Maths

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

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

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

(a, 4a, 7). (3, 1, 2b)= 0

3a + 4a – 14b = 0 a – 2b = 0 ……. (i)

(a, 4a, 7) . (b, a, 2) = 0

ab – 4a² + 14 = 0……… (ii)

(i) & (ii) =>2b² – 16b²+ 14 = 0

$\Rightarrow b^2=1\Rightarrow a=2b=\pm 2$

Plane : x – y + z = 0

P (4, 5, 1)

4 + 5 + 1 = 10

 $\overline{x}=15\Rightarrow {\displaystyle \sum _{i=1}^{20}{x}_i=300}$

$\frac{{\displaystyle \sum _{i=1}^{20}{\left(x_i-15\right)}^2}}{20}=9\Rightarrow {\displaystyle \sum _{i=1}^{20}{\left(x_i-15\right)}^2=180}$

${\displaystyle \sum _{i=1}^{20}{\left(x_i+\alpha \right)}^2=178*20=3560}$

$4680+2\alpha \left(300\right)+20{\alpha }^2=3560$

${\alpha }^2+30\alpha +234-178=0$

= 2, 28

${\alpha }_{max}2={\left(-2\right)}^2=4$

 $\frac{7}{2}\left(1+cos2\theta \right)-\frac{3}{2}\left(1-cos2\theta \right)-2co{s}^{2}2\theta =2$

Put cos 2 $\theta$= t

Equation 2t²– 5t = 0, t (2t – 5) = 0

$t=0,\frac{5}{2}$

cos 20 = 0, 0 < 2 < 4

$x_1+x_2=2\left(ta{n}^2\theta +co{t}^2\theta \right)\Rightarrow ?$

$=2\left(1+1\right)+2\left(1+1\right)+2\left(1+1\right)+2\left(1+1\right)=16$