Class 12th

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

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

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

Integrating

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

Passes (1, 0)

⇒ 1 = c

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

$y^2\le 8xy>\sqrt[]{2}x$

$2x^2=8x\Rightarrow x\left(x-4\right)=0\Rightarrow x=0,x=4x=0,x=y$

Required area = ${\displaystyle \underset{1}{\overset{4}{\int }}\left(2\sqrt[]{2}\sqrt[]{x}-\sqrt[]{2}x\right)}$ dx

$=\frac{28\sqrt[]{2}}{3}-\frac{15\sqrt[]{2}}{2}=\frac{11\sqrt[]{2}}{6}$

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 Þ 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) =  $-\frac{D}{4a}=-\frac{36-32}{-32}=\frac{1}{8}$  

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

= $\frac{\sqrt[]{17}-13}{8}$  

|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

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

R₁ – 2 R₂, R₃ – R₂

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

System of equation will have no solution for = -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

f is one and onto.

Organic compound → AgBr (s)

  0.5 g                    0.40 g

Here; Moles of Br in organic compound = Moles of Br in Ag Br

= Moles of AgBr

=  $\frac{0.40}{188}mol$

Mass of Br in organic compound =  $\frac{0.4}{188}*80g$  

$\begin{array}{l}\mathrm{\%}ofBr=\frac{0.4*80}{188*0.5}*100\mathrm{\%}\\ =34\mathrm{\%}\end{array}$

$f\left(x\right)=a{x}^2+bx+c$

g (x) = px + q

$f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$  

$\Rightarrow 8x^2-2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$  

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

$9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$  

->4x² + 6x + 1 = apx² + bpx + cp + q

->ap = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

->b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3