Class 12th

$2x^2\frac{dy}{dx}-2xy+3y^2=0$

$\frac{dy}{dx}=\frac{y}{x}-\frac{3}{2}{\left(\frac{y}{x}\right)}^2$

Put y = vx

  $\frac{dy}{dx}=v+x\frac{dv}{dx}$

$Point\left(e,\frac{e}{3}\right)$

$\Rightarrow \frac{-e}{\left(\frac{e}{3}\right)}+\frac{3}{2}\mathcal{l}ne=C$

$C=\frac{3}{2}-3=\frac{-3}{2}$

$\frac{3}{2}\mathcal{l}n\left|x\right|\frac{-x}{y}+\frac{3}{2}=0$

$x=1\Rightarrow y=\frac{2}{3}$              

$b_n={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{co{s}^{2}nx}{sinx}dx}$

$={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{1+cos2nx}{2sinx}dx}$

$b_n-b_{n-1}={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{cos\left(2nx\right)-cos\left(2\left(n-1\right)x\right)}{2sinx}}dx$

$={\frac{cos\left(2n-1\right)x}{\left(2n-1\right)}]}_0^{\pi /2}=\frac{1}{2n-1}\left[0-1\right]=\frac{-1}{2n-1}$

$b_{n-1}-b_n=\frac{1}{2n-1}$

$b_1-b_2=\frac{1}{3}\left(A\right)-\frac{1}{5},......$

$b_2-b_3=\frac{1}{3}\left(B\right)-5,-7,-9,.....$

$b_3-b_4=\frac{1}{7}\left(C\right)$

$S=\pi r\mathcal{l}$

$=\pi x\sqrt[]{x^2+y^2}$

$=\pi x\sqrt[]{x^2+25x^2}$

$\frac{ds}{dt}=\pi \sqrt[]{26}.2x.\frac{dx}{dt}$  

${\left(\frac{ds}{dt}\right)}_{y=10}\to x=2$                

$5\pi .3x^2\frac{dx}{dt}=1$                

$\left(\frac{dx}{dt}\right)=\frac{1}{15\pi \left(4\right)}=\frac{1}{60\pi }$

According to the John-Tollen distortion, there is a unsymmetrical filling of eq set of Cu⁺= ions but there is symmetrical filling of t₂g set of ions.

  $\left(x+3y-z-5\right)+\lambda \left(2x-y+z-3\right)=0$

$\Rightarrow \left(1+2\lambda \right)x+\left(3-\lambda \right)y+\left(\lambda -1\right)z-5-3\lambda =0$                

Point $\left(2,1,-2\right)\Rightarrow \left(1+2\lambda \right)\left(2\right)+\left(3-\lambda \right)\left(1\right)+\left(\lambda -1\right)\left(-2\right)-5-3\lambda =0$  

$-2\lambda +2=0\Rightarrow \lambda =1$                

 P : 3x + 2y + 0.z = 8

X (1, -2, 4)

Y (5, -1, 2)

X + Y = (6, -3, 6)

Y – X = (4, 1, -2)

Following are the $\Delta H°$  hydration of given ions

Cr²⁺ - 1926 KJ/mole

Mn²⁺ - 1860 KJ/mole

Fe²⁺ - 2000 KJ/mole

Co²⁺ - 2078 KJ/mole

Cobalt (II) meta borate is a blue colour of Bead.

Biuret test is given by proteins and dipeptide & compounds having following groups

Hence positive test given by tripeptide & biuret.

Monomer of novolac is

Molecular mass of monomer unit

= 12 * 7 + 1 * 8 + 16 * 2

= 84 + 8 + 32

= 124 g

Molecular mass of polymer = 963 g

$\therefore$ n * mass of monomer = mass of polymer + (n – 1) * 18

(n – 1) unit of H₂O is removed during formation

$\therefore 124*n=963+18\left(n-1\right)$        

106n = 945

$\therefore n=\frac{945}{106}=8.91\approx 9$

$\underset{x\to \frac{\pi }{2}}{lim}\frac{si{n}^{2}x\left(si{n}^{2}x-3sinx+2\right)}{co{s}^{2}x.\left(\sqrt[]{2si{n}^{2}x+3sinx+4}+\sqrt[]{si{n}^{2}x+6sinx+2}\right)}$  

$\downarrow$                                                       $\downarrow$                            

3                                                       3

$\underset{x\to \frac{\pi }{2}}{lim}\frac{1}{9}.\frac{\left(sinx-1\right)\left(sinx-2\right)}{\left(1-sinx\right)\left(1+sinx\right)}=\frac{1}{6}\left(-1\right).\frac{-1}{2}=\frac{1}{12}$