Maths

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

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

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

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

= 5m – 2

Remainder = 5 – 2 = 3

$x^2+y^2-x+2y=\frac{11}{4}$

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

or $\Delta PQRPR=QRsin2\ge \frac{1}{3}$

$=4.6sin\frac{\pi }{8}$

$as\Delta PQR=\frac{1}{2}PR*PQ$

$=4sin\frac{\pi }{4}=\frac{4}{\sqrt[]{2}}=2\sqrt[]{2}$

$y^2=2x-3$ ……. (i)

Equation of chord of contact

PQ : r = O

(y * 1) = (x + 0) – 3

y = x – 3              ……… (ii)

From (i) and (ii)

y = 1 or 3

$MPQ=\frac{4}{4}=1$

  $MQR=\frac{2}{6}=\frac{1}{3}$              

$M_{PQ}*M_{PR}=-1\Rightarrow PQ\perp PR$      

Orthocentre = P (2, -1)

If y = y(x), $x\in \left(0,\frac{\pi }{2}\right)$ $\frac{}{}$ be the solution curve of the different equation

$\frac{}{}$ $\therefore \frac{dy}{dx}+\left(8+4cot2x\right)y=\frac{2e^{-4x}}{si{n}^{2}2x}\left(2sinx+cos2x\right)$  which is a linear different equation

I.F = $e^{{\displaystyle \int \left(8+4cot2x\right)dx}}=e^{8x+2cos\left(sin2x\right)}=e^{8x}.si{n}^{2}2x$ ${}^{{\displaystyle }}{}^{}{}^{}{}^{}$

Given, $y\left(\frac{\pi }{4}\right)=e^{-\pi }\Rightarrow c=0$ $\frac{}{}{}^{}$

$\therefore y=\frac{e^{-4x}}{sin2x}$

$\therefore y\left(\frac{\pi }{6}\right)=\frac{e^{-4\frac{\pi }{6}}}{sin\left(2.\frac{\pi }{6}\right)}=\frac{2}{\sqrt[]{3}}{e}^{\frac{2\pi }{3}}$

  $\frac{dy}{dx}=\frac{x+y-2}{x-y}$  

Let x – 1 = X, y – 1 = Y

then DE: $\frac{dY}{dX}=\frac{X+Y}{X-Y}=\frac{1+\frac{Y}{X}}{1-\frac{Y}{X}}$  

Put y = vx

then  $\frac{dY}{dX}=V+X\frac{dV}{dX}$  

V + $X\frac{dV}{dX}=\frac{1+V}{1-V}$

$X\frac{dV}{dX}=\frac{1+V}{1-V}-V$

$=\frac{1+V^2}{1-V}$

$\frac{1-V}{1+V^2}dV=\frac{dX}{X}$

$\frac{V-1}{V^2+1}dV+\frac{dX}{X}=0$

$\frac{1}{2}\mathcal{l}n\left|V^2+1\right|-ta{n}^{-1}V+\mathcal{l}n\left|X\right|=c$

$\mathcal{l}n\left(\sqrt[]{1+{\left(\frac{Y-1}{X-}\right)}^2}\cdot \left|X-1\right|\right)ta{n}^{-1}\frac{Y-1}{X-1}=c$

$\left(2,1\right)\Rightarrow \mathcal{l}n\left(\sqrt[]{1+0}\cdot 1\right)-0=c$

c = 0  

$\mathcal{l}n\sqrt[]{{\left(X-1\right)}^2+{\left(Y-1\right)}^2}=ta{n}^{-1}\frac{Y-1}{X-1}$

point (k + 1, 2) $\mathcal{l}n\sqrt[]{k^2+1}=ta{n}^{-1}\frac{1}{k}$

$\Rightarrow \frac{1}{2}\mathcal{l}n\left(k^2+1\right)=ta{n}^{-1}\frac{1}{k}$              

Probability that chosen candidate is female $=\frac{40}{60}=\frac{2}{3}$

Kindly consider the following figure

I(x) = ${\displaystyle \int \frac{se{c}^{2}x-2022}{si{n}^{2022}x}dx}$  

$={\displaystyle \int si{n}^{-2022}x\cdot se{c}^{2}xdx-{\displaystyle \int 2022si{n}^{-2022}xdx}}$

$=si{n}^{-2022}x\cdot tanx-{\displaystyle \int \left(-2022\right)si{n}^{-2023}x\cdot cosx\cdot tanxdx-{\displaystyle \int 2022si{n}^{-2022}xdx}}$

$=tanx\cdot si{n}^{-2022}x+2022{\displaystyle \int si{n}^{-2022}-2022{\displaystyle \int si{n}^{-2022}xdx}}$

 I(x) = $tanx\cdot si{n}^{-2022}x+c$  

Given,   $I\left(\frac{\pi }{4}\right)=2^{1011}$

-> $2^{1011}=\frac{1}{{\left(\frac{1}{\sqrt[]{2}}\right)}^{2022}}+c\Rightarrow c=0$

$I\left(x\right)=\frac{tanx}{si{n}^{2022}x},I\left(\frac{\pi }{3}\right)=\frac{\sqrt[]{3}}{\left(\frac{\sqrt[]{3}}{2}\right)}=\frac{2^{2022}}{{\left(\sqrt[]{3}\right)}^{2021}}$  

S  $\equiv sint,C\equiv cost$ $$

Let orthocenter be (h, k)

Since it if an equilateral triangle hence orthocenter coincides with centroid.

$\therefore a+s+c=3h,b+s-c=3k$

$\frac{a}{3}=1,\frac{b}{3}=\frac{1}{3}\Rightarrow a=3,b=1$

$\therefore a^2-b^2-8$

${\displaystyle \sum _{r=1}^{20}\left(r^2+1\right)r!}$

t_r = (r² + 1)r!

= r²r! + r!

= r (r + 1 – 1)r! + r!

= r (r + 1)! – (r – 1)r!

= V_r – V_r-1

_{${\displaystyle \sum _{r=1}^{20}\left(V_r-V_{r-1}\right)}$}

= V₁ – V₀

_{$+V_2-V_1$}

_{$+V_3-V_2$}

_{$+V_{20}-V_{19}$}

_{$=V_{20}-V_0=20\left(21!\right)-0$}

_{$\left(22-2\right)\left(21!\right)=22!-2\left(21!\right)$}