Maths

$2021\equiv -2$ mod (7)

$\Rightarrow {\left(2021\right)}^{2023}\equiv {\left(-2\right)}^{2023}mod\left(7\right)$ …… (i)

Now,  ${\left(-2\right)}^3\equiv -1mod\left(7\right)$

$\Rightarrow {\left(-2\right)}^{2023}\equiv \left(-2\right)mod\left(7\right)\equiv 5mod\left(7\right)$ ……. (ii)

(i) & (ii)

$\Rightarrow {\left(2021\right)}^{2023}\equiv 5mod\left(7\right)$

$\therefore$ Remainder = 5

System of equation can be written as

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -8\hfill & \hfill 9\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 3\\ -1\end{array}\right)$

$\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 15\hfill & \hfill -24\hfill & \hfill 27\hfill \\ \hfill 6\hfill & \hfill 3\hfill & \hfill 3a\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}b\\ 9\\ -3\end{array}\right)$

$R_3-2R_1,R_2-5R_1$

for no solution

3a + 9 = 0 but $\frac{3}{2}-\frac{9b}{2}\ne 0$

$\Rightarrow a=-3\Rightarrow b\ne \frac{1}{3}$

$\left|adj\left(24A\right)\right|=\left|adj\left(3adj\left(2A\right)\right)\right|$

$\Rightarrow {\left|24A\right|}^2={\left|3adj\left(2A\right)\right|}^2$

$24^6{\left|A\right|}^2=3^6.{\left(2^3\right)}^4{\left|A\right|}^4$

${\left|A\right|}^2=\frac{24^6}{3^6.2^{12}}=\frac{2^{18}.3^6}{3^6.2^{12}}=2^6$

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

$f^3\left(x\right)=-\frac{x+1}{x-1}\Rightarrow f^4\left(x\right)=\frac{\frac{x-1}{x+1}+1}{\frac{x-1}{x+1}-1}=x$

$So,f^6\left(6\right)+f^7\left(7\right)=f^2\left(6\right)+f^3\left(7\right)$

= $-\frac{1}{6}-\frac{7+1}{7-1}=-\frac{9}{6}=-\frac{3}{2}$

Three consecutive integers belong to 98 sets and four consecutive integers belongs to 97 sets.

Þ Number of permutations of b₁ b₂ b₃ b₄ = number of permutations when b₁ b₂ b₃ are consecutive + number of permutations when b₂, b₃, b₄ are consecutive – b₁ b₂ b₃ b₄ are consecutive = 98 * 97 * 98 * 97 – 97 = 18915

2x + y – 3z = 4

${\pi }_2=\left|\begin{array}{ccc}\hfill x-2\hfill & \hfill y+3\hfill & \hfill z-2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -5\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right|=0$

${\pi }_2:-5x+5y+z+23=0$

now line lying in both the planes have DR.

$\frac{a}{1+15}=\frac{b}{15-2}=\frac{c}{10+5}$

$\frac{a}{16}=\frac{b}{13}=\frac{c}{15}$

So direction ratio's a : b : c = 16 : 13 : 15

$x+f\text{'}\left(x\right)=f\text{'}\left(0\right)$

$f\text{'}\left(0\right)=1-c^{2}fromequation\left(i\right)$

$x+f\text{'}\left(x\right)=1-c^2$

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

f(0) = 0

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

c = 3/2

$f"\left(x\right)=-1=2\left(c+1\right)$

$f\left(x\right)=\frac{-x^2}{2}+\left(\frac{-5}{4}\right)x$

now $\left(f\left(1\right)+f\left(2\right)+.......f\left(20\right)\right)$

$=\frac{-1}{2}\left(1^2+2^2+......+20^2\right)-\frac{5}{4}\left(1+2+....+20\right)$

$=\frac{20\cdot 21}{2}\cdot \frac{\left(-82-15\right)}{12}$

now $2\left|f\left(1\right)+f\left(2\right)+....+f\left(20\right)\right|$

$=\frac{20\cdot 21\cdot 97}{12}=3395$

Given 2a + 2b = $4\left(2\sqrt[]{2}+\sqrt[]{14}\right)$

$=4\sqrt[]{2}\left(2+\sqrt[]{7}\right)$

$b^2=a^2\left(\frac{11}{4}-1\right)$

4b2 = 7a2

b = $\frac{\sqrt[]{7}}{2}a$