Maths

$\left[\frac{x}{\sqrt[]{x^2-y^2}}+e^{y/x}\right]x\frac{dy}{dx}=x+\left[\frac{x}{\sqrt[]{x^2-y^2}}+e^{y/x}\right]y$

$\Rightarrow e^{y/x}\left[xdy-ydx\right]=xdx+\frac{x}{\sqrt[]{x^2-y^2}}\left(ydx-xdy\right)$

$\Rightarrow e^{y/x}d\left(y/x\right)=\frac{dx}{x}-\frac{d\left(y/x\right)}{\sqrt[]{1-{\left(y/x\right)}^2}}$

Integrating

$e^{y/x}=\mathcal{l}nx-si{n}^{-1}\left(\frac{y}{x}\right)+c$

Passes (1, 0)

1 = c

$\Rightarrow \alpha =\frac{1}{2}exp\left(\sqrt[]{e}-1+\frac{\pi }{6}\right)$

For x < 0 0 < e^x < 1 [e^x] = 0

$0\le x<1a{e}^x+\left[x-1\right]$               

$=a{e}^x+\left[x\right]-1$               

= a e^x – 1             b + [sin px]

$\therefore f\left(x\right)=\left[\begin{array}{l}0x<0\\ a{e}^x-10\le x<1\\ b-11\le x<2\\ -cx\ge 2\end{array}\right]$               

For f to be continuous at x = 0

a – 1 = 0 ⇒ a = 1

$\begin{array}{l}\therefore a+b+c=1+e+1-e=2\\ a+b+c\ne 1\end{array}$               

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

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.