Maths

$\alpha +\beta =-\frac{\lambda }{3},\alpha \beta =-\frac{1}{3}$

$\therefore \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=15\Rightarrow \lambda =\pm 3$

$\therefore 6{\left({\alpha }^3+{\beta }^3\right)}^2=6{\left({\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right)\right)}^2=6*{\left(-\left(1+1\right)\right)}^2=24$

For inconsistent, $\Delta =0,\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 2a\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\alpha \hfill & \hfill 5\hfill \end{array}\right|=0$

a = 1

$\therefore {\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right|=1\left(1\right)+1\left(12+5\right)+1\left(-3-8\right)=7\ne 0$              

Equation of tangent to the circle at (2, 4) is

(4 + A) x + (8 + B) y + 2A + 2B + 2C = 0……. (i)

Also equation of tangent to parabola y = x² at (2, 4) is

4x – y = 4 ………. (ii)

Comparing (i) and (ii) we get, A + C = 16

  $\frac{6}{11}$ = Required probability         

After solving, we get n = 4

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 =

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

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