Maths

For R? let a = 1+√2, b=1-√2, c = 8¹/?
aR? b => a² + b² = (1+√2)² + (1-√2)² = 6 ∈ Q
aR? c => b² + c² = (1-√2)² + (8¹/? )² = 3 ∈ Q
aR? c => a² + c² = (1+√2)² + (8¹/? )² = 3 + 4√2 ∉ Q
∴ R? is not transitive.
For R? let a = 1+√2, b=√2, c=1-√2
aR? B => a² + b² = (1+√2)² + (√2)² = 5+2√2 ∉ Q
bR? b => b² + c² = (√2)² + (1-√2)² = 5-2√2 ∉ Q
aR? c => a² + c² = (1+√2)² + (1-√2)² = 6 ∈ Q
∴ R? is not transitive.

C? = 2C? S = 2√20-4 = 8

$\mathrm{}a\mathrm{c}\mathrm{o}\mathrm{s}?\theta =b\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)=c\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)=k$

$a=\frac{k}{\mathrm{c}\mathrm{o}\mathrm{s}?\theta },b=\frac{k}{\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)},c=\frac{k}{\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)}$

$ab+bc+ca=k^2\frac{\left[\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)+\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)\right]}{\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)\mathrm{c}\mathrm{o}\mathrm{s}?\theta \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)}$

$={\mathrm{k}}^2\left[\frac{\mathrm{c}\mathrm{o}\mathrm{s}?\theta +2\mathrm{c}\mathrm{o}\mathrm{s}?(\theta +\pi )\cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\frac{\pi }{3}\right)}{\mathrm{c}\mathrm{o}\mathrm{s}?\theta \cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)\cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)}\right]$

$={\mathrm{k}}^2\left[\frac{\mathrm{c}\mathrm{o}\mathrm{s}?\theta -2\mathrm{c}\mathrm{o}\mathrm{s}?\theta \cdot \frac{1}{2}}{\mathrm{c}\mathrm{o}\mathrm{s}?\theta \cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)\cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)}\right]=0$

$\mathrm{c}\mathrm{o}\mathrm{s}??=\frac{(a\hat{i}+b\hat{j}+c\hat{k})\cdot (b\hat{i}+c\hat{j}+a\hat{k})}{\sqrt[]{a^2+b^2+c^2}\cdot \sqrt[]{b^2+c^2+a^2}}=ab+bc+ca=0$

$?=\frac{\pi }{2}$

$\left|z_1-1\right|=\mathrm{R}\mathrm{e}?\left(z_1\right)$ Let $z_1{x}_1+i{y}_1$and $z_2=x_2+i{y}_2$

${\left({\mathrm{x}}_1-1\right)}^2+{\mathrm{y}}_1^2={\mathrm{x}}_1^2$

${\mathrm{y}}_1^2-2{\mathrm{x}}_1+1=0$

$\left|z_2-1\right|=\mathrm{R}\mathrm{e}?\left(z_2\right)$

${\left(x_2-1\right)}^2+y_2^2=x_2^2$

${\mathrm{y}}_2^2-2{\mathrm{x}}_2+1=0$

$\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\left({\mathrm{y}}_1+{\mathrm{y}}_2\right)=2\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)$

${\mathrm{y}}_1+{\mathrm{y}}_2=2\left(\frac{{\mathrm{x}}_1-{\mathrm{x}}_2}{{\mathrm{y}}_1-{\mathrm{y}}_2}\right)$

$\mathrm{a}\mathrm{r}\mathrm{g}?\left(z_1-z_2\right)=\frac{\pi }{6}$

${\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?\left(\frac{{\mathrm{y}}_1-{\mathrm{y}}_2}{{\mathrm{x}}_1-{\mathrm{x}}_2}\right)=\frac{\pi }{6}$

$\frac{{\mathrm{y}}_1-{\mathrm{y}}_2}{{\mathrm{x}}_1-{\mathrm{x}}_2}=\frac{1}{\sqrt[]{3}}$

$\therefore {\mathrm{y}}_1+{\mathrm{y}}_2=2\sqrt[]{3}$

$\Rightarrow \mathrm{}\mathrm{I}\mathrm{m}?\left(z_1+z_2\right)=2\sqrt[]{3}$

$\mathrm{}488=\frac{n}{2}\left. [2\left(\frac{100}{5}\right)+(n-1)\left(\frac{2}{5}\right)\right]$  

$488=\frac{n}{2}(101-n)$

$\Rightarrow \mathrm{}{\mathrm{n}}^2-101\mathrm{n}+2440=0$

$\Rightarrow \mathrm{n}=61$ or 40

For $n=40\mathrm{}\Rightarrow \mathrm{}{T}_n>0$  

For $\mathrm{n}=61\mathrm{}\Rightarrow \mathrm{}{\mathrm{T}}_{\mathrm{n}}<0$

${\mathrm{T}}_{\mathrm{n}}=\frac{100}{5}+(61-1)\left(-\frac{2}{5}\right)=-4$

First, express area as a function of t. Suppose there is a triangle whose vertices are A (0,0), B (t,0) and C (0, t). Here, we can use the determinant formula for the area of a triangle. $\text{Area}=\frac{1}{2}{x}_1(y_2$
$\left|{\mathrm{-\; y}}_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ Let us substitute the coordinates in the above equation: $\begin{array}{l}\text{Area}=\frac{1}{2}\left|0(0-t)+t(t-0)+0(0-0)\right|\\ =\frac{1}{2}\left|t*t\right|\end{array}$

$=\frac{1}{2}{t}^2$

$\text{So, the area as a function of}t\text{is:}$

$\mathrm{Area}\left(t\right)=\frac{1}{2}{t}^2$ Now, let us calculate area when t = 4 and substitute t = 4 into function: $\mathrm{Area}\left(4\right)=\frac{1}{2}*4^2=\frac{1}{2}*16=8$

Below are a few important tips to remember the integrals of some particular functions:
1. Know the derivatives for each integral.
2. Make yourself familiar with the standard formulas first.
3. Practice daily for better memory.
4. Group similar formulas