Maths

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

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

 = $\frac{5}{16}$

$l=\underset{n\to \infty }{lim}{\left(u_n\right)}^{\frac{-4}{n^2}}=\underset{n\to \infty }{lim}{\left\{\left(1+\frac{1}{n^2}\right){\left(1+\frac{2^2}{n^2}\right)}^2{\left(1+\frac{3^2}{n^2}\right)}^3......{\left(1+\frac{n^2}{n^2}\right)}^n\right\}}^{\frac{-4}{n^2}},$ Taking log on both the sides

$\underset{n\to \infty }{lim}\frac{-4}{n^2}{\displaystyle \sum _{r=1}^{n}rlog\left(1+\frac{r^2}{n^2}\right)}=\underset{n\to \infty }{lim}\frac{-4}{n}{\displaystyle \sum _{r=1}^n\frac{r}{n}log\left(1+{\left(\frac{r}{n}\right)}^2\right)}$

$\therefore logl=-4{\displaystyle \underset{0}{\overset{1}{\int }}xlog\left(1+x^2\right)dx}$

$logl=-2\left[log4-1\right]=-2log\frac{4}{e}=log\frac{e^2}{16}$       

$l=\frac{e^2}{16}$

x + y + z = 4

3x + 2y + 5z = 3

$9x+4y+\left(28+\left[\lambda \right]\right)z=\left[\lambda \right]$           

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 9\hfill & \hfill 4\hfill & \hfill 28+\left[\lambda \right]\hfill \end{array}\right|=56+2\left[\lambda \right]-20-\left(84+3\left[\lambda \right]-45\right)+\left(-6\right)$

$=-\left[\lambda \right]-9$              

$If\left[\lambda \right]+9\ne 0$ then unique solution

& if  $\left[\lambda \right]+9=0then{\Delta }_1={\Delta }_2={\Delta }_3=0$ so infinite solution will exist

Hence $\lambda \in R$  

Required probability of obtaining a (sum = 7) = $\frac{13}{96}\left(given\right)$  

    $i.e.2\left(\left(\frac{1}{6}+x\right)\left(\frac{1}{6}-x\right)+\frac{1}{6}*\frac{1}{6}+\frac{1}{6}*\frac{1}{6}\right)=\frac{13}{96}$

$x=\frac{1}{8}$