Maths

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

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

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

$\Delta =0$

$\left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 5\hfill & \hfill \alpha \hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right|=0$

15 -2α +α- 6 – 1 = 0

α = 8

For = 8, equations are

x + y + 3 = 6

2x + 5y + 8z =β

x + 2y + 3z = 14

$\left(2,5,8\right)=\mathcal{l}\left(1,1,1\right)+m\left(1,2,3\right)$

$\begin{array}{cc}\hfill 2=\mathcal{l}+m\hfill & \hfill 5=\mathcal{l}+2m\hfill \end{array}]\to 3=m,\mathcal{l}=-1$

8 = $\mathcal{l}+3m$

$\beta =6\mathcal{l}+14m$

=- 6 + 42 = 36

α + β = 8 + 36 = 44

$A=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$ $EA=\left[\begin{array}{cc}\hfill a\hfill & \hfill c\hfill \\ \hfill b\hfill & \hfill d\hfill \end{array}\right]\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill -a+c\hfill & \hfill 2a-c\hfill \\ \hfill -b+d\hfill & \hfill 2b-d\hfill \end{array}\right]$

For a = c For $\begin{array}{cc}\hfill -a+c=0\hfill & \hfill 2a-c=1\hfill \end{array}]\to a=1,c=1E=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

d = b + 1, d = 1, b = 0

$\begin{array}{cc}\hfill b+d=1\hfill & \hfill 2b-d=-1\hfill \end{array}]\to b=0,d=1R_1\to R_1\to R_2\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

For $\begin{array}{cc}\hfill -a+c=1\hfill & \hfill 2a-c=1\hfill \end{array}]\to a=0,c=1$

$For\begin{array}{cc}\hfill -a+c=-1\hfill & \hfill 2a-c=2\hfill \end{array}]\to a=1,c=0$

$\begin{array}{cc}\hfill -b+d=-2\hfill & \hfill 2b-d=7\hfill \end{array}]\to b=5,d=3\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 3\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

R₂ → 5R₁ + 3R₂

For $For\begin{array}{cc}\hfill -a+c=-1\hfill & \hfill 2a-c=2\hfill \end{array}]\to a=1,c=1$

$\begin{array}{cc}\hfill -b+d=-1\hfill & \hfill 2b-d=3\hfill \end{array}]\to b=2,d=1$

(A) R₁ → R₁ + R₂

(B) R₂ → R₂ + 2R₁ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

(C) R₂ → 3R₂ + 5R₁

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

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

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

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

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

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

${\mathcal{l}}_1:$

$\overrightarrow{r}=\frac{1}{8}\widehat{i}+\lambda \left(-\frac{1}{8},\frac{1}{4\sqrt[]{2}},0\right)$

$\overrightarrow{r}=-\frac{1}{8}\widehat{i}+\mu \left(\frac{1}{8},0,-\frac{-1}{6\sqrt[]{3}}\right)$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \frac{-1}{8}\hfill & \hfill \frac{1}{4\sqrt[]{2}}\hfill & \hfill 0\hfill \\ \hfill \frac{1}{8}\hfill & \hfill 0\hfill & \hfill \frac{-1}{6\sqrt[]{3}}\hfill \end{array}\right|$

$=\left(-\frac{1}{24\sqrt[]{6}},\frac{-1}{48\sqrt[]{3}},-\frac{1}{32\sqrt[]{2}}\right)$

$d=projectionofA\overrightarrow{C}on\overrightarrow{n}=\left(-\frac{2}{8},0,0\right).\frac{\left(4,2\sqrt[]{2},3\sqrt[]{3}\right)}{\sqrt[]{16}+8+27}=\frac{-1}{\sqrt[]{16+8+27}}$

d²⁼ $\frac{1}{51}$                                       

$e=\frac{5}{4}$

$b^2=a^2\left(e^2-1\right)$

$\Rightarrow b=\frac{3a}{4}$

$p\left(\frac{8}{\sqrt[]{5}},\frac{12}{5}\right)$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\Rightarrow \frac{64}{5a^2}-\frac{144}{25b^2}=1$

$\Rightarrow \frac{\sqrt[]{5}x}{3}+\frac{5}{8}y=\frac{8}{3}+\frac{3}{2}=\frac{25}{6}$

$f\left(x\right)=\left|\left(x-1\right)\left(x+1\right)\left(x-3\right)\right|+x-3$  

For    $x\in \left[-1,1\right]\cup \left[3,\infty \right)$

y = x² (x – 3)

  $\frac{dy}{dx}=2x\left(x-3\right)+x^2=3x^2-6x=3x\left(x-2\right)$

Number of max = 2

Number of min = 1

$a\in \left\{1,2,3,....,9\right\}$  

  $b\in \left\{0,1,2,...,9\right\}$              

9 * 9 = 81

81 + 81 + 81 = 243