Maths

Circumcentre (D) $\equiv \left(5,\frac{\alpha }{4}\right)$

${\left(5-\alpha \right)}^2+{\left(\frac{\alpha }{4}+2\right)}^2={\left(5-\alpha \right)}^2+{\left(\frac{\alpha }{4}-6\right)}^2...............\left(i\right)$

${\left(5-\frac{\alpha }{4}\right)}^2+{\left(\frac{\alpha }{4}+2\right)}^2....................\left(ii\right)$

(i) $\frac{\alpha }{4}+2=\pm \left(\frac{\alpha }{4}-6\right)$

(ii) 9 + 16 = 9 + 16

$\oplus \to x$

$\left(-\right)\to \frac{\alpha }{2}=4\Rightarrow \alpha =8$

ar $\left(ABC\right)=24$

2S = 24

R = 5, r = $\frac{\Delta }{s}=\frac{24}{12}=2$

2cos $\left(\frac{x^2+x}{6}\right)=4^x+4^{-x}$

$-2\le LHS\le 2LHS=2\&RHS=2\Rightarrow x=0only\to thenLHS=2also$

RHS $\ge$ 2

$m_1{m}_2=-1,$ for square a,b,c,d let

$A\left(10\left(cos\alpha -sin\alpha \right),10\left(sin\alpha +cos\alpha \right)\right)$

Diagonal : (cosα - sinα)x + (sinα + cosα)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{\left|10{\left(cos\alpha -sin\alpha \right)}^2+10{\left(sin\alpha +cos\alpha \right)}^2-10\right|}{\sqrt[]{2}}=\frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}}=\frac{a}{\sqrt[]{2}}\Rightarrow a=10$

Also, a² + 11a + 3 $\left(m_1^2+m_2^2\right)=220$

210 + 3 $\left(c{m}_1^2+m_2^2\right)=220$

$m_1^2+m_2^2=\frac{10}{3}$

Also, m₁ m₂ = -1

m² + $\frac{1}{m^2}=\frac{10}{3}$

or $-\sqrt[]{3},\frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3},\frac{-1}{\sqrt[]{3}}$

$m^4-\frac{10}{3}{m}^2+1=0\Rightarrow m^2=\frac{\frac{10}{3}\pm \sqrt[]{\frac{100}{9}-4}}{2}-\frac{\frac{10}{3}\pm \frac{8}{3}}{2}=3,\frac{1}{3}$

m = $\pm \sqrt[]{3},\pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$\left(sin\alpha +cos\alpha \right)x-\left(cos\alpha -sin\alpha \right)y$

=10 cos2α - 10cos2α = 0

Slope of AC = $\frac{sin\alpha +cos\alpha }{cos\alpha -sin\alpha }=\frac{tan\alpha +1}{1-tan\alpha }=tan\left(\alpha +\frac{\pi }{4}\right)\alpha =30°$

FIGURE

? = $72\left(\frac{1}{16}+\frac{9}{16}\right)+100-30+13=\frac{720}{16}+83=128$

 $\frac{dy}{dx}+\left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y=\frac{x+3}{x+1},x>-1$

IF = $e^{{\displaystyle \int pdx}}=\frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}$

${\displaystyle \int Pdx={\displaystyle \int \frac{2x^2+11x+13}{x^3+6x^2+11x+6}dn={\displaystyle \int \left(\frac{2}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}\right)dx}}}$

$=\mathcal{l}n\left({\left(x+1\right)}^2\cdot \left(x+2\right)/\left(x+3\right)\right)$

$\frac{2x^2+11x+13}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}$

$2x^2+11x+13=A\left(x+2\right)\left(x+3\right)+B\left(x+1\right)\left(x+3\right)+C\left(x+1\right)\left(x+2\right)$

x = -1

⇒ 4 = 2A ⇒ A = 2

x = -2

⇒ -1 = -B Þ B = 1

x₂ – 3 ⇒ -2 = 2c

c = -1

$y\cdot \frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}={\displaystyle \int \frac{x+3}{x+1}\cdot \frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}dx}$

= ${\displaystyle \int \left(x+1\right)\left(x+2\right)dx}$

= $\frac{x^3}{3}+\frac{3x^2}{2}+2x+c$

$\left(0,1\right)\Rightarrow 1\cdot \frac{2}{3}=c$

x = 1 $y\cdot \left(3\right)=\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}=\frac{3}{2}+3=\frac{9}{2}$

y = $\frac{3}{2}$

$\frac{dy}{dx}=\frac{x+y-2}{x-y}$

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 + $X\frac{dV}{dX}=\frac{1+V}{1-V}$

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

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

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

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

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

$\mathcal{l}n\sqrt[]{V^2+1}\cdot X-ta{n}^{-1}V=c$

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

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

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

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

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

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

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

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

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

$+V_2-V_1$

$+V_3-V_2$

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

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

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

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

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

y = 2x $\left|3x^2-5x+2\right|,0\le x\le 1$

$=\{\begin{array}{cc}\hfill -3x^2+7x-2,\hfill & \hfill 3x^2-3x+2,\hfill \end{array}\begin{array}{cc}\hfill 0\le x\le \frac{2}{3}\hfill & \hfill \frac{2}{3}<x\le 1\hfill \end{array}$

3x² – 5x + 2 = 0

$x=\frac{+5\pm \sqrt[]{25-24}}{6}$

= $\frac{5+1}{6}=1,\frac{2}{3}$

3x² – 7x + 3 = 0

x = $\frac{7\pm \sqrt[]{49-39}}{6}=\frac{7\pm \sqrt[]{13}}{6}$

$-3x^2+7x-2$

$\frac{-7\pm \sqrt[]{49-24}}{-6}=\frac{7+5}{6}=2,\frac{1}{3}$

3x² + 7x – 2 = 1

3x² – 7x + 1 = 0

x = $\frac{7\pm \sqrt[]{49-12}}{6}=\frac{7\pm \sqrt[]{37}}{6}$

I = ${\displaystyle \underset{0}{\overset{6}{\int }}\left(\begin{array}{cc}\hfill \left(-2\right)\hfill & \hfill +1\hfill \end{array}\right)dx+{\displaystyle \underset{\frac{7-\sqrt[]{37}}{6}}{\overset{\frac{1}{3}}{\int }}\left(\begin{array}{cc}\hfill \left(-1\right)\hfill & \hfill +1\hfill \end{array}\right)dx}+{\displaystyle \underset{-\frac{1}{3}}{\overset{\frac{7-\sqrt[]{13}}{6}}{\int }}\left(\begin{array}{cc}\hfill 0\hfill & \hfill +1\hfill \end{array}\right)dx+{\displaystyle \underset{\frac{7-\sqrt[]{13}}{6}}{\overset{\frac{2}{3}}{\int }}\left(1+1\right)dx}}}+{\displaystyle \underset{\frac{2}{3}}{\overset{1}{\int }}\left(1+1\right)dx}$

$=\frac{\sqrt[]{37}+\sqrt[]{13}-4}{6}$