Maths

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,  $\Delta =\Rightarrow$

x² = 1 – 2i

so 2 = 1 – 2i = 2

8 = 8

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

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

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

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

$=2*25$

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

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

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

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

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

 $\overrightarrow{a}\cdot \widehat{c}=\frac{\alpha +6+2}{\sqrt[]{1}+4+4}$

$\frac{10}{3}=\frac{8+\alpha }{3}\Rightarrow \alpha =2$

$\overrightarrow{b}*\overrightarrow{c}=\widehat{i}\left(2\beta -8\right)+\widehat{j}\left(10\right)+\widehat{k}\left(6+\beta \right)$

$=-6\widehat{i}+10\widehat{j}+7\widehat{k}$

So = 1

Draw g(t) = t³ – 3t

g'(t) = 3(t² – 1)

g(1) is maximum in (-2, 2)

So, maximum (t³ – 3t) = $\{\begin{array}{cc}\hfill t^3-3t\hfill & \hfill ;-2<t<-1\hfill \\ \hfill 2\hfill & \hfill ;-1<t<2\hfill \end{array}$

$I={\displaystyle \underset{-2}{\overset{2}{\int }}f}\left(x\right)dx$

= ${\displaystyle \underset{-2}{\overset{-1}{\int }}\left(t^3-3t\right)dt}+{\displaystyle \underset{-1}{\overset{2}{\int }}2dt}$

I = $\frac{27}{4}$

again rewrite the f(x)

$f\left(x\right)=\left\{\begin{array}{ccc}\hfill \begin{array}{cc}\hfill x^3-3x\hfill & \hfill 2\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{cc}\hfill x\le -1\hfill & \hfill -1<x\le 2\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\hfill x^2+2x-6\hfill & \hfill 9\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{cc}\hfill 2<x<3\hfill & \hfill 3\le x<4\hfill \end{array}\hfill \\ \hfill \begin{array}{ccc}\hfill 10\hfill & \hfill 11\hfill & \hfill 2x+1\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{ccc}\hfill 4\le x<5\hfill & \hfill x=5\hfill & \hfill x>5\hfill \end{array}\hfill \end{array}\right\}$

$f\text{'}\left(x\right)=\left\{\begin{array}{cccccc}\hfill 3x^2-3\hfill & \hfill ;\hfill & \hfill x<-1\hfill & \hfill 0\hfill & \hfill ;\hfill & \hfill -1<x<2\hfill \\ \hfill 2x+2\hfill & \hfill ;\hfill & \hfill 2<x<3\hfill & \hfill 0\hfill & \hfill ;\hfill & \hfill 3<x<4\hfill \\ \hfill 0\hfill & \hfill ;\hfill & \hfill 4<x<5\hfill & \hfill 2\hfill & \hfill ;\hfill & \hfill x>5\hfill \end{array}\right\}$

So f(x) is not differentiable at x = 2, 3, 4, 5

so m = 4

Image of pt (2,4,7) in the plane 3x – y + 4z = 2 is

$\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-2\left(6-4+28-2\right)}{3^2+{\left(-1\right)}^2+4^2}$

Let $\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-28}{13}=\lambda$

$x=3\lambda +2y=-\lambda +4z=4\lambda +7$

Now according to the question

(a, b, c) = $\left(3\lambda +2,-\lambda +4,4\lambda +7\right)$

Now $2a+b+2c=6\lambda +4-\lambda +4+8\lambda +14$

= $13\lambda +22$

= 28 + 22 [Use $\lambda =\frac{-28}{13}$ ]

= 6

 $\frac{dy}{dx}=\frac{y}{x}+\frac{\sqrt[]{y^2+16x^2}}{x}$

Put y = $V\cdot x$

differentiable worst = x

$\frac{dy}{dx}=V+x\cdot \frac{dV}{dx}$

V + $x\frac{dV}{dx}=V+\sqrt[]{V^2+16}$

$x\frac{dV}{dx}=\sqrt[]{V^2+16}$

apply variable separable method

${\displaystyle \int \frac{dV}{\sqrt[]{V^2+16}}={\displaystyle \int \frac{dx}{x}+\mathcal{l}nC}}$

$\Rightarrow \mathcal{l}n\left|V+\sqrt[]{V^2+16}\right|=\mathcal{l}nCx$

$\frac{y}{x}+\frac{\sqrt[]{y^2+16x^2}}{x}=Cx$

Given y(1) = 3 C = 8

Now at x = 2

$\frac{y}{2}+\frac{\sqrt[]{y^2+16.4}}{2}=8.2$

y2 + 64 = (32 – y)2

y = 15