Class 12th

x + y + z = 48

x, y, z  $\in$ {2, 4, 8, 16, 32, 32}

16 + 16 + 16 -> P_{1 $={\left(\frac{1}{16}\right)}^3$}  

 $32+8+8\to +P_2=\frac{1}{32}.{\left(\frac{1}{8}\right)}^2.\frac{3!}{2!}$

$P=\frac{1}{2^{12}}+\frac{3}{2^{11}}$

$=\frac{1+6}{2^{12}}=\frac{7}{2^{12}}$  

$2sin12°-sin72°$

$=sin12°-\left(sin72°-sin12°\right)$

$=sin12°-2cos42°sin30°$

$=2cos30°\left(-sin18°\right)$

$=-\sqrt[]{3}sin18°$

$=-\sqrt[]{3}\left(\frac{\sqrt[]{5}-1}{4}\right)$  

$\frac{dx}{dt}=12\left(1+cos2t\right)$

$\frac{dy}{dt}=24\left(1+sint\right)cost$

$\frac{1+sint}{cost}=\sqrt[]{3}$

$\Rightarrow \frac{1}{2}=\frac{\sqrt[]{3}}{2}cost-\frac{1}{2}sint$

$y_0=12{\left(1+\frac{1}{2}\right)}^2=12*\frac{9}{4}=27$

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

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

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.