Maths

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

$EA=\left[\begin{array}{cc}\hfill a\hfill & \hfill c\hfill \\ \hfill b\hfill & \hfill d\hfill \end{array}\right]\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill -a+c\hfill & \hfill 2a-c\hfill \\ \hfill -b+d\hfill & \hfill 2b-d\hfill \end{array}\right]$

For a = c For $\begin{array}{cc}\hfill -a+c=0\hfill & \hfill 2a-c=1\hfill \end{array}]\to a=1,c=1E=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$  

d = b + 1, d = 1, b = 0

$\begin{array}{cc}\hfill b+d=1\hfill & \hfill 2b-d=-1\hfill \end{array}]\to b=0,d=1R_1\to R_1\to R_2\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

For  $\begin{array}{cc}\hfill -a+c=1\hfill & \hfill 2a-c=1\hfill \end{array}]\to a=0,c=1$

$For\begin{array}{cc}\hfill -a+c=-1\hfill & \hfill 2a-c=2\hfill \end{array}]\to a=1,c=0$

$\begin{array}{cc}\hfill -b+d=-2\hfill & \hfill 2b-d=7\hfill \end{array}]\to b=5,d=3\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 3\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

R₂ -> 5R₁ + 3R₂

For $For\begin{array}{cc}\hfill -a+c=-1\hfill & \hfill 2a-c=2\hfill \end{array}]\to a=1,c=1$

$\begin{array}{cc}\hfill -b+d=-1\hfill & \hfill 2b-d=3\hfill \end{array}]\to b=2,d=1$

(A) ® R₁ ® R₁ + R₂

_{(B) ® R2 ® R2 + 2R1 $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$}

(C) ® R₂ ® 3R₂ + 5R₁

$\left|z-\frac{1}{z}\right|=2$

${\left|z\right|}_{max}=?$

$\left|z-\frac{1}{2}\right|\ge \left|\left|z\right|-\frac{1}{\left|z\right|}\right|$

$2\ge \left|r-\frac{1}{r}\right|$

0 $\le r^2+2r-1\&r^2-2r-1\le 0$

$r=\frac{-2\pm \sqrt[]{8}}{2}$ $r=\frac{2\pm \sqrt[]{8}}{2}$

$=-1\pm \sqrt[]{2}$ $=1\pm \sqrt[]{2}$

r $\ge \sqrt[]{2}-1\&0\le r\le 1+\sqrt[]{2}$

$\sqrt[]{2}-1\le r\le \sqrt[]{2}+1$

L₁ : 3x – 4y + 12 = 0

L₂ : 8x + 6y + 11 = 0

$\left(\alpha ,\beta \right)$ lies on that angle which contain origin

$\therefore$ Equation of angle bisector of that angle which contain origin is

$\frac{3x-4y+12}{5}=\frac{8x+6y+11}{10}$

$\Rightarrow 2x+14y-13=0$

$\left(\alpha ,\beta \right)$ lies on it

$\Rightarrow 2\alpha +14\beta -13=0$ …… (i)

$\Rightarrow 3\alpha -4\beta +7=0$ ……. (ii)

Solving (i) & (ii)

$\alpha =-\frac{23}{25}\&\beta =\frac{53}{50}$

$\therefore \alpha +\beta =\frac{7}{50}$

$\Rightarrow 100\left(\alpha +\beta \right)=14$

$l={\displaystyle {\int }_{-3}^{103}\left(\left[sin\left(\pi x\right)\right]+e^{\left[cos\left(2\pi x\right)\right]}\right)}dx$

$\left[sin\pi x\right]$ is periodic with period 2 and

$e^{\left[cos2\pi x\right]}$ is periodic with period 1.

So,

$I=52{\displaystyle {\int }_0^2\left(\left[sin\left(\pi x\right)\right]+e^{\left[cos2\pi x\right]}\right)}dx$

= $\frac{52}{e}$

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$\Rightarrow \frac{1}{2}=\frac{1}{3}+\frac{1}{5}-P\left(A\cap B\right)$

$P\left(\frac{A}{B\text{'}}\right)+P\left(\frac{B}{A\text{'}}\right)$

$=\frac{P\left(A\right)-P\left(A\cap B\right)}{1-P\left(B\right)}+\frac{P\left(B\right)-P\left(A\cap B\right)}{1-P\left(A\right)}=\frac{5}{8}$

$l=\underset{n\to x}{lim}\frac{1}{2^n}\left(\frac{1}{\sqrt[]{1-\frac{1}{2^n}}}+\frac{1}{\sqrt[]{1-\frac{2}{2^n}}}+\frac{1}{\sqrt[]{1-\frac{3}{2^n}}}+.....+\frac{1}{\sqrt[]{1-\frac{2^n-1}{2^n}}}\right)$

Let 2n = t and if n $\infty$ then t $\infty$

$l=\underset{n\to x}{lim}\frac{1}{t}\left({\displaystyle \sum _{r=1}^{t=1}\frac{1}{\sqrt[]{1+\frac{r}{t}}}}\right)$

$={\left[2x^{\frac{1}{2}}\right]}_0^1=2$

$\underset{x\to \frac{x}{4}}{lim}\frac{8\sqrt[]{2}-{\left(cosx+sinx\right)}^7}{\sqrt[]{2}-\sqrt[]{2}sin2x}\left(\frac{0}{0}form\right)$

$=\underset{x\to \frac{x}{4}}{lim}\frac{-7{\left(cosx+sinx\right)}^6\left(-sinx+cosx\right)}{-2\sqrt[]{2}cos2x}$ $\underset{x\to \frac{\pi }{4}}{Lt}\frac{{\left(cosx+sinx\right)}^5\left(co{s}^{2}x-si{n}^{2}x\right)}{2\sqrt[]{2}cos2x}=\frac{7{\left(\sqrt[]{2}\right)}^5}{2\sqrt[]{2}}=14$

${\displaystyle \sum _{n=1}^{21}\frac{3}{\left(4n-1\right)\left(4n+3\right)}}$

$=\frac{3}{4}{\displaystyle \sum _{n=1}^{21}\frac{\left(4n+3\right)-\left(4n-1\right)}{\left(4n-1\right)\left(4n+3\right)}}$

$=\frac{3}{4}\frac{4n+3-2}{3\left(4n+3\right)}=\frac{n}{4n+3}$

for n = 21

$S_{21}=\frac{21}{84+3}=\frac{21}{87}=\frac{7}{29}$

$11\equiv 2mod\left(9\right)$

${\left(11\right)}^{1011}\equiv 2^{1011}mod\left(9\right)$

Again 23 $\equiv$ 1 mod (9)

$\Rightarrow {\left(2^3\right)}^{337}\equiv {\left(-1\right)}^{337}mod\left(9\right)$

$\therefore {\left(11\right)}^{1011}+{\left(1011\right)}^{11}\equiv 8mod\left(9\right)$

$\therefore$ Remainder = 8

f : {1, 3, 5, 7, ….,99} {2, 4, 6, 8, …….100}

$f\left(3\right)\ge f\left(9\right)\ge f\left(15\right)......\ge f\left(99\right)$

$3+\left(n-1\right)6=99\Rightarrow n=17$

cases f (3) > f (9) > f (15) ……. > f (99)

$\therefore$ from the set {2, 4, 6, …., 100}

17 distinct numbers can be selected in 50C17 ways again remaining {1, 5, 7, 11, ….} can map in 33! ways

$\therefore$ total number of such required functions

${=}^{50}{C}_{17}*33!$

$=\frac{50!}{33!17!}*33!{=}^{50}{P}_{33}$