Class 12th

Due to equal and similar charge particle will repel each other, hence will never precipitate.

Due to common ion effect solubility of AgCl will decreases in KCl, AgCl and AgNO₃ but in deionized water, no common ion effect will takes place so maximum solubility.

Covalent character is $LiC\mathcal{l}>NaC\mathcal{l}>KC\mathcal{l}>CsC\mathcal{l}.$  As the cationic size increases polarization decreases.

(A) Cr = [Ar]3d⁵4s¹

              (B) m =   $-\mathcal{l}to+\mathcal{l}$

              (C) According to Aufbau principle, orbital are filled in order of their increasing energies.

              (D) Total nodes = n – 1

Moles of Fe₃O₄ =   $\frac{4.64*10^3}{232}=20$

              Moles of CO =   $\frac{2.52*10^3}{28}=90$

              So limiting reagent = Fe₃O₄

              So moles of Fe formed = 60

              Weight of Fe = 60 * 56 = 3360

Draw y = cos²x and y = 2x + $\text{exception 25:}$

3cos²2θ + 6cos2θ -   $10\frac{\left(1+cos2\theta \right)}{2}+5=0$

$cos2\theta \left(3cos2\theta +1\right)=0$

 Þ cos2θ = 0,     $\frac{-1}{3}$

Draw y = cos2θ, y = 0 and y =  $\frac{-1}{3},$  find the pt. of intersection.

Let $\pi \equiv -x+y+z-1=0$

1 2 > 0 as both pt.lies on same side

now $P_1=\left|\frac{-1+2-1-1}{\sqrt[]{3}}\right|=\frac{1}{\sqrt[]{3}}$

P2 = $\left|\frac{-2+\left(-1\right)+3-1}{\sqrt[]{3}}\right|=\frac{1}{\sqrt[]{3}}$

as P₁ = P₂ so distance between foot of perpendicular will be same as distance between the points

d = $\sqrt[]{{\left(1-2\right)}^2+{\left(2+1\right)}^2+{\left(-1-3\right)}^2}$

= $\sqrt[]{1+9+16}$$=$

2cos⁴ x – cos²x = $2{\left(\frac{1+cos2x}{2}\right)}^2-cos2x$

$=\frac{1+co{s}^{2}2x}{2}$

$\sqrt[]{2}{\displaystyle \int \frac{2dx}{1+co{s}^{2}2x}=\sqrt[]{2}{\displaystyle \int 2}\cdot \frac{se{c}^{2}2xdx}{2+ta{n}^{2}2x}}$

$=\sqrt[]{2}\cdot \frac{1}{\sqrt[]{2}}\cdot ta{n}^{-1}\left(\frac{tanx}{\sqrt[]{2}}\right)$

now IF = $e^{{\displaystyle \int Pdx}}$

$e^{\sqrt[]{2}{\displaystyle \int 2\frac{se{c}^{2}2xdx}{2+ta{n}^{2}2x}}}=e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}$

Solution: $y\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}={\displaystyle \int x.e^{ta{n}^{-1}\left(\sqrt[]{2}\cdot cot2x\right)}\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}dx}$

$y\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}=\frac{x^2}{2}\cdot e^{\pi /2}+C$

at $x=\frac{\pi }{4},y=\frac{{\pi }^2}{32},C=0$

at $x=\frac{\pi }{3},y=\frac{{\pi }^2}{18}{e}^{ta{n}^{-1}\alpha }$

$y\cdot e^{ta{n}^{-1}}\frac{\left(tan2\frac{\pi }{3}\right)}{\sqrt[]{2}}=\frac{{\pi }^2}{18}\cdot e^{\pi /2}$

$\frac{{\pi }^2}{18}\cdot e^{ta{n}^{-1}\alpha }\cdot e^{ta{n}^{-1}}\left(-\frac{\sqrt[]{3}}{2}\right)=\frac{{\pi }^2}{18}\cdot e^{\pi /2}$

tan^-1 a + tan^-1 $\left(-\sqrt[]{3}/2\right)=\frac{\pi }{2}$

cot^-1a = tan^-1 $\left(-\frac{\sqrt[]{3}}{2}\right)$

tan^-1 $\frac{1}{\alpha }=ta{n}^{-1}\left(-\frac{\sqrt[]{3}}{2}\right)$

$\frac{1}{\alpha }=-\sqrt[]{3}/2$

2 = $\frac{2}{3}$

Let z = x + iy

$\left|z-2\right|\le 1\Rightarrow \left|x-2+iy\right|\le 1$

(x – 2)² + y² $\le 1$

$z\left(1+i\right)+\overline{z}\left(1-i\right)\le 2$

$\left(x+iy\right)\left(1+i\right)+\left(x-iy\right)\left(1-i\right)\le 2$

$x+ix+iy-y+x-ix-iy-y\le 2$

x – y $\le 1$

PD will be least as CP – r = DP, general pts on circle with centre (2, 0) is (2 + r cos, 0 + r sin )

here r = 1, (2 + cos , sin ) now slope of CP is $\frac{4-0}{0-2}=-2$

tan = 2

so D point will be $\left(2-\frac{1}{\sqrt[]{5}},\frac{2}{\sqrt[]{5}}\right)AP$ will be the greatest. A(1, 0)

now $\left|z_1^2\right|+\left|z_2^2\right|$

$={\left|1+0i\right|}^2+{\left|2-\frac{1}{\sqrt[]{5}}+\frac{2i}{\sqrt[]{5}}\right|}^2$

$=1+{\left(2-\frac{1}{\sqrt[]{5}}\right)}^2+\frac{4}{5}$

$=6-\frac{4}{\sqrt[]{5}}$

now $5\left({\left|z_1\right|}^2+{\left|z_2\right|}^2\right)=\alpha +\beta \sqrt[]{5}$

$5\left(6-\frac{4}{\sqrt[]{5}}\right)=\alpha +\beta \sqrt[]{5}$

= 30, = 4

+ = 26