Maths

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

A = ∫? ² lnx dx = 2ln2 – 1
A' = 4 - 2 (2ln2 – 1) = 6 – 4ln2

  $\left(\lambda ,2\lambda ,3\lambda \right)$

$cos\theta =\frac{6}{\sqrt[]{42}}\Rightarrow sin\theta =\frac{\sqrt[]{6}}{\sqrt[]{42}}$ (where q is the angle between L₁ & L₂)

$Area\Delta OAB=\frac{1}{2}\left(OA\right)\left(OB\right)sin\theta$

$=\frac{1}{2}\left(\sqrt[]{3}\right)\left|\lambda \right|\left(\sqrt[]{14}\right)\frac{\sqrt[]{6}}{\sqrt[]{42}}=\sqrt[]{6}\Rightarrow \lambda =\pm 2$

dy/dx = (e? – x)? ¹
dx/dy = e? - x
dx/dy + x = e? ⇒ I.F. = e^ (∫dy) = e?
∴ xe? = ∫e? dy + c
xe? = e²? /2 + c
y (0) = 0 ⇒ c = -1/2
∴ x = e? /2 - (1/2)e? ⇒ e? – e? = 2x
e²? – 2xe? - 1 = 0
e? = x ± √ (x² + 1) ⇒ e? = x + √ (x² + 1)
y = ln (x + √ (x² + 1)

(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)

(5³)^740 * 5² = 5² (125)^740 = 5² (126 – 1)^740
So remainder = 4.

C.V. = (σ/x? ) * 100 ⇒ σ = (C.V. * x? ) / 100
∴ σ? = (50*30)/100 = 15 and σ? = (60*25)/100 = 15 ⇒ σ? - σ? = 0