Maths

$C_2=a_2+b_2=\left(a_1-3\right)+2b_1=12$

$a_1+2b_1=15.............\left(i\right)$

$a_1+4b_1=19............\left(ii\right)$              

Solving (i) & (ii), b₁ = 2, a₁ = 11

$=5\left[22+9\left(-3\right)\right]+2\left(\frac{2^{10}-1}{2-1}\right)$

$=2^{11}-27=2^6*2^5-27=2021$               

${\overrightarrow{n}}_1=6\widehat{i}+7\widehat{j}+8\widehat{k}{\overrightarrow{n}}_2=3\widehat{i}+5\widehat{j}+7\widehat{k}$

${\overrightarrow{n}}_1*{\overrightarrow{n}}_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 6\hfill & \hfill 7\hfill & \hfill 8\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill 7\hfill \end{array}\right|=9\widehat{i}-18\widehat{j}+9k$              

$\therefore$ the normal to required plane is $\widehat{i}-2\widehat{j}+\widehat{k}$

Equation of plane 1(x + 1) – 2(y – 1) + (z – 3) = 0

x – 2y + z = 0

P (7, -2, 13)

$\therefore PQ=\left|\frac{7+4+13}{\sqrt[]{1+4+1}}\right|=\frac{24}{\sqrt[]{6}}$

${\left(PQ\right)}^2=\frac{24*24}{6}=96$

$\frac{3+7+x+y}{4}=5$              

$\Rightarrow x+y=10.........\left(i\right)$

$\frac{1}{4}\left(9+49+x^2+y^2\right)-25=10$              

x² + y² = 82 .(ii)

$\therefore \overline{x}=\frac{\left(3+2x\right)+\left(7+2y\right)+\left(x+y\right)+\left(x-y\right)}{4}$              

$=\frac{10+4x+2y}{4}$              

Solving (i) & (ii), x = 9, y = 1

$\therefore \overline{x}=\frac{48}{4}=12$          

$\frac{{\left(2i\right)}^n}{{\left(1-i\right)}^{n-2}}$

$=\frac{{\left(2i\right)}^n{\left(1+i\right)}^{n-2}}{{\left(1+1\right)}^{n-2}}$            

$=-4{\left(-1+i\right)}^{n-2}$              

  $now{\left(-1+i\right)}^2=-2i$

$\therefore -4{\left(-1+i\right)}^{n-2}\in \left(+\right)ve$  integer for n – 2 = 4

->n = 6

$\overrightarrow{a}=\widehat{i}+2\widehat{j}+\widehat{k}$                

$\overrightarrow{b}=2\widehat{i}+4\widehat{j}-5\widehat{k}\overrightarrow{c}=-\lambda \widehat{i}+2\widehat{j}+3\widehat{k}$              

$\overrightarrow{b}+\overrightarrow{c}=\left(2-\lambda \right)\widehat{i}+6\widehat{j}-2\widehat{k}$

${\left(12-\lambda \right)}^2={\left(2-\lambda \right)}^2+40$

$\lambda =5$

The 1^st such digit is 11 * 19 = 209

Sum = [209 + 220 + 231 + .+ 495] - [231 + 319 + 341 + 418 + 451] = 7744

$\left|2A\right|=2^3\left|A\right|$

replace A by adj 2A

  $\left|2adj2A\right|=2^3\left|adj2A\right|$           

$=2^3{\left|2A\right|}^2=2^3{\left(2^3\left|A\right|\right)}^2$

$=2^9{\left|A\right|}^2$              

Again replace A by (adj A)

|2adj 2 (adj (adj 2A)| = 2⁹ |adj 2A|⁴

= 2⁹ (|2A|²)⁴

->|A²| = 4

a + b = 1   $\alpha +\gamma =\frac{10}{3}$

$\alpha \beta =2\lambda$ $\alpha \gamma =9\lambda$

$\frac{\beta }{\gamma }=\frac{2}{9},$ $\beta -\gamma =1-\frac{10}{3}=-\frac{7}{3}$

$\Rightarrow \beta =\frac{2}{3}$ $\gamma =3$

$\alpha =\frac{1}{3},\lambda =\frac{1}{9}$

$\therefore \frac{\beta \gamma }{\lambda }=18$                                  

$f\left(x\right)=2x^3-3x^2-12x$  

  $f\text{'}\left(x\right)=6x^2-6x-12$             

= 6 (x – 2) (x + 1)

a = -1, b = 2

$A={\displaystyle \underset{-1}{\overset{0}{\int }}\left(2x^3-3x^2-12x\right)dx-{\displaystyle \underset{0}{\overset{2}{\int }}\left(2x^3-3x^2-12x\right)dx=\frac{57}{2}}}$

->4A = 114

 = coefficient of x⁴ in (1 + x)²¹ + coefficient of x⁴ in (1 + x)²¹

${=}^{21}{C}_4{+}^{21}{C}_4=2{.}^{21}{C}_4$              

$A_3$ = coefficient of x³ in (1 + x)²¹ + coefficient of x³ in (1 + x)²¹