Class 12th

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,  $\Delta =\Rightarrow$

I= $\frac{15}{3}=5A$

$V_B+I*1?15=v_A$

$v_B?v_A=15?5*1$

= 10v

x² = 1 – 2i

so 2 = 1 – 2i = 2

8 = 8

now $\left|{\alpha }^8+{\beta }^8\right|=2\left|{\alpha }^8\right|$

$=2{\left|\left({\alpha }^2\right)\right|}^4$

$=2{\left|{\alpha }^2\right|}^4$

$=2{\left|1-2i\right|}^4$

$=2*25$

When object is at infinity

u = $\infty ,{\mu }_1=1,R=15cm$

v = ? ${\mu }_2=1.5$

$\frac{-{\mu }_1}{u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-{\mu }_1}{R}$

$\frac{-1}{\infty }+\frac{1.5}{v}=$ $\frac{1.5-1}{15}$

$\frac{1.5}{v}=$ $\frac{0.5}{15}$

v =

 45cm

Now for Refraction through C2

u = + 15cm

v = ?

${\mu }_1=1.5$

${\mu }_2=1$

R = 15

$\frac{-{\mu }_1}{u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-{\mu }_1}{R}$

$\frac{-1.5}{15}+\frac{1}{v}=\frac{1-1.5}{-15}$

$\frac{-1}{10}+\frac{1}{v}=\frac{1}{30}$

$\frac{1}{v}=\frac{1}{30}+\frac{1}{10}=\frac{4}{10}$

$v=\frac{30}{4}=+7.5cm$

from centre = 15 + 7.5 = 22.5 cm = 225mm

$=\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill \delta \hfill \end{array}\right|=0\Rightarrow \delta =-3$

and ${\Delta }_1=\left|\begin{array}{ccc}\hfill 7\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill 4\hfill & \hfill -3\hfill \end{array}\right|=0\Rightarrow k=6$

Area of $\Delta ABC$

= $\frac{1}{2}AB\cdot BC$

$=\frac{1}{2}\cdot \sqrt[]{2}\cdot 1$

$=\frac{1}{\sqrt[]{2}}$

now required area

$={\displaystyle \underset{0}{\overset{4}{\int }}\left(2\sqrt[]{2}-\sqrt[]{2}x\right)dx-\frac{1}{\sqrt[]{2}}}$

$=\frac{32\sqrt[]{2}}{3}=8\sqrt[]{2}-\frac{1}{\sqrt[]{2}}$

$=\frac{13\sqrt[]{2}}{6}$