Class 12th

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

  $\frac{x_1+x_2+x_3+x_4}{4}=\frac{7}{2}$

$x_1+x_2+x_3+x_4=14$

and $\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$

x₅ = 10

Variance $\frac{{\displaystyle \sum _{i=1}^4{x}_i^2}}{4}-{\left(\frac{\Sigma xi}{4}\right)}^2=a$

$\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}-\frac{49}{4}=a$

$x_1^2+x_2^2+x_3^2+x_4^2=4a+49$

and $\frac{x_1^2+x_2^2+x_5^2+x_4^2+x_5^2}{5}-{\left(\frac{24}{5}\right)}^2=\frac{194}{25}$

$\frac{4a+49+x_5^2}{5}=\frac{576+194}{25}$

49 + 49 + $x_5^2=\frac{770}{5}$

49 $+x_5^2+49=154$

4a + 149 = 154

4a = 5

now 4a + x₅ = 15

Tangent to C₁ at (-1, 1) is T = 0                                                           

 x(-1) + 4(1) = 2

-x + y = 2

find OP by dropping  from (3, 2) to centre

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

AP = $\sqrt[]{r^2-O{P}^2}$

$=\sqrt[]{5-\frac{9}{2}}=\frac{1}{\sqrt[]{2}}$

$tan\theta =\frac{OP}{AP}=\frac{3/\sqrt[]{2}}{1/\sqrt[]{2}}=3$

area of $\Delta ABN=\frac{1}{2}A{N}^2\cdot sin2\theta$

AN = $\frac{\sqrt[]{5}}{3}$

$=\frac{1}{2}\cdot \frac{5}{9}\cdot \left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)$

$=\frac{5}{2.9}*\frac{2.3}{1+3^2}=\frac{1}{6}$

sin = $\frac{AP}{AN}$

Kindly consider the following figure

 $AP\perp BP$

M₁ M₂ = 1

$\frac{2t}{t^2-3}*\frac{2/6}{3-\frac{1}{t^2}}=-1$

t = 1

So, A (1, 2) and B (1, 2) they must be end pts of focal chord.

Length of latus rectum $=\frac{2b^2}{a}$

$4=\frac{2b^2}{a}$

b2 = 2a and ae = 1

Eccentricity of ellipse (Horizontal)

b² = a² (1 – e²)

2a = a² (1 – e²)

2 = $\frac{1}{e}\left(1-e^2\right)$

e² + 2e – 1 = 0

$e=\frac{-2\pm \sqrt[]{4+4}}{2}$

$e=-1+\sqrt[]{2}$

now $\frac{1}{e^2}$$=$$3$$+$$2$

No. of compounds containing asymmetric carbon are

Millie q. of H₂SO₄ used by NH₃ = 12.5 * 1 * 2 = 25

So millimoles of N = 25

Moles of N = 25 * 10^-3

Wt of N = 14 * 25 * 10^-3

% of N =    $\frac{14*25*10^{-3}}{0.55}*100=63.66\cong 64$

 $I={\displaystyle \underset{0}{\overset{5}{\int }}cos}\left(\pi x-\pi \left[\frac{x}{2}\right]\right)dx$

$I={\displaystyle \underset{0}{\overset{2}{\int }}cos\left(\pi x\right)}dx+{\displaystyle \underset{2}{\overset{4}{\int }}cos}\left(\pi x-\pi \right)dx+{\displaystyle \underset{4}{\overset{5}{\int }}cos\left(\pi x-2\pi \right)dx}$

$I={\frac{sin\pi x}{\pi }|}_0^2+{\frac{sin\left(\pi x-\pi \right)}{\pi }|}_2^4+{\frac{sin\left(\pi x-2\pi \right)}{\pi }|}_4^5=0$