Maths

$\left(2x-10y^3\right)dy+ydx=0$              

$\frac{dx}{dy}+\frac{2x}{y}-10y^2=0$              

$\frac{dx}{dy}+\frac{2}{y}x=10y^2\to$ Linear differential equation

$P=\frac{2}{y},Q=10y^2$              

$Nowputx=2,y=\beta then2{\beta }^2=2{\beta }^5-2$              

or ${\beta }^5-{\beta }^2-1=0$  

So B will be roots of $y^5-y^2-1=0$  

Required equation of plane will be (x – y – z – 1) + $\lambda$ (2x + y – 3z + 4) = 0

Given ${\perp }^r$ distance of (i) from origin = $\sqrt[]{\frac{2}{21}}$  

$\frac{\left|4\lambda -1\right|}{\sqrt[]{{\left(2\lambda +1\right)}^2+{\left(\lambda -1\right)}^2+{\left(3\lambda +1\right)}^2}}=\sqrt[]{\frac{2}{21}}$   

$\Rightarrow \lambda =\frac{1}{2}or\frac{15}{154}$

So plane be 4x – y – 5z + 2 = 0 for $\lambda =\frac{1}{2}$

number of elements in $A\cap B=5$ which is

(0, 0) (1, 0) (1, 1) (1, -1) (2, 0)

Similarly number of elements in $A\cap C=5$ which is

(2, 0) (2, 2) (1, 1) (2, 1) (3, 1)

Hence number of relation from

$\left(A\cap B\right)to\left(A\cap C\right)=2^{5*5}=2^{25}$            

->P = 25

$\frac{z-i}{z+2i}\in R$

So, $\frac{z-i}{z+2i}=\left(\frac{\overline{z-i}}{z+2i}\right)$

$\frac{z-i}{z+2i}=\frac{\overline{z}+i}{\overline{z}-2i}orz\overline{z}-i\overline{z}-2iz-2=z\overline{z}+2i\overline{z}+iz-2$       

$\Rightarrow z+\overline{z}=0$    

=> z is purely imaginary

i.e. x = 0 if z = x + iy

so, z = iy

=> S is a straight line in complex plane

$\left(p\wedge \left(p\to q\right)\wedge \left(q\to r\right)\right)\to r$

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

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

$=\sim p\vee \sim q\vee \sim r\vee r$

=> tautology

Breadth = b – 2x

& height = x

Let volume V = $\left(a-2x\right)\left(b-2x\right)x$  

For minimum volume $\frac{dv}{dx}=0$  

$\left(a-2x\right)\left(b-2x\right)-2\left(a-2x\right)x-2\left(b-2x\right)x=0$              

Since x =  $\left\{\left(a+b\right)+\sqrt[]{a^2+b^2-ab}\right\}/6$ not possible because maxima occurs

Given curve is f(x) + xf'(x) = x² i.e. y + x $\frac{dy}{dx}=x^2$

where P = $\frac{1}{x},Q=x$

$I.F.=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{1}{x}dx}}=e^{lnx}=x$

Solution be y.x = ${\displaystyle \int x.xdx}$

$xy=\frac{x^3}{3}+c.........\left(i\right)$

(i) passes through (-2, 2) then $-4=\frac{-8}{3}+c$

$c=\frac{-4}{3}$        

(i) -> 3xy = x³ – 4

or x³ – 3xf(x) – 4 = 0

In $\Delta BCD,tan?=\frac{x}{a+b}........\left(i\right)$  

In $\Delta APC,tan\left(\theta +?\right)=\frac{x}{b}..........\left(ii\right)$  

Now tan θ = tan $\left(\theta +?-?\right)$  

$=\frac{tan\left(\theta +?\right)-tan?}{1+tan\left(\theta +?\right)tan?}$  

given , $tan\theta =\frac{1}{2}so\frac{ax}{b\left(a+b\right)+x^2}=\frac{1}{2}$  

-> x² – 2ax + b (a + b) = 0

Equation of ${\perp }^r$ bisector of

$AB:y-3=\frac{t}{3}\left(x-t\right)$

For C put x = 0 so C $\left(0,3-\frac{t^2}{3}\right)$

$h=\frac{t}{2}\&K=\frac{6-\frac{t^2}{3}}{2}$

$2k=6-\frac{1}{3}*4h^2$          

$2x^2+3y-9=0$    

Required probability = probability of both getting 0 head or 1 head or 2 head or 3 head

 = $\frac{5}{16}$