Maths

 $AP\perp BP$

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

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

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

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

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

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

Use aRb = a is related to b, belongs to A iff a belongs to A.

In simple terms, aRb is true if both a & b belongs to the same set.

For reflexive

aRa, a $\in$ A, so it is true.

For symmetric

Let aRb be true

Þ a & b belongs to the same set.

Þ b & a also belongs to the same set

Þ bRa will be true

For transitive

Let aRb and bRc be true.

aRb Þ a, b belongs to the same set

bRc Þ b, c belongs to the same set

Þ (a, c) belongs to the same set

Þ so aRc will be true.

So R is an equivalence relation.

 $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill \frac{1}{4}\hfill & \hfill \frac{1}{2}\hfill & \hfill 1\hfill \end{array}\right]$

$A^2=$ $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill \frac{1}{4}\hfill & \hfill \frac{1}{2}\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill \frac{1}{4}\hfill & \hfill \frac{1}{2}\hfill & \hfill 1\hfill \end{array}\right]$

$=3\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill \frac{1}{4}\hfill & \hfill \frac{1}{2}\hfill & \hfill 1\hfill \end{array}\right]$

A² = 3A

 A³ = 3A²

A³ = 3²A

A⁴ = 3³A

Aⁿ = 3^n-1A

now, A² + A³ +….+A¹⁰

= 3A + 3² A +…. + 3⁹A

= 3A (1 + 3 +….+ 3⁸

$=3A\cdot \frac{\left(3^9-1\right)}{3-1}$

$=\frac{3^{10}-3}{2}A$

$\left(p\vee q\right)\Rightarrow q$

= $\sim \left(p\vee q\right)\vee q$

= $\left(\sim p\wedge \sim q\right)\vee q$

= $\left(\sim p\vee q\right)\wedge \left(\sim q\vee q\right)$

= $\sim p\vee q$

now $\left(p\wedge q\right)\Delta \left(\sim p\vee q\right)istautolog\text{?}y$

pq $p\wedge q$ $\sim p\vee q$ $\left(p\wedge q\right)\Delta \left(\sim p\vee q\right)$

TTTTT

TFFFT

FTFTT

FFFTT

So,  $\Delta =\Rightarrow$