Class 11th

A = {1,2,3,4,5} B = {0,1,2,3,4}
No. of elements common in A&B = 4.
∴ No. of elements common in A * B and B * A = 4 * 4 = 16

Centre of C₃ will lie on the radical axis of C₁ and C₂ which is 10x + 6y + 26 = 0.

Let center of C₃ is (h, k).

Equation of chord of contact through (h, k) to the C₁ may be given as hx + ky = 25 … (I)

Let the mid point of the chord is (x₁, y₁) the equation of the chord with the help of mid point may be given as  

$x{x}_1+y{y}_1=x_1^2+y_1^2$

Since (I) and (II) represents same straight line 10x + 6y + 26 = 0

$\Rightarrow \frac{h}{25}=\frac{x_1}{x_1^2+y_1^2},\frac{k}{25}=\frac{y_1}{x_1^2+y_1^2}$            

The locus of (x₁, y₁) is

$5x+3y+\frac{13}{25}\left(x^2+y^2\right)=0$          

a², b², c² . A.P.
a² + 1, b² + 1, c² + 1 . P.
a² + ab + bc + ca, b² + ab + bc + ca, c² + ab + bc + CA . P.
⇒ (a + b) (a + c), (a + b) (b + c), (b + c) (c + a) . P.
⇒ 1/ (b+c), 1/ (c+a), 1/ (a+b) . A.P.
⇒ b + c, c + a, a + b . P.

  $56=\frac{50n+60m}{m+n}$

56 n + 56 m = 50 n + 60 m

6n = 4m

$\frac{n}{m}=2/3$            

$\therefore 9.\frac{n}{m}=9*\frac{2}{3}=6$            

Let S = z + 2z² + 3z³ + . + 18z¹?
zS = z² + 2z³ + . + 18z¹?
(1 − z)S = z + z² + z³ + . - 18z¹?
(1 − z)S = z (z¹? - 1)/ (z-1) - 18z¹?
Now z = cos 20° + isin 20° ⇒ z¹? = 1
Also |z| = 1
⇒ (1 − z)S = -18z
⇒ S = -18z / (1-z)
|S|? ¹ = | (1-z)/ (-18z)| = |1-z|/18
= (1/18) |1 – cos 20° - isin 20°| = (1/18) |2sin² 10° — 2isin 10°cos 10°|
= (1/18) |2sin 10° (sin 10° – icos 10°)| = (1/9)sin 10°

The two altitudes are

$\left(x-\sqrt[]{3}y+1\right)\left(x+\sqrt[]{3}y-5\right)=0$            

Point of int. of the 2 altitudes is  $\left(2,\sqrt[]{3}\right)$  

Let slope of 3^rd altitude be 'm'

then   $\left|\frac{m-\frac{1}{\sqrt[]{3}}}{1+\frac{1}{\sqrt[]{3}}}\right|=\sqrt[]{3}\Rightarrow \frac{\sqrt[]{3}m-1}{\sqrt[]{3}+m}=\pm \sqrt[]{3}$

$\sqrt[]{3}m-1=\pm 3\pm \sqrt[]{3}m$            

  $m=\infty ,=\frac{-2}{2\sqrt[]{3}}=\frac{-1}{\sqrt[]{3}}$          

$\therefore$ The third altitude is x = 2

Kindly consider the following figure

Amphiprotic species are those which behave like acid and base both. They can donate and accept a proton.

(6!/ (5!1!) * 2! + (6!/ (4!2!) * 2! + (6!/ (3!)²2!) * 2! = 62

First element in 20^th group is equal to '20'.

(20, 21, ….upto 39 times)

$\therefore Sum=\frac{39}{2}\left(2*20+\left(39-1\right)*1\right)$