Maths

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

_{$+V_2-V_1$}

_{$+V_3-V_2$}

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

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

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

$\left|\overrightarrow{a}\right|=9\&\left(x\overrightarrow{a}+y\overrightarrow{b}\right).\left(6y\overrightarrow{a}-18x\overrightarrow{b}\right)=0$

$\Rightarrow 6xy{\left|\overrightarrow{a}\right|}^2-18x^2\left(\overrightarrow{a}.\overrightarrow{b}\right)+6y^2\left(\overrightarrow{a}.\overrightarrow{b}\right)-18xy{\left|\overrightarrow{b}\right|}^2=0$

This should hold $\forall x,x,y\in R*R$

$\therefore {\left|\overrightarrow{a}\right|}^2=3{\left|\overrightarrow{b}\right|}^2\&\left(\overrightarrow{a}.\overrightarrow{b}\right)=0$

$\therefore \left|\overrightarrow{a}*\overrightarrow{b}\right|=\frac{{\left|\overrightarrow{a}\right|}^2}{\sqrt[]{3}}=\frac{81}{\sqrt[]{3}}=27\sqrt[]{3}$

(A) $\left(\wedge ,\vee \right):\left(P\vee Q\right)\wedge \left(P\wedge \sim Q\right)$ $$ not a tautology

(B) $\left(\vee ,\vee \right):\left(P\vee Q\right)\vee \left(P\vee \sim Q\right)=\left(P\vee T\right)$ $$ in a tautology

$$  $\left(p\vee q\right)\vee \left(p\vee \sim q\right)=$ T using venn diagrams

a₀ = 0, a₁ = 0

a_n+2 = 3a_n+1 – 2a_n + 1

a₂₅ a₂₃ – 2a₂₅a₂₂ – a₂₃a₂₄ + 4a₂₂a₂₄ =?

a₂ = 3a₁ – 2a₀ + 1

a₃ = 3a₂ – 2a₁ + 1

a₄ = 3a₃ – 2a₂ + 1

a₅ = 3a₄ – 2a₃ + 1

a_n+2  = 3a_n+1 – 2a_n + 1

$\left(+\right)\Rightarrow \left(a_2+a_3+a_4+....+a_{n+1}+a_{n+2}\right)$

-> a_n+2 = 2 (a₂ + a₃ + …. + a_n + a_n+1) –2 (a₁ + a₂ + ….+ a_n) + n + 1

a_n+2 = 2a_n+1 + n + 1

a₂₅ a₂₃ -2a₂₅ a₂₂ -a₂₃ a₂₄ + 4a₂₂ a₂₄

= a₂₅ (a₂₃ – 2a₂₂) -2a₂₄ (a₂₃ – 2a₂₂)

As a_n+2 = 2a_n+1 + n + 1

-> a_n+2 – 2a_n+1 = n + 1

-> a_n+1 -2a_n = n

-> 24 * 22 = 528