Maths

A = $\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)$

$A^2=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill a^2+bc\hfill & \hfill ab+bd\hfill \\ \hfill ac+dc\hfill & \hfill ac+d^2\hfill \end{array}\right)$

a2 + bc = bc + d2 = 1 ………. (i)

and b (a + d) = c (a + d) = 0 ……… (ii)

Case 1

b = c = 0

then possible ordered pair of

(a, d) $\equiv$  (1, 1) (-1, -1) (-1, 1) (1, -1) total 4 possible case

Case 2

a = -d

then (a, d) $\equiv$  (-1, 1) (1, -1)

then bc = 0

now if b = 0

then possible choice for {-1, 0, 1, 2, …….10} = 12

Similarly if c = 0 then possible choice for $b\in \left\{-1,0,1,2,......10\right\}$ is = 12

but (0, 0) counted twice

$\therefore$ bc = 0 in (12 + 12 – 1) = 23 ways

$\therefore$ total number of ways = 2 * 23 = 46

$\therefore$ total number of required matrices = 46 + 4 = 50

Factors of 36 = 2².3².1

Five-digit combinations can be

(1, 2, 3, 3), (1, 4, 3, 1), (1, 9, 2, 1), (1, 4, 9, 11), (1, 2, 3, 6, 1), (1, 6, 1, 1)

i.e., total numbers $\frac{5!5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{3!}+\frac{5!}{2!}+\frac{5!}{3!2!}=\left(30*3\right)+20+60+10=180.$

$d_1=\frac{199-100}{2}\cancel{\in }l$

$d_2=\frac{199-100}{3}=33$

$d_3=\frac{199-100}{4}\cancel{\in }l$

$d_n=\frac{199-100}{i+1}\in l$

$\Rightarrow di=33+11or9$

$\therefore$ sum of common differences = 33 + 11 + 9 = 53

Let and are the roots of $\left(p^2+q^2\right)x^2-2q\left(p+r\right)x+q^2+r^2=0$

$\therefore \alpha +\beta >0and\alpha \beta >0$ Also, it has a common root with x² + 2x – 8 = 0

$\therefore$ The common root between above two equations is 4.

$\Rightarrow 16\left(p^2+q^2\right)-8q\left(p+r\right)+q^2+r^2=0\Rightarrow \left(16p^2-8pq+q^2\right)+\left(16q^2-8qr+r^2\right)=0$

$\Rightarrow {\left(4p-q\right)}^2+{\left(4q-r\right)}^2=0\Rightarrow q=4pandr=16p\therefore \frac{q^2+r^2}{p^2}=\frac{16p^2+256p^2}{p^2}=272$

 $\frac{dy}{dx}-y=2-e^{-x}$

$I.F.=e^{-{\displaystyle \int dx}}=e^{-x}$

$\therefore$ soln

$y{e}^{-x}={\displaystyle \int \left(2e^{-x}-e^{-2x}\right)dx}$

$y=-2+\frac{e^{-x}}{2}+C{e}^x$

as for x $\infty$ y finite c = 0

$\therefore y=\frac{e^{-x}}{2}-2$

$\begin{array}{l}\Rightarrow x+2y=-3\\ \Rightarrow a=-3b=\frac{-3}{2}\end{array}$

$\therefore a=4b=-3+6=3$

(D)

$\sim \left(p\iff \sim q\right)\wedge q=\left(p\iff \sim q\right)\wedge qis:$

$\therefore \left(\sim \left(P\iff \sim Q\right)\right)\wedge$ q is equivalent to $\left(p\Rightarrow q\right)\wedge$ p

$L_1:\mathcal{l}x-y+3\left(1-z\right)=1,x+2y-z=2$

plane containing the line P : 3x – 8y + 7z = 4

If $\overrightarrow{n}$ be vector parallel to L.

then $\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \mathcal{l}\hfill & \hfill -1\hfill & \hfill 3\left(1-\mathcal{l}\right)\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(6\mathcal{l}-5\right)\widehat{i}+\left(3-2\mathcal{l}\right)\widehat{j}+\left(2\mathcal{l}+1\right)\widehat{k}$ as P containing the line

$\therefore 3\left(6\mathcal{l}-5\right)-8\left(3-2\mathcal{l}\right)+7\left(2\mathcal{l}+1\right)=0$

$\Rightarrow \mathcal{l}=\frac{2}{3}$

If be the acute angle between line L & Y axis then cos = $\frac{5/3}{\sqrt[]{1+\frac{25}{9}+\frac{49}{9}}}=\frac{5}{\sqrt[]{83}}$

$\therefore 415co{s}^2\theta =125$

 $tan\left(2ta{n}^{-1}\frac{1}{5}+se{c}^{-1}\frac{\sqrt[]{5}}{2}+2ta{n}^{-1}\frac{1}{8}\right)tan\left(2ta{n}^{-1}\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}*\frac{1}{8}}+se{c}^{-1}\frac{\sqrt[]{5}}{2}\right)$

$=tan\left(ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{1}{2}\right)=tan\left(ta{n}^{-1}\frac{\frac{3}{4}+\frac{1}{2}}{1-\frac{3}{8}}\right)$

$=tan\left(ta{n}^{-1}\frac{\frac{5}{4}}{\frac{5}{8}}\right)=2$

 $\overline{x}=\frac{\sum xi}{40}=30\Rightarrow \sum xi=1200$ ………. (i)

${\alpha }^2=\frac{1}{40}\sum x{i}_i^2-{\left(30\right)}^2=25$

$\Rightarrow \sum x_i^2=37000$

after omitting two wrong observations

$\sum y_i^2=37000-144-100=36756$

$\therefore 38a^2=36756-36158=238$

 $S=\left\{\theta \in \left[0,2\pi \right]:8^{2si{n}^{2}x}+8^{2co{s}^{2}x}=16\right\}$

Now apply AM $\ge GM$ for $\frac{8^{2si{n}^{2}x}+8^{2co{s}^{2}x}}{2}\le {\left(8^{2si{n}^{2}x+2co{s}^{2}x}\right)}^{\frac{1}{2}}\Rightarrow 8^{2si{n}^{2}x}=8^{2co{s}^{2}x}$

$si{n}^2\theta =co{s}^2\theta$ $\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

$=4+\left[cosec\left(\frac{\pi }{2}+\pi \right)+cosec\left(\frac{\pi }{2}+3\pi \right)+cosec\left(\frac{\pi }{2}+5\pi \right)+cosec\left(\frac{\pi }{2}+7\pi \right)\right]$

$=4-2\left(4\right)=-4$