Ncert Solutions Maths class 11th

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

(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)

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

(k-3)/ (h-2) * (k-0)/ (h-0) = -1
⇒ k (2k – 3) = -2 (h – 2)h
⇒ 2h² + 2k² – 4h – 3k = 0
2x² + 2y² – 4x – 3y = 0
(0,0)

Equation of tangent is (x/a) (1/2) + (y/b) (√3/2) = 1
Equation of auxiliary circle is x² + y² = a²
Homogenising (ii) with (i) and making coefficient of x² + coefficient of y² = 0
⇒ (3a²/4b²) - (7/4) = 0 ⇒ e = 2/√7

log? (abc) = 6 ⇒ abc = 6?
a = b/r & c = br
⇒ b = 36 and a = 36/r
⇒ r = 2,3,4,6,9,12,18
Also b - a = 36 (1 - 1/r) is a perfect cube.
∴ r = 4 ⇒ a + b + c = 9 + 36 + 144 = 189

β² – 6β + 12 = 0 β − 6 = -12/β
∴ Reqd. Expression is (α – 2)³)? + (-12)¹² / (αβ)¹²) - 1
= (α³ – 8 – 6α (α – 2)? + (12)¹² / (12)¹²) - 1 = (α (α² – 6α + 12) – 8)? = 8? = 2¹²

Solving, x² – 9 = kx² ⇒ x² (k − 1) + 9 = 0 ⇒ x? + x? = 0 and x? = 9 / (k-1)
|x? - x? | = 10 = √ (x? + x? )² - 4x? x? ) ⇒ k = 16/25