Maths

Surface area, S = 4pr²

$?\frac{ds}{dt}=4?.2r\frac{dr}{dt}=8?\frac{dr}{dt}=constant=k\left(say\right)$                

$?\frac{ds}{dt}=k?s=kt+c$

$?4?r^2=kt+c$        

Initially t = 0, r = 3

c = 36 p

When t = 5, r = 7, k = 32p

When t = 9, r = r, r = 9

(3 * 3 * 3 * ………2022 times) ¸ 5

Remainder = (-1) (-1) (-1) ….1011 times)

Remainder = -1 + 5 = 4

$A=\left\{z\in c:1\le \left|z-\left(1+i\right)\right|\le 2\right\}$

$\left|z-\left(1+i\right)\right|\ge 1$

and $$ $\left|z-\left(1+i\right)\right|\le 2$

and also $\left|z-\left(1-i\right)\right|=1$  $$

hence B has infinite set.

S (4, 4) and V (3, 2)

$\therefore$ point of intersection of directrix with axis of parabola is A (2, 0)

Image of A (2, 0) with respect to line

$\therefore x+2y=6isB\left(x_2,y_2\right)$

$\therefore \frac{x_2-2}{1}=\frac{y_2-0}{2}\Rightarrow \frac{-2\left(2+0-6\right)}{5}$

$\therefore B\left(\frac{18}{5},\frac{16}{5}\right)$

Point B is point of intersection of directrix with axes of parabola P₂.

$\therefore x+2y=\lambda$

$B\left(\frac{18}{5},\frac{16}{5}\right)$ lies on the line x + 2y = $\lambda$  $\therefore \lambda =\frac{18}{5}+\frac{32}{5}=10$

 $\frac{x^2}{a^2}-\frac{y^2}{1}=1$

Length of latus rectum = $\frac{2}{a}$

$and\frac{x^2}{4}+\frac{y^2}{3}=1$

length of latus rectum = $\frac{6}{2}$ = 3

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

$12\left(e_H^2+e_H^2\right)=12\left[\left(1+\frac{9}{4}\right)+\left(1-\frac{3}{4}\right)\right]=12\left[\frac{13}{4}+\frac{1}{4}\right]$ = 12 * $\frac{14}{4}=42$

Since student guesses only two wrong. So there are three possibilities

(i) both wrong in section A

(ii) both wrong in section B

(iii) one wrong in each section A and B.

$\therefore$ Required possibilities =

${=}^4{C}_4{*}^6{C}_4{\left(\frac{3}{4}\right)}^4*{\left(\frac{1}{4}\right)}^4{\left(\frac{3}{4}\right)}^2{+}^4{C}_3{*}^6{C}_5{\left(\frac{3}{4}\right)}^3{\left(\frac{1}{4}\right)}^5*\frac{1}{4}*\frac{3}{4}$ ${+}^4{C}_2{*}^6{C}_6*{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^2*{\left(\frac{1}{4}\right)}^6$

$=\frac{27}{4^{10}}\left[15*27+24*3+2\right]=\frac{27*479}{4^{10}}$

OM² = OP² = PM²

^{$\left|\frac{1+r}{\sqrt[]{2}}\right|=r^2-1$}

^{$\therefore r=3$}

^$\therefore$ equation of circle is ${\left(x-1\right)}^2+{\left(y-3\right)}^2=3^2$

$\therefore h+k+r=7$

Required area

= ${\displaystyle {\int }_{-4}^2\left(4-y-\frac{y^2}{2}\right)dy}$

= ${\left(4y-\frac{y^2}{2}-\frac{y^3}{6}\right)}_{-4}^2=18squnits$

S = 1 + 3 + 3² + 3³ + ….+ 3²⁰²¹   $=\frac{3^{2022}-1}{2}=\frac{1}{2}\left[a^{1011}-1\right]$

^{$=\frac{1}{2}\left[9^{1011}-1\right]=\frac{1}{2}\left[100k+10110-1-1\right]$}                          

= 50k₁ + 4

Remainder = 4

$\left\{a\in \left(1,2,3,......,100\right):HCF\left(a,24\right)=1\right\}$

HCF of (a, 24) = 1 $\therefore$ a = 1, 5, 7, 11, 13, 17, 19, 23 sum of these numbers = 96

$\therefore$ There are four such blocks and a number 97 is there upto 100.

$\therefore$ complete sum = 96 + (24 * 8 + 96) + (48 * 8 + 96) + (72 * 8 + 96) + 97 = 1633