Maths

Kindly go through the solution

Use characteristic equation = 0

$V={\left|z\right|}^2+{\left|z-3\right|}^2+{\left|z-6i\right|}^2$  

Put z = x + iy

$V=3\left[{\left(x-1\right)}^2+{\left(y-2\right)}^2+10\right]$                

For minimum x = 1, y = 2, V₀ = 30

So, z₀ = 1 + 2i

f (a) = a, a is max of powers of prime P such that P^a divides a.

f (2) = 1; g (2) = 3

f (3) = 1; g (3) = 4

f (4) = 2; g (4) = 5

f (5) = 1; g (5) = 6

f (2) + g (2) = 4

f (3) + g (3) = 5

f (4) + g (4) = 7

f (5) + g (5) = 7

So, f (x) + g (x) is many-one-and an into function. We won't get 'I' in the range.

For R₁, take a = 1, b = 0, c = -1

$ab\ge 0,bc\ge 0$ but AC < 0

So R₁ is not an equivalence relation.

For R₂, it will not be symmetric.

So R₂ is also not an equivalence relation.

  $\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-2\left(3a-4b+12c+19\right)}{3^2+{\left(-4\right)}^2+12^2}$

$\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-6a+8b-24c-38}{169}$               

$\left(x,y,z\right)\equiv \left(a-6,\beta ,\gamma \right)$               

  $\frac{\left(a-b\right)-a}{3}=\frac{\beta -b}{-4}=\frac{\gamma -c}{12}=\frac{-6a+8b-24c-38}{169}$              

  $\frac{\beta -b}{-4}=-2$              

$\Rightarrow \beta =8+b$               

$\Rightarrow 3a-4b+12c=150$                   ….(i)

a + b + c = 5

$\Rightarrow 3a+3b+3c=15$ ….(ii)

Applying (i) – (ii), we get :

= 56 + 216 + 7b – 9c = 56 + 216 – 135 = 137

Let the equation of circle be

$x\left(x-\frac{1}{2}\right)+y^2+\lambda y=0$               

$\Rightarrow x^2+y^2-\frac{1}{2}x+\lambda y=0$

Radius = $\sqrt[]{\frac{1}{16}+\frac{{\lambda }^2}{4}}=2$  

$\Rightarrow {\lambda }^2=\frac{63}{4}\Rightarrow {\left(x-\frac{1}{4}\right)}^2+{\left(y+\frac{\lambda }{2}\right)}^2=4$   

$?$ This circle and parabola

$y-\alpha ={\left(x-\frac{1}{4}\right)}^2$ touch each other, so

$\alpha =-\frac{\lambda }{2}+2\Rightarrow \alpha -2=-\frac{\lambda }{2}\Rightarrow {\left(\alpha -2\right)}^2=\frac{{\lambda }^2}{4}=\frac{63}{16}$  

${\left(4\alpha -8\right)}^2=63$  

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side

  $?a=\sqrt[]{{\left(3?1\right)}^2+{\left(6?2\right)}^2}=\sqrt[]{20}\text{?}\text{?}and$

$\frac{b}{2}=\frac{4}{\sqrt[]{5}}?b=\frac{8}{\sqrt[]{5}}$

Area

$ab=2\sqrt[]{5}.\frac{8}{\sqrt[]{5}}=16$

  $T_r=\frac{r}{{\left(2r^2\right)}^2+1}$

$=\frac{r}{{\left(2r^2+1\right)}^2-{\left(2r\right)}^2}$

$=\frac{1}{4}\frac{4r}{\left(2r^2+2r+1\right)\left(2r^2-2r+1\right)}$

$S_{10}=\frac{1}{4}{\displaystyle \sum _{r=1}^{10}\left(\frac{1}{\left(2r^2-2r+1\right)}-\frac{1}{\left(2r^2+2r+1\right)}\right)}$

$\Rightarrow S_{10}=\frac{1}{4}.\frac{220}{221}=\frac{55}{221}=\frac{m}{n}$

$\therefore m+m=276$

  $T_{r+1}{=}^{60}{C}_r{\left(x^{\frac{1}{2}}\right)}^{60-r}{\left(x^{-\frac{1}{3}}\right)}^r{\left(5^{\frac{-1}{4}}\right)}^{60-r}{\left(5^{\frac{1}{2}}\right)}^r$

for

$x^{10}\frac{60-r}{2}-\frac{r}{3}=10$    

$\Rightarrow 180-3r-2r=60$             

->r = 24

k = 3 + exponent of 5 in 

= $3+\left(\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]-\left[\frac{24}{5}\right]-\left[\frac{24}{5^2}\right]-\left[\frac{36}{5}\right]-\left[\frac{36}{5^2}\right]\right)$  

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5