Complex Numbers and Quadratic Equations

From properties of nth root of unity

$1^{2021}+{\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}=0$

$\Rightarrow {\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}=-1$

Use binomial theorem

2023 = 7 × 289

$2021^{2022}+2022^{2021}={\left(2022-2\right)}^{2022}+{\left(2023-1\right)}^{2021}$

$=7P_1+2^{2022}+7P_2-1$

$=7\left(P_1+P_2\right)+1+7P_3-1$                          

Let e^x = t then equation reduces to

$t^2-11t-\frac{45}{t}+\frac{81}{2}=0$                

$\Rightarrow 2t^3-22t^2+81t-45=0$                     ……. (i)

If roots of

$e^{2x}-11e^x-45e^{-x}+\frac{81}{2}=0$            

$\Rightarrow {\alpha }_1+{\alpha }_2+{\alpha }_3=\mathcal{l}n45\Rightarrow p=45$             &nb

  $\overline{z}=i{z}^2$

Let z = x + iy

x – iy = I (x² – y² + 2xiy)

Case-I

x = 0

-y² = -y

y = 0, 1

Case – II

$y=-\frac{1}{2}$

$\Rightarrow x^2-\frac{1}{4}=\frac{1}{2}\Rightarrow x=\pm \frac{\sqrt[]{3}}{2}$

Area of polygon

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \frac{\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \\ \hfill \frac{-\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left|-\sqrt[]{3}-\frac{\sqrt[]{3}}{2}\right|=\frac{3\sqrt[]{3}}{4}$  

$7^{2022}+3^{2022}$

$={\left(49\right)}^{1011}+{\left(9\right)}^{1011}$

$={\left(50-1\right)}^{1011}+{\left(10-1\right)}^{1011}$

$=5\lambda -1+5k-1$

= 5m – 2

Remainder = 5 – 2 = 3

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

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

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

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

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

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

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

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

Put $1+x^2=t^2\Rightarrow 2xdx=2tdt$  

$\therefore {\displaystyle \underset{1}{\overset{2}{\int }}\frac{15\left(t^2-1\right)tdt}{\sqrt[]{t^2+t^3}}dt}$ Put (1 + t) = u²

$30{\displaystyle \underset{\sqrt[]{2}}{\overset{\sqrt[]{3}}{\int }}\left(u^4-2u^2\right)du}$ dt = 2u du

$=-6\sqrt[]{3}+16\sqrt[]{2}=\alpha \sqrt[]{2}+\beta \sqrt[]{3}$

$\begin{array}{l}\alpha =16,\beta =-6\\ \therefore \alpha +\beta =10\end{array}$      

$\frac{2}{?4}=\frac{?1}{2}$

y – 4 = 2 (x – 3)

y = 2x – 2

x² + (2x – 2)² = 25

$5x^2?8x?21=0$

$z\left(\frac{?7}{5},\frac{?24}{5}\right)$

Let the number of chocolates given to C₁, C₂, C₃ & C₄ be a, b, c, d respectively.

Given 4   $\le b\le 7$

$2\le c\le 6$

Now using these the maximum number of chocolates that can be given to C₁ or C₄ is 24 (where b & c are given 2 & 4 chocolates).

$\therefore 0\le a\le 24$

$0\le d\le 24$

& a + b + c + d = 30

So, total possible solution to the above equ