$

$=\frac{x}{x-1}=1+\frac{1}{x-1}$

Range of $f(g(x)= R -\{1\}$

Range of $f(g(x))$ is not onto

$\&$ $f(g(x))$ is one-one

So $f(g(x))$ is one-one but not onto.

$\Rightarrow f(n)=n f(1)$

$\Rightarrow f$ is one-one

Now, Let $f \left( g \left( x _{2}\right)\right)= f \left( g \left( x _{1}\right)\right)$

$\Rightarrow g\left(x_{2}\right)=g\left(x_{1}\right)($ as $f$ is one-one $)$

$\Rightarrow x _{1}= x _{2}$ (as fog is one-one)

$\Rightarrow g$ is one-one

Now, $f(g(n))=g(n) f(1)$

may be many-one if

$g ( n )$ is many-one

$\because g : A \rightarrow A$ such that $g (f( x ))=f( x )$

$\Rightarrow$ If $x$ is even then $g(x)=x ....(1)$

If $x$ is odd then $g(x+1)=x+1....(2)$

from $(1)$ and $(2)$ we can say that

$g(x)=x$ if $x$ is even

$\Rightarrow$ If $x$ is odd then $g(x)$ can take any value in set $A$

so number of $g(x)=10^{5} \times 1$

$\Rightarrow \lg ( x ) l \leq 1 \quad, \quad g (2)=\frac{3}{7}$

$\left|\frac{ x ^{2}- x -2}{2 x ^{2}- x -6}\right| \leq 1$

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

$\frac{x+1}{2 x+3} \leq 1$ and $\frac{x+1}{2 x+3} \geq-1$

$\frac{x+1-2 x-3}{2 x+3} \leq 0$ and $\frac{x+1+2 x+3}{2 x+3} \geq 0$

$\frac{x+2}{2 x+3} \geq 0$ and $\frac{3 x+4}{2 x+3} \geq 0$

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

$5 x+3=6 x y-\alpha y$

$x(6 y-5)=\alpha y+3$

$x=\frac{\alpha y+3}{6 y-5}$

$f^{-1}(x)=\frac{\alpha x+3}{6 x-5}\,....(ii)$

fo $f(x)=x$

$f(x)=f^{-1}(x)$

From eq $^{n}$ $(i)\, \& \,(ii)$

Clearly $\alpha=5$

$\Rightarrow^{\prime} \mathrm{f}^{\prime}$ must be one-one and $'^{\prime}\mathrm{g}^{\prime}$ must be $ONTO$

Interchanging $x \& y$ for inverse

$x=5^{\left(\log _{a} y\right)}=y^{\left(\log _{a} 5\right)}$

option $(1)$ or option $(2)$

Further, from given relation

$\log _{5} y =\log _{ a } x$

$\Rightarrow x=a^{\left(\log _{5} y\right)}=y^{\left(\log _{5} a\right)}$

$\Rightarrow x=y^{\left(\frac{1}{\log _{a} 5}\right)}=f^{-1}(y)$

$(a, b)=(c, d) \quad \Rightarrow \quad a d=b c$

$(4,3)=( c , d ) \quad \Rightarrow \quad 4 d =3 c$

$\Rightarrow \frac{4}{3}=\frac{c}{d}$

$\frac{c}{d}=\frac{4}{3}$ and $c,d\, \in \, \{2,3, \ldots \ldots, 30\}$

$( c , d )=\{(4,3),(8,6),(12,9),(16,12),(20,15), (24,18),(28,21)\}$

No. of ordered pair $=7$

there exists a non singular matrix $P (\operatorname{det}( P ) \neq 0)$

such that $PAP ^{-1}= B$

For reflexive

$ARA \Rightarrow PAP ^{-1}= A \quad \ldots(1)$ must be true for $P = I ,$ Eq.(1) is true so $'R'$ is reflexive

For symmetric

$ARB \Leftrightarrow PAP ^{-1}= B \quad \ldots(1)$ is true

for $BRA$ iff $PBP ^{-1}= A \ldots$. (2) must be true

$\because PAP -1= B$

$P ^{-1} PAP ^{-1}= P ^{-1} B$

$IAP ^{-1} P = P ^{-1} BP$

$A = P ^{-1} BP \ldots(3)$

from $(2)$ And $(3) PBP ^{-1}= P ^{-1} BP$

can be true some $P = P ^{-1} \Rightarrow P ^{2}= I (\operatorname{det}( P ) \neq 0)$

So $'R'$ is symmetric

For trnasitive

$ARB \Leftrightarrow PAP ^{-1}= B \ldots$ is true

$BRC \Leftrightarrow PBP ^{-1}= C \ldots$ is true

now $\quad PPAP ^{-1} P ^{-1}= C$

$P ^{2} A \left( P ^{2}\right)^{-1}= C \Rightarrow ARC$

So $'R'$ is transitive relation

$\Rightarrow$ Hence $R$ is equivalence

$\Rightarrow x(x-y)(x+y)-3 y(x-y)(x+y)=0$

$\Rightarrow(x-3 y)(x-y)(x+y)=0$

Now, $x=y \forall(x, y) \in N \times N$ so reflexive

But not symmetric \& transitive

See, $(3,1)$ satisfies but $(1,3)$ does not.

