Relations and Functions

 $f\left(x\right)=[\begin{array}{ccc}\hfill 2n\hfill & \hfill \hfill & \hfill n=2,4,6....\hfill \\ \hfill n-1\hfill & \hfill \hfill & \hfill n=3,7,11,15....\hfill \\ \hfill \frac{n+1}{2}\hfill & \hfill \hfill & \hfill n=1,5,9,13....\hfill \end{array}$

for n = 2, 4, 6 ……

f (n) = 4, 8, 12, ….4 (n) form

for n = 3, 7, 11, 15, ….

f (n) = 1, 3, 5, 7, …. (4n + 1) or, (4n + 3) from

$\therefore$ f is one and onto.

 $\frac{1}{\left(20-a\right)\left(40-a\right)}+\frac{1}{\left(40-a\right)\left(60-a\right)}+\frac{1}{\left(60-a\right)\left(80-a\right)}+.....+\frac{1}{\left(180-a\right)\left(200-a\right)}=\frac{1}{256}$

LHS = $\frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{40-a}\right)+\frac{1}{20}\left(\frac{1}{40-a}-\frac{1}{60-a}\right)+...+\frac{1}{20}\left(\frac{1}{180-a}-\frac{1}{200-a}\right)$

$=\frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{200-a}\right)=\frac{1}{20}.\frac{180}{\left(20-a\right)\left(200-a\right)}$

$\Rightarrow a^2-220a+4000-2304=0\Rightarrow a^{2}220a+1696=0$

$a=\frac{220\pm 204}{8}=212^2$

R = { (a, b): b = pq, where p, q $\ge 3$ are prime}

$\Rightarrow 60*11=660$

p, q $\in$  {3, 5, 7, 11, 13, 17, 19, 23, 29, 3, 37, 41, 43, 47, 53, 59} total 16

p, q $\in$  {3, 5, 7, 11, 13, 17, 19}

{3, 5, 7, 11, 13, 17, 19}

7 + 3 + 1 = 11

. $?f\left(n\right)=\{\begin{array}{c}\hfill 2n,n=1,2,3,4,5\hfill \end{array}$

$\begin{array}{c}\hfill \mathrm{ \; \; \; \; \; \; \; \; \; \; 2}n-11,n=6,7,8,9,10\hfill \end{array}$

$\therefore f\left(1\right)=2,f\left(2\right)=4,....,f\left(5\right)=10$

And f(6) = 1, f(7) = 3, f(8) = 5, …., f(10) = 9

Now,

$f\left(g\left(n\right)\right)=\{\begin{array}{cc}\hfill n+1\hfill & \hfill ifnisodd\hfill \\ \hfill n-1,\hfill & \hfill ifniseven\hfill \end{array}$

$\therefore f\left(g\left(10\right)\right)=9\Rightarrow g\left(1\right)=1$

$\therefore g\left(10\right)\left(g\left(1\right)+g\left(2\right)+g\left(3\right)+g\left(4\right)+g\left(5\right)\right)=190$

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\text{'}+\text{'}isabinaryoperationonthesetNbutithasnoidentityelement.\\ Hence,thestatementis\text{'}False\text{'}.\end{array}$

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Onlybijectivefunctionsareinvertible.\\ Hence,thestatementis\text{'}False\text{'}.\end{array}$

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Letf(x)=2x,g(x)=x-1andh(x)=2x+3\\ fo\{goh(x)\}=fo\{g(2x+3)\}\\ =f(2x+3-1)=f(2x+2)=2(2x+2)=4x+4\\ and(fog)oh(x)=(fog)\{h(x)\}\\ =fog(2x+3)\\ =f(2x+3-1)=f(2x+2)=2(2x+2)=4x+4\\ \therefore fo\{goh(x)\}=(fog)oh(x)=4x+4\\ Hence,thestatementisTrue.\end{array}$

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

This is a True or False Type Question as classified in NCERT Exemplar

Sol:

This is a True or False Type Question as classified in NCERT Exemplar

Sol: