Maths

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 = $\frac{AP}{AN}$

 $AP\perp BP$

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$

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

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

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

${\left(3x^3-2x^2+\frac{5}{x^5}\right)}^{10}$

General term $10!\frac{{\left(3x^3\right)}^{\alpha }\cdot {\left(-2^x\right)}^{\beta }\cdot {\left(5x^{-5}\right)}^{\gamma }}{\alpha !\beta !\gamma !}$

$=10!\frac{3^{\alpha }\cdot {\left(-2\right)}^{\beta }\cdot 5^{\gamma }\cdot x^{3\alpha +2\beta -5\gamma }}{\alpha !\beta !\gamma !}$

Now for constant term 3 + 2 $-5\gamma =0$ ………….(i)

$\alpha +\beta +\gamma =10$ …………(ii)

From equation (i) & (ii)

$3\alpha +2\left(10-\alpha -\gamma \right)=5\gamma$

$\alpha +20=7\gamma$

= 1, $\gamma$ = 3, = 6

Constant term $10!\cdot \frac{3^1\cdot {\left(-2\right)}^6\cdot 5^3}{3!6!}=2^9\cdot 3^2\cdot 5^4\cdot 7^1$

Case 1: 1 $\le \frac{1}{4x^2-1}\le 1$

4x2 – 1 $\ge 1or4x^2-1\le -1$

$x^2\ge \frac{2}{4}or{x}^2\le 0$

So x $\in (-\infty ,-\frac{1}{\sqrt[]{2}}]\cup [\frac{1}{\sqrt[]{2}},\infty )\cup \left\{0\right\}$

Let perimeter of $\Delta$ is x and that of square is 22 – x

now area $=\frac{\sqrt[]{3}}{4}{\left(\frac{x}{3}\right)}^2+{\left(\frac{22-x}{4}\right)}^2$

for maximum or minimum,  $\frac{dA}{dx}=0$

x $=\frac{22\sqrt[]{3}}{\sqrt[]{3}+4}$

now side of a $\Delta =\frac{x}{3}$

$=\frac{22\sqrt[]{3}}{3\left(\sqrt[]{3}+4\right)}$

$=\frac{66}{}$$\frac{9}{}$$\frac{+}{}$$\frac{4}{}$

Let A, A' be (, 2) AB and A'B subtends $\frac{\pi }{4}$ angle at (0, 0) slope of OA = $\frac{2}{\alpha }$

slope of OB = $\frac{3}{2}$

$tan\frac{\pi }{4}=\left|\frac{\frac{2}{\alpha }-\frac{3}{2}}{1+\frac{2}{\alpha }\cdot \frac{3}{2}}\right|$

$\Rightarrow 1+\frac{3}{\alpha }=\pm \left(\frac{4-3\alpha }{2\alpha }\right)$

$\Rightarrow \frac{\alpha +3}{\alpha }=\pm \left(\frac{4-3\alpha }{2\alpha }\right)\Rightarrow \alpha =10,-\frac{2}{5}$

now distance between A'A, (10, 2) & $\left(\frac{-2}{5},2\right)is\frac{52}{5}$

$a_{n+2}=2\cdot a_{n+1}-a_{n+1}$

$a_2=2a_1-a_0+1$

a₂ = 1

a₃ = 3

a₄ = 6

So for n $\ge 2,a_n=\frac{n\left(n-1\right)}{2}$

${\displaystyle \sum _{n=2}^{}\frac{n\left(n-1\right)}{2\cdot 7^n}=\frac{1}{7^2}+\frac{3}{7^3}+\frac{6}{7^4}+......}$

Let S = $\frac{1}{7^2}+\frac{3}{7^3}+\frac{6}{7^4}+.....$

$\frac{\frac{S}{7}=\frac{1}{7^3}+\frac{3}{7^4}+....}{S-\frac{S}{7}=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+....}$

$\frac{\frac{6}{7}S\cdot \frac{1}{7}=\frac{1}{7^3}+\frac{2}{7^3}+....}{\frac{6}{7}S-\frac{6}{49}S=\frac{1}{7^2}+\frac{1}{7^3}+....}$

$\frac{36}{49}S=\frac{\frac{1}{7^2}}{1-\frac{1}{7}}$

$S=\frac{1}{7^2}*\frac{7}{6}*\frac{49}{36}$

$S=\frac{7}{216}$