Class 12th

$A=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]then{A}^2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2k\hfill & \hfill -2\hfill \\ \hfill -k\hfill & \hfill 2k+1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{cc}\hfill 2k\hfill & \hfill -2\hfill \\ \hfill -k\hfill & \hfill 2k+1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill -2k\hfill & \hfill 4k+2\hfill \\ \hfill 2k^2+k\hfill & \hfill -4k-1\hfill \end{array}\right]$              

Now, A(A³ + 3l) = 2l gives A³ + 3l = 2A^-1

$\Rightarrow -2k+3=\frac{1}{k}\Rightarrow 2k^2-3k+1=0$    

  $k=\frac{1}{2},1$            

& $4k+2=\frac{2}{k}or2k+1-\frac{1}{k}soor2k^2+k-1=0$

=> $k=\frac{1}{2}$

Compound react with S_N¹ mechanism in presence of polar protic solvent, which follows carbocation forming path.

Hence correct order is.

$\frac{dy}{dx}=2\left(y+2sinx-5\right)x-2cosx$

where P = -2x, Q = 4x sin x – 10x – 2 cosx

Solution be $y.e^{-x^2}={\displaystyle \int \left(4xsinx-10x-2cosx\right).e^{-x^2}dx+c}$

$y.e^{-x^2}{\displaystyle \int \left(4xsinx-10x-2cosx\right).e^{-x^2}dx+c}$     

$y.e^{-x^2}=e^{-x^2}\left(5-2sinx\right)+c........\left(i\right)$  

Put x = 0, y = 7 then 7 = 5 + c i.e. c = 2

Put x = p then $y.e^{-{\pi }^2}=5.e^{-{\pi }^2}+2$

y = 5 + $2e^{x^2}$

$y=lo{g}_{10}x+lo{g}_{10}{x}^{1/3}+lo{g}_{10}{x}^{1/9}+........upto\infty terms$

$=lo{g}_{10}x\left(1+\frac{1}{3}+\frac{1}{9}+......\right)$   

y = log₁₀ x * $\frac{1}{1-\frac{1}{3}}=\frac{3}{2}lo{g}_{10}x$  

Now, $\frac{2+4+6+....+2y}{3+6+9+....+3y}=\frac{4}{lo{g}_{10}x}$

$\Rightarrow \frac{2*y\frac{\left(y+1\right)}{2}}{3y\frac{\left(y+1\right)}{2}}=\frac{4}{lo{g}_{10}x}solo{g}_{10}x=6$

$\Rightarrow y=\frac{3}{2}*6=9$

So, (x, y) = (10⁶, 9)

Even after using all, the heaviest is not clear.

A3 is 5th in terms of both height and heaviness.

$y=\frac{x^2}{2}+\frac{2}{3}{x}^3+\frac{3}{4}{x}^4+.....$

$=\left(x^2+x^{}+x^4+.....\right)+\left(-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}........\right)$

$y=\frac{x^2}{1-x}+log\left(1-x\right)+x=\frac{x}{1-x}+log\left(1-x\right)$

$atx=\frac{1}{2},y=1+ln\frac{1}{2}=1-ln2$

$e^{y+1}=e^{1-ln2+1}=e^{2-ln2}=\frac{e^2}{2}$

Biuret is

Denticity is 2 of terminal -NH₂ only, because middle –NH will undergo in cross conjugation.

Ascending order in weight

A2 < A5 < A6 < A7 < A3 < A4 or A1 < A1 or A4

Ascending order in weight

A2 < A5 < A6 < A7 < A3 < A4 or A1 < A1 or A4

Ascending order in height

A6 < A7 < A5 < A4 < A3 < A1 < A2

According to heaviness, either A1 or A4 come second from the last.

$\left(2x-10y^3\right)dy+ydx=0$              

$\frac{dx}{dy}+\frac{2x}{y}-10y^2=0$              

$\frac{dx}{dy}+\frac{2}{y}x=10y^2\to$ Linear differential equation

$P=\frac{2}{y},Q=10y^2$              

$Nowputx=2,y=\beta then2{\beta }^2=2{\beta }^5-2$              

or ${\beta }^5-{\beta }^2-1=0$  

So B will be roots of $y^5-y^2-1=0$