Also $(3,1) \,\&\,(1,-1)$ satisfies but $(3,-1)$ does not

$P (1)=f(1)+ g (1) .....(1)$

Now $P ( x )$ is divisible by $x ^{2}+ x +1$

$\Rightarrow P ( x )= Q ( x )\left( x ^{2}+ x +1\right)$

$P ( w )=0= P \left( w ^{2}\right)$ where $w , w ^{2}$ are non-real cube roots of units

$P ( x )=f\left( x ^{3}\right)+ xg \left( x ^{3}\right)$

$P ( w )=f\left( w ^{3}\right)+ wg \left( w ^{3}\right)=0$

$f(1)+\operatorname{wg}(1)=2 .....(2)$

$P \left( w ^{2}\right)=f\left( w ^{6}\right)+ w ^{2} g \left( w ^{6}\right)=0$

$f(1)+w^{2} g(1)=0 .....(3)$

$(2)+(3)$

$\Rightarrow 2 f(1)+\left(w+w^{2}\right) g(1)=0$

$2 f(1)= g (1) .....(4)$

$(2)-(3)$

$\Rightarrow\left( w - w ^{2}\right) g (1)=0$

$g(1)=0=f(1) \quad$ from (4)

from $1) \,P (1)=f(1)+ g (1)=0$

replace $x$ by $\frac{1}{x}$

af $\left(\frac{1}{x}\right)+\alpha f(x)=\frac{b}{x}+\beta x.....(2)$

$(1)+(2)$

$(a+\alpha) f(x)+(a+\alpha) f\left(\frac{1}{x}\right)=x(b+\beta)+(b+\beta) \frac{1}{x}$

$\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}=\frac{b+\beta}{a+\alpha}=\frac{2}{1}=2$

$\text { Put } \mathrm{m}=1, \mathrm{n}=1$

$f(2)=2 f(1)$

$\text { Put } \mathrm{m}=2, \mathrm{n}=1$

$f(3)=f(2)+f(1)=3 f(1)$

$\text { Put } \mathrm{m}=3, \mathrm{n}=3$

$f(6)=2 f(3) \Rightarrow f(3)=9$

$\Rightarrow f(1)=3, f(2)=6$

$f(2) \cdot f(3)=6 \times 9=54$

$(3+\cos \left(\frac{3 \pi}{4}+x\right)+\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)$

$-\cos \left(\frac{3 \pi}{4}-x\right))$

$f(x)=\log _{\sqrt{5}}\left[3+2 \cos \left(\frac{\pi}{4}\right) \cos (x)-2 \sin \left(\frac{3 \pi}{4}\right) \sin (x)\right]$

$f(x)=\log _{\sqrt{5}}[3+\sqrt{2}(\cos x-\sin x)]$

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

$\Rightarrow \log _{\sqrt{5}}\left[3+\sqrt{2}(-\sqrt{2}) \leq f(x) \leq \log _{\sqrt{5}}[3+\sqrt{2}(\sqrt{2})]\right]$

$\Rightarrow \log _{\sqrt{5}}(1) \leq f(x) \leq \log _{\sqrt{5}}(5)$

So Range of $f(x)$ is $[0,2]$

put $\mathrm{x}=0$

we get $\lambda=\frac{1}{60}$

Now put $\lambda$ in equation $(1)$

$\Rightarrow \mathrm{k\,f}(\mathrm{k})+2=\frac{1}{60}(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)(\mathrm{x}-5)$

Put $\mathrm{x}=10$

$\Rightarrow 10 \,\mathrm{f}(10)+2=\frac{1}{60}(8)(7)(6)(5)$

$\Rightarrow 52-10 \,\mathrm{f}(10)=52-26=26$

$\frac{|[x]|-2}{|[x]|-3} \geq 0$

Case $I:$ When $|[x]|-2 \geq 0$

and $|[x]|-3\,>\,0$

$\therefore x \in(-\infty,-3) \cup[4, \infty] \ldots . .(1)$

Case $II:$ When $|[x]|-2 \leq 0$

and $|[x]|-3\,<\,0$

$\therefore \mathrm{x} \in[-2,3) \quad \ldots(2)$

So, from $(1)$ and $(2)$

We get

Domain of function

$=(-\infty,-3) \cup[-2,3) \cup[4, \infty)$

$\therefore(a+b+c)=-3+(-2)+3=-2(a\,<\,b\,<\,c)$

$\Rightarrow x^{2}-x \leq 0$

$\Rightarrow x \in[0,1]$

$\text { Also, } 0\,<\,\sin ^{-1}\left(\frac{2 x-1}{2}\right) \leq \frac{\pi}{2}$

$\Rightarrow 0\,<\,\frac{2 x-1}{2} \leq 1$

$\Rightarrow 0\,<\,2 x-1 \leq 2$

$1\,<\,2 x \leq 3$

$\frac{1}{2}\,<\,x \leq \frac{3}{2}$

Taking intersection

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

$\Rightarrow \alpha=\frac{1}{2}, \beta=1$

$\Rightarrow \alpha=\frac{1}{2}, \beta=1$

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

The only possibility is: $0+1+2=3$

$\Rightarrow$ Elements $1,2,3$ in the domain can be mapped with $0,1,2$ only.

So number of bijective functions.

$=\lfloor 3 \times\lfloor 5=720$

Put $m=1 f(n)=f(1) \cdot f(n) \Rightarrow f(1)=1$

Put $m=n=2$

$f(4)=f(2) \cdot f(2)$

$f(2)=1 \Rightarrow f(4)=1 \text { or } f(2)=2 \Rightarrow f(4)=4$

Put $m=2, n=3$

$f(6)=f(2) \cdot f(3)$

$\text { when } f(2)=1  \Rightarrow f(3)=1 \text { to } 7$

$f(2)=2 \Rightarrow f(3)=1 \text { or } 2 \text { or } 3$

$f(5), f(7)$ can take any value

Total $=(1 \times 1 \times 7 \times 1 \times 7 \times 1 \times 7)$

$+(1 \times 1 \times 3 \times 1 \times 7 \times 1 \times 7)$

$=490$

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

So, $-1=\cos \frac{a}{2}$

$\Rightarrow a=2(2 n+1) \pi$

Thus $f(x)=\cos 2(2 n+1) \pi x$

Now k is natural number

Thus $f(k)=1$

$\sum_{k=1}^{20} \frac{1}{\sin k \sin (k+1)}=\frac{1}{\sin 1} \sum_{k=1}^{20}\left[\frac{\sin ((k+1)-k)}{\sin k \cdot \sin (k+1)}\right]$

$=\frac{1}{\sin 1} \sum_{k=1}^{20}(\cot k-\cot (k+1))$

$=\frac{\cot 1-\cot 21}{\sin 1}=\operatorname{cosec}^{2} 1 \cos \operatorname{ec}(21) \cdot \sin 20$

$g[f(x)]=\left[\begin{array}{cc}x+a+1 & x+a<0 \;and\; x<0 \\ |x-1|+1 & |x-1|<0 \;and\; x \geq 0 \\ (x+a-1)^{2}+b & x+a \geq 0 \;and\; x<0 \\ (|x-1|-1)^{2}+b & |x-1| \geq 0 \;and\; x \geq 0\end{array}\right.$

$g[f(x)]=\left[\begin{array}{cc}x+a+1 & x \in(-\infty,-a) \;and\; x \in(-\infty, 0) \\ |x-1|+1 & x \in \phi \\ (x+a-1)^{2}+b & x \in[-a, \infty) \;and\; x \in(-\infty, 0) \\ (|x-1|-1)^{2}+b & x \in R \;and\; x \in[0, \infty)\end{array}\right.$

$g[f(x)]=\left[\begin{array}{cc}x+a+1 & x \in(-\infty,-a) \\ (x+a-1)^{2}+b & x \in[-a, 0) \\ (|x-1|-1)^{2}+b & x \in[0, \infty)\end{array}\right.$

$g ( f ( x ))$ is continuous

at $x=-a$ And at $x=0$

$1=b+1$ And $(a-1)^{2}+b=b$

$b =0$ And $a =1$

$\Rightarrow \quad a+b=1$

$g(3 n+2)=3 n+3$

$g(3 n+3)=3 n+1, n \geq 0$

For $x=3 n+1$

$(1)$ $\operatorname{gogog}(3 n+1)=\operatorname{gog}(3 n+2)=g(3 n+3)=3 n+1$

Similarly

$\operatorname{gogog}(3 n+2)=3 n+2$

$\operatorname{gogog}(3 n+3)=3 n+3$

So gogog $(x)=x \quad \forall x \in N$

$(2)$ If $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ and $\mathrm{f}$ is an onto function such that $\mathrm{f}(\mathrm{g}(\mathrm{x}))=\mathrm{f}(\mathrm{x})$ then

