Maths

${\displaystyle \underset{0}{\overset{1}{\int }}\left[-8x^2+6x-1\right]dx}$  

Let f (x) = -8x² + 6x – 1

f (0) = -1, f (1) = -3

max f (x) = $-\frac{D}{4a}=-\frac{36-32}{-32}=\frac{1}{8}$  

= $-\frac{1}{4}-1\left(\frac{3}{4}-\frac{\lambda }{2}\right)-2\left(\frac{\sqrt[]{17}-3}{8}-\frac{3}{4}\right)-3\left(1-\frac{\sqrt[]{17}-3}{8}\right)$  

= $\frac{\sqrt[]{17}-13}{8}$

Let   $\frac{A_2}{A_1}=\frac{A_3}{A_2}=...=r$

$\therefore A_1{A}_3{A}_5{A}_7=\frac{1}{1296}$              

$\Rightarrow A_1{r}^3=\frac{1}{6}.......\left(i\right)$               

Again, A₂ + A₄ = $\frac{7}{36}$

$A_{1}r=\frac{7}{36}-\frac{1}{6}=\frac{1}{36}........\left(ii\right)$

$\left(i\right)\&\left(ii\right)r=\sqrt[]{6}\&A_1=\frac{1}{36\sqrt[]{6}}$

$\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$             

Sum of digits

1 + 2 + 3 + 5 + 6 + 7 = 24

So, either 3 or 6 rejected at a time

Case 1 Last digit is 2

……….2

no. of cases = ²C₁ * 4! = 48

Case 2 Last digit is 6

……….6

= 4! = 24

Total cases = 72

|A| = 2

$\left|\left|A\right|adj\left(5adj{A}^3\right)\right|$

= $\left|A{P}^3\right|\left|adj\left(5adj\left(A^3\right)\right)\right|$

$\begin{array}{l}={\left|A\right|}^{15}.5^6\\ =2^{15}*5^6=2^9*10^6\end{array}$

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