Sets

Equation of plane ${\perp }^r$ to line $\frac{x-3}{2}=\frac{x-1}{1}=\frac{x-2}{1}$ and passes through the point

(2, 3, -1) is 2(x – 2) + 1(y – 3) + 1 (z + 1) = 0

=> 2x + y + z – 6 = 0         .(i)

Hence point (1, 2, 2) satisfies equation (i)

$z=\frac{1}{1-cos\theta +2isin\theta }=\frac{1-cos\theta -2isin\theta }{{\left(1-cos\theta \right)}^2-{\left(2isin\theta \right)}^2}=\frac{2si{n}^2\frac{\theta }{2}-2isin\theta }{{\left(1-cos\theta \right)}^2+4si{n}^2\theta }$      

$\therefore Re\left(z\right)=\frac{2si{n}^2\frac{\theta }{2}}{4si{n}^2\frac{\theta }{2}\left(si{n}^2\frac{\theta }{2}+4co{s}^2\frac{\theta }{2}\right)}=\frac{1}{2\left(1+3co{s}^2\frac{\theta }{2}\right)}=\frac{1}{5}\Rightarrow co{s}^2\frac{\theta }{2}=\frac{1}{2}$   

$\Rightarrow \theta =\frac{\pi }{2}\therefore {\displaystyle {\int }_0^{\frac{\pi }{2}}sinxdx=1}$

$2cosx\left(4sin\left(\frac{\pi }{4}+x\right)sin\left(\frac{\pi }{4}-x\right)-1\right)=1$  

->2cos x (2 cos 2x – 1) = 1

$\Rightarrow cos3x=\frac{1}{2},For0\le x\le \pi ,x=\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9}$                

S = {1, 2, 3, 4, 5, 6, 9}

Elements of type 3n -> 3, 6, 9

Type 3n + 1 ->1, 4

3n + 2 -> 2, 5

Number of subset of S containing one element which are not divisible by  $3{=}^2{C}_1{+}^2{C}_1=4$ number of subset of S containing two numbers whose sum is not divisible $3{=}^3{C}_1{*}^2{C}_1{+}^3{C}_1{*}^2{C}_1{+}^2{C}_2{+}^2{C}_2=14$ by

Number of subset of S containing 3 elements whose sum is not divisible by

Number of subset containing 4 elements whose sum is not divisible by 3

Number of subset of S containing 6 elements = 4

Hence total subset = 80

  $\left|x-2\right|>1givesx-2<-1orx-2>1$ $$

i.e. x < 1 or x > 3 . (i) represent set A

$\sqrt[]{x^2-3}>1gives{x}^2-3>1or{x}^2-4>0$

x < 2 or x > 2 . (ii) represent set B

$\left|x-4\right|\ge 2givesx-4\le -2orx-4\ge 2$

$$  $x\le 2orx\ge 6$ . (iii)respect set C

so number of subset = 28 = 256

$A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$ $B^2=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$                  

$B=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ $=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]=B$

$A^2=\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$ $\Rightarrow .....B^3=B^2=B\Rightarrow B^n=B$

$=\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]=A$

$\Rightarrow ...=A^3=A^2=A\Rightarrow A^n=A$

$n{A}^n+m{B}^m=l\Rightarrow nA+mB=l$

$\left[\begin{array}{cc}\hfill 2n\hfill & \hfill -2n\hfill \\ \hfill n\hfill & \hfill -n\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill -m\hfill & \hfill 2m\hfill \\ \hfill -m\hfill & \hfill 2m\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow 2n-m=1$  n – m = 0            ->n = m = 1

-2n + 2m = 0     -n + 2m = 1

Kindly go through the solution

 $\begin{array}{l}-2<x+1<2\\ -3<x<1\end{array}|\begin{array}{cc}\hfill x-1\le -2orx-1\ge 2\hfill & \hfill x\le -1orx\ge 3\hfill \end{array}$

2cos  $\left(\frac{x^2+x}{6}\right)=4^x+4^{-x}$  

$-2\le LHS\le 2LHS=2\&RHS=2\Rightarrow x=0only\to thenLHS=2also$                

RHS $\ge$ 2

$A=\left\{z\in c:1\le \left|z-\left(1+i\right)\right|\le 2\right\}$

$\left|z-\left(1+i\right)\right|\ge 1$

and $$ $\left|z-\left(1+i\right)\right|\le 2$

and also $\left|z-\left(1-i\right)\right|=1$  $$

hence B has infinite set.