Maths

$\frac{dy}{dx}+\frac{2x}{x-1}\cdot y=\frac{1}{{\left(x-1\right)}^2}$

IF $=e^{{\displaystyle \int \frac{2x}{x-1}dx}}$

= $e^{2x}\cdot {\left(x-1\right)}^2$

$y\cdot e^{2x}{\left(x-1\right)}^2=\{e^{2x}\frac{{\left(x-1\right)}^2}{{\left(x-1\right)}^2}dx+C$

y = $\frac{e^{2x}}{2{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^2}$

y(2) = $\frac{1+e^4}{2e^4},\Rightarrow C=\frac{1}{2}$ $\frac{{}^{}}{{}^{}}\frac{}{}$

y(3) = $\frac{e^{\alpha }+1}{\beta e^{\alpha }}=\frac{e^6+1}{8e^6}$

Data contradiction.

$\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\overrightarrow{c}$

Kindly consider the following figure

pva $\left(\sim rvp\right)$ $$

$\sim \left(p\vee q\right)\vee \left(\sim rvp\right)$

its negation as asked in question

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

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

= $\left(p\wedge r\wedge \sim p\right)\left[asp\wedge \sim pisfalse\right]$

Let base = b

$tan60°=\frac{h}{b}$

$tan30°=\frac{h-20}{b}$

$Mean=\frac{3+12+7+a+\left(43-a\right)}{5}=13$  

Variance = $\frac{3^2+12^2+7^2+a^2+{\left(43-a\right)}^2}{5}-{\left(13\right)}^2$  

$\Rightarrow \frac{2a^2-a+1}{5}\to Naturalnumber$      

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not $4\lambda or4\lambda +1from.$  

As each square form is $4\lambda or4\lambda +1$  

Total number of possible relation = $2^{n^2}=2^4=16$ ${}^{{}^{}}{}^{}$

Favourable relations = $?,\left\{\left(x,x\right)\right\},\left\{\left(y,y\right)\right\}$ $$

$\left\{\left(x,x\right),\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right),\left(x,y\right),\left(y,x\right)\right\}$

Probability = $\frac{5}{16}$

Consider the equation of plane,

$P:\left(2x+3y+z+20\right)+\lambda \left(x-3y+5z-8\right)=0$  

$?$  Plane P is perpendicular to 2x + 3y + z + 20 = 0

So,  $4+2\lambda +9-9\lambda +1+5\lambda =0$  

0  $\lambda =7$  

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

$\left(2,-\frac{1}{2},2\right)$  

In plane P is (a, b, c) then

$\frac{a-2}{1}=\frac{b+\frac{1}{2}}{-2}=\frac{c-2}{4}$  

and  $\left(\frac{a+2}{2}\right)-2\left(\frac{b-\frac{1}{2}}{2}\right)+4\left(\frac{c+2}{2}\right)=4$     

clearly 

$a=\frac{4}{3},b=\frac{5}{6}andc=-\frac{2}{3}$  

So, a : b : c = 8 : 5 : -4

$?{\mathcal{l}}_{1}and{\mathcal{l}}_2$ are perpendicular, so

$3*1+\left(-2\right)\left(\frac{\alpha }{2}\right)+0*2=0$

⇒a = 3

Now angle between  ${\mathcal{l}}_{2}and{\mathcal{l}}_3$ ,

$cos\theta =\frac{1\left(-3\right)+\frac{\alpha }{2}\left(-2\right)+2\left(4\right)}{\sqrt[]{1+\frac{{\alpha }^2}{4}+4}.\sqrt[]{9+4+16}}$

$\Rightarrow cos\theta =\frac{\frac{2}{29}}{2}\Rightarrow \theta =co{s}^{-1}\left(\frac{4}{29}\right)=se{c}^{-1}\left(\frac{29}{4}\right)$

Let P (at², 2 at) where

$a=\frac{3}{2}$

T : yt = x + at² so point Q is

$\left(-a,at-\frac{a}{t}\right)$

N : y = -tx + 2at + at³ passes through (5, -8)

$-8=-5t+3t+\frac{3}{2}{t}^3$

$3t^3-4t+16=0$

⇒ t = -2

  So ordinate of point Q is $-\frac{9}{4}$