Class 12th

Common potential after connection.

${\mathrm{V}}_{\text{common}}=\frac{{\mathrm{C}}_1{\mathrm{V}}_1+{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{60*20+0}{120}=10\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{t}$

Loss of energy $=\frac{1}{2}{\mathrm{C}\mathrm{V}}^2-\frac{1}{2}\left(2\mathrm{C}\right)*{\mathrm{V}}_{\text{Common}}^2$

$\begin{array}{rr}& =\frac{1}{2}*60*{10}^{-12}*(20{)}^2-60*{10}^{-12}*(10{)}^2\\ & =60*{10}^{-12}(200-100)\\ & =6000*{10}^{-12}\\ & =6\mathrm{n}\mathrm{J}\end{array}$

$\mathrm{}?=\stackrel{?}{B}\cdot \stackrel{?}{A}=(3\hat{i}+4\hat{k})\cdot (25\hat{i}+25\hat{k})$

$?=(3*25)+(4*25)=175\text{weber}$

$\frac{{\left(x-3\right)}^2}{16}+\frac{{\left(y-4\right)}^2}{9}\le 1,x,y\in N,{\left(x-7\right)}^2+{\left(y-4\right)}^2\ge 36,x,y\in R$        

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

Given : ${\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2\&\overrightarrow{a}\cdot \overrightarrow{b}=3$  

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta =3$

$\left|\overrightarrow{a}\right|cos\theta =\frac{3}{\sqrt[]{6}}=\sqrt[]{\frac{9}{6}}=\sqrt[]{\frac{3}{2}}$

${\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^{2}si{n}^2\theta =75$

$\left|\overrightarrow{a}\right|sin\theta =\sqrt[]{\frac{75}{6}}=\frac{5}{\sqrt[]{2}}$

$\left|\overrightarrow{a}\right|cos\theta =\sqrt[]{\frac{3}{2}}$

${\left|\overrightarrow{a}\right|}^2=\frac{25}{2}+\frac{3}{2}=\frac{28}{2}=14$               

y = 5x² + 2x – 25

P(2, -1)

T_(p) : T = 0

->y – 1 = 10x(2) + 2(x + 2) – 50

$\Rightarrow y=22x-45$ is also tangent to y = x³ – x² + x at point (a, b)

For y = x³ -x² + x

$\frac{dy}{dx}=3x^2-2x+1>0$

y = 22x  $-\frac{1939}{27}$ which is not tangent to the curve.

$3a^2-2a+1=22$  (slope of tangent)

$\Rightarrow 3a^2-2a-21=0\to a=\frac{2\pm \sqrt[]{4+252}}{6}=\frac{2\pm 16}{6}=3,\frac{-7}{3}$

b = 27 – 9 + 3 = 21

tangent : y – 21 = 22(x – 3)

->y = 22x – 45

a = 3, b = 21

2a + 9b = 6 + 189 = 195

Also,   $a^3-a^2+a=b$

For a = $\frac{-7}{3}$  

b =   $\frac{-343}{27}-\frac{49}{9}-\frac{7}{3}$

=   $\frac{-343-147-63}{27}=\frac{-553}{27}$

$f\left(x\right)=4\left|2x+3\right|+9\left[x+\frac{1}{2}\right]-12\left[x+20\right],-20<x<20$ doubtful points for differentiability :   $x=\frac{-3}{2},$

$f\left(x\right)=4\left(2x+3\right)+9\left(-1\right)-12\left(-2\right)-240=8x-213forx=\frac{-3}{2}+h$         

= $-8x-230$  for x = $\frac{-3}{2}-h$  

Not diff. at x = $\frac{-3}{2}$  

other doubtful points : $x+\frac{1}{2}=integer$  

$-20+\frac{1}{2}<x+\frac{1}{2}<20+\frac{1}{2}$

$x+\frac{1}{2}=-19,-18,....,19,20$

$x=-19.5,-18.5,-17.5,.......,18.5,19.5\to$ total 40 numbers.

 No. of number = 19.5 – $\left(-19.5\right)+1=40\left(-1.5\right)included$  

  $-20<x<90\Rightarrow x=-19,-18,.....,18,15\to 39points$

No. of number = 19 - (-19) + 1 = 39

Total : 40 + 39 = 79

$r{\cdot }^n\text{?}{C}_r=n{\cdot }^{n-1}\text{?}{C}_{r-1}$

${\displaystyle \sum _{k=1}^{10}{\left(k{\cdot }^{10}\text{?}{C}_k\right)}^2}={\displaystyle \sum _{k=1}^{10}{\left(10{\cdot }^9\text{?}{C}_{K-1}\right)}^2}$

= $100{\cdot }^{18}\text{?}{C}_9$  

22000L =   $100{\cdot }^{18}\text{?}{C}_9$

 L =

$18!=2^{16}\cdot 3^8\cdot 5^3\cdot 7^2\cdot 11^1\cdot 13\cdot 17$

$9+4+2+1$ $9!=2^7\cdot 3^4\cdot 5^1\cdot 7^1$

6 + 2                    4 + 2 + 1   $\frac{18!}{{\left(9!\right)}^2}=2^2\cdot 5\cdot 11\cdot 13\cdot 17$            

3 +                       3 + 1

 = 221

$A=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$

$A^2=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=I+B,B=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$A^4=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]B^2\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=0$

$A^{2n}={\left(I+B\right)}^n$

I + nB + 0 + 0 +……….

$A^{2n}=I+\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 6n\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6n\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$X\text{'}{A}^{k}x=\left[33\right]$ $=\left[111\right]A^k\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$

$\left[111\right]A^{2n}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left(k=2n\right)$

$\left[1+1+6n+1\right]=\left[6n+3\right]=\left[33\right]n=5$

k = 10

np + npq = 82.5 Þ np (1 + q) = 82.5

$np\cdot npq=1350\Rightarrow {\left(np\right)}^{2}q=1350$

n = ?

$\Rightarrow \frac{{\left(1+q\right)}^2}{q}=\frac{82.5*82.5}{1350}$

$q^2+2q+1=\frac{121}{24}q$

-> 24q² – 73q + 24 = 0

q $=\frac{73\pm \sqrt[]{73^2-4*576}}{48}$  

$=\frac{73\pm 55}{48}=\frac{128}{48},\frac{18}{48}=\frac{8}{3},\frac{3}{8}$

$p=\frac{5}{8},q=\frac{3}{8}$

                $n\cdot \frac{5}{8}\cdot \frac{11}{8}=\frac{165}{2}$

n = 96

Structure of Bithinol is;