One of its possibilities is by taking $f(x)$ as onto function

$f(x)=a \quad x=3 n+1$

$\quad\quad \quad a \quad x=3 n+2, a \in N$

$\quad\quad \quad a \quad x=3 n+3$

$(3)$ If $f: N \rightarrow N$ and $f$ is a one-one function such that $f(g(x))=f(x)$ then

$g(x)=x$ but $g(x) \neq x$

$(4)$ As $f: N \rightarrow N, f=3 n+1$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad=3 \mathrm{n}+2$

So, $g(3 n+1)=3 n+2, g(3 n+2)=3 n+3, g(3 n+3)=3 n+1$

So $g(f(x)) \neq f(x)$

$\therefore x=\frac{3 y-2}{y-1}$

$\therefore f^{-1}(x)=\frac{3 x-2}{x-1}$

$ g(x)=y=2 x-3$

$\therefore x=\frac{y+3}{2}$

$\therefore g^{-1}(x)=\frac{x+3}{2}$

$\because f^{-1}(x)+g^{-1}(x)=\frac{13}{2}$

$\therefore x^{2}-5 x+6=0$

$\therefore$ sum of roots

$x_{1}+x_{2}=5$

For domain of $\mathrm{R}^{-1}$

Collection of all integral of y's

For $x=0, \quad 3 y^{2} \leq 8$

$\Rightarrow \mathrm{y} \in\{-1,0,1\}$

Case-I : $\quad$ If $f(x)=2 \forall x \in A$ then number of function $=1$

Case-II : $\quad$ If $f(x)=2$ for exactly two elements then total number of many-one function $={ }^{3} C _{2}{ }^{3} C _{1}=9$

Case-III : If $f(x)=2$ for exactly one element then total number of many-one functions $={ }^{3} C _{1}{ }^{3} C _{1}=9$

Total $=19$

put $x = y =1 \quad f(2)=(f(1))^{2}=3^{2}$

put $x=2, y=1 \quad f(3)=(f(1))^{3}=3^{3}$ :

Similarly $f(x)=3^{x}$

$\sum_{i=1}^{n} f(i)=363 \Rightarrow \sum_{i=1}^{n} 3^{i}=363$

$\left(3+3^{2}+\ldots+3^{n}\right)=363$

$\frac{3\left(3^{n}-1\right)}{2}=363$

$3^{n}-1=242 \Rightarrow 3^{n}=243$

$\Rightarrow n =5$

$f(\mathrm{x})$ is decreasing function

$f(x) \in\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]$

$\Rightarrow f(n)=n f(1)$

$f(n)=2 n$

$g(n)=\sum_{k=1}^{n-1} 2 n=2\left(\frac{(n-1) n}{2}\right)=n(n-1)$

$g(n)=20 \Rightarrow n(n-1)=20$

$n=5$

eg. $\quad a =2+\sqrt{3} \& b =2-\sqrt{3}$

$a^{2}+b^{2}=14 \in Q$

Let $\quad c =(1+2 \sqrt{3})$

$b ^{2}+ c ^{2}=20 \in Q$

But $\quad a^{2}+c^{2}=(2+\sqrt{3})^{2}+(1+2 \sqrt{3})^{2} \notin Q$

for $R _{2}$ Let $a ^{2}=1, b ^{2}=\sqrt{3} \& c ^{2}=2$

$a^{2}+b^{2} \notin Q \& b^{2}+c^{2} \notin Q$

But $a^{2}+c^{2} \in Q$

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

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

$\Rightarrow\left(a^{2}-a\right)+x(a+1)=\left(a^{2}+a\right) x+x^{2}(a-1)$

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

$\Rightarrow a=1$

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

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

$\sum_{x=1}^{\infty} f(x)=2 \quad$ where $x, y \in N$

$f(1)+f(2)+f(3)+\ldots . \infty=2 \ldots .(1)($ Given $)$

Now for $f(2)$ put $x=y=1$

$f(2)=f(1+1)=f(1) \cdot f(1)=(f(1))^{2}$

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

Now put these values in equation

$f(1)+(f(1))^{2}+\left[f(1)^{2}+\ldots \infty=2\right]$

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

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

Now $f(2)=\left(\frac{2}{3}\right)^{2}$

$f(4)=\left(\frac{2}{3}\right)^{4}$

then the value of $\frac{f(4)}{f(2)}=\frac{\left(\frac{2}{3}\right)^{4}}{\left(\frac{2}{3}\right)^{2}}=\frac{4}{9}$