Maths

System of equation is

$\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill \left|\lambda \right|\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left[\begin{array}{l}-2\\ 4\\ 4\lambda -4\end{array}\right]$

R₁ – 2 R₂, R₃ – R₂

$\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -2\hfill & \hfill \left|\lambda \right|-1\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}-10\\ 4\\ 4\lambda -8\end{array}\right)$

System of equation will have no solution for $\lambda$ = 7.

 $f\left(x\right)=[\begin{array}{ccc}\hfill 2n\hfill & \hfill \hfill & \hfill n=2,4,6....\hfill \\ \hfill n-1\hfill & \hfill \hfill & \hfill n=3,7,11,15....\hfill \\ \hfill \frac{n+1}{2}\hfill & \hfill \hfill & \hfill n=1,5,9,13....\hfill \end{array}$

for n = 2, 4, 6 ……

f (n) = 4, 8, 12, ….4 (n) form

for n = 3, 7, 11, 15, ….

f (n) = 1, 3, 5, 7, …. (4n + 1) or, (4n + 3) from

$\therefore$ f is one and onto.

${\left(x-1\right)}^2+{\left(y-3\right)}^2=10-\alpha$

$\begin{array}{l}x-y=-1\\ x+y=a+b\end{array}\}\to m\left(\frac{a+b-1}{2},\frac{a+b+1}{2}\right)$

A' = 2m – A = (b – 1, a + 1)

$C_2:x^2+y^2+2gx+2fy+\frac{38}{5}=0$

r  $\sqrt[]{g^2+f^2-c}$

$=\sqrt[]{4+4-\frac{38}{5}}=\sqrt[]{\frac{2}{5}}$

$C_1:\sqrt[]{\frac{2}{5}}=\sqrt[]{1+9-\alpha }=\sqrt[]{10-\alpha }$

$\alpha +6r^2=\frac{48}{5}+6\left(\frac{2}{5}\right)=\frac{60}{5}=12$

T = {9, 10, 11, 12, …., 1000}

S = {4, 6, 9}

A = $\left\{a_1+a_2+....+a_k:k\in N,a_i\in S\right\}$

Let 4 appear x no. of times

6 appear y no. of times

9 appear z no. of times

10 then set A has n element 4x + 6y + 9z

Concept : Let a and b are co-prime numbers, then members from (a - ) (b – 1) and more can be expressed in the form ax + by where x, y $\in$  {0, 1, 2, …}

So, all the number of the form

2y + 3z are (2 1). (3 – 1), ….

i.e., 6, 7, 8, 9, 10, 11, ….

form : 2 + t, t = 0, 1, 2, 3, ….

So, 6y + 9z = 3 (2y + 3z) = 3 (2 + t) = 6 + 3t, t = 0, 1, 2, 3, …

sum of element of T – A is 11

 $t_n=\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)},n=1,2,3,....,99$

$=\frac{1}{2}\left(\frac{1}{\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+3\right)}\right)=V_n-V_{n+1}$

$=\frac{1}{12}\left(\frac{1716}{101*17}\right)=\frac{143}{101*17}=\frac{k}{101}\Rightarrow k=\frac{143}{17}$

34 k = 286

$\Delta =p!\left(p+1\right)!\left(p+2\right)!{\Delta }_1=2p!\left(p+1\right);\left(p+2\right)!=2{\left(p!\right)}^3.{\left(p+1\right)}^2.{\left(p+2\right)}^1\to \alpha +\beta =3+1=4$

${\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill p+1\hfill & \hfill \left(p+2\right)\left(p+1\right)\hfill \\ \hfill 1\hfill & \hfill p+2\hfill & \hfill \left(p+3\right)\left(p+2\right)\hfill \\ \hfill 1\hfill & \hfill p+3\hfill & \hfill \left(p+4\right)\left(p+3\right)\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill \left(p+2\right)\left(-2\right)\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill \left(p+3\right)\left(-2\right)\hfill \\ \hfill 1\hfill & \hfill p+3\hfill & \hfill \left(p+4\right)\left(p+3\right)\hfill \end{array}\right|$

= 2(p + 3) – 2 (p + 2) = 2

 $A=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right],a_2,b_2,c_2\in \left\{0,1\right\}$

S = $a_1+a_2+a_3+b_1+b_2+b_3+c_1+c_2+c_3$ is prime

$0\le s\le 9$

Prime value = 2,3, 5, 7

$ForS=2,1+1+0+0+0+0+0+0+0\to \frac{9!}{2!7!}=36$

Total number of matrices

= 36 + 84 + 126 + 36 = 282

 $T_5{=}^n{C}_4{\left(2^{\frac{1}{4}}\right)}^{n-4}.{\left({\left(\frac{1}{3}\right)}^{\frac{1}{4}}\right)}^4{=}^n{C}_4.2^{\frac{n-4}{4}}.\frac{1}{3}$

$T_6{=}^9{C}_5{\left(2^{\frac{1}{2}}\right)}^4.{\left({\left(\frac{1}{3}\right)}^{\frac{1}{4}}\right)}^5{=}^9{C}_5.2.\frac{1}{3^{5/4}}=\frac{2}{3}.\frac{1}{3^{\frac{1}{4}}}.\frac{9*8*7*6}{4*3*2*1}$

 ${\displaystyle \sum _{r=1}^{\infty }\frac{a+\left(r-1\right)d}{2^r}=4,4\left(a+d\right)=?}$

$\Rightarrow a{\displaystyle \sum _{r=1}^{\infty }{\left(\frac{1}{2}\right)}^r+d{\displaystyle \sum _{r=1}^{\infty }\left(r-1\right){\left(\frac{1}{2}\right)}^r=4}}$

$={\left(\frac{1}{2}\right)}^2{\displaystyle \sum _{k=0}^{\infty }k{\left(\frac{1}{2}\right)}^{k-1}}$

$=\frac{1}{4}.\frac{1}{{\left(1-\frac{1}{2}\right)}^2}$

= 1

a + d = 4

4 (a + d) = 16