Class 11th

Equation of given line x – y + 1 = 0 .(i),

equation of per perpendicular line PP' is

-x – y + $\lambda =0$  

As line passing through (3, 5)

$\therefore -3-5+\lambda =0$   

$\lambda =8$  

Equation of line PP' is –x – y + 8 = 0 .(ii)

Solving (i) and (ii)

$\begin{array}{l}-2y=-9\\ y=\frac{9}{2}\end{array}\}Q=\left(\frac{7}{2},\frac{9}{2}\right)$

$y=\frac{9}{2}$

$\therefore x-\frac{9}{2}+1=0$

$P\text{'}=\left(4,4\right)$           

$\frac{5+y_1}{2}=\frac{9}{2}$           

y₁ = 4

According to KTG, the gas exerts pressure because its molecule :

suffer change in momentum when impinge on the walls of container.

$x^2-2\left(3k-1\right)x+8k^2-7>0\Rightarrow a>0,\&D<0,$       

a = 1 > 0 and D < 0

4 (3k – 1)² – 4 (8k² – 7) < 0

$\left(9k^2+1-6k-8k^2+7<0\right)$            

$k(k-4)-2(k-4)<0$    

$k\in \left(2,4\right)$           

K = 3

  $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=r$

For 'B', x = 2r + 1, y = 3r – 1, z = -2r + 1

As AB is perpendicular to the line,

$\left\{\left(2r+1\right)-0\right\}2+\left\{\left(3r-1\right)-1\right\}3+\left\{\left(-2r+1\right)-2\right\}$

$r=\frac{2}{17}\Rightarrow B\left(\frac{2}{17},\frac{-11}{7},\frac{13}{17}\right)$ ->

direction ratios of AB

(2r + 1, 3r – 2, -2r – 1)

Equation of AB

$\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$  

l + m – n = 0

l + m = n . (i)

l² + m² = n²

Now from (i)

l² + m² = (l + m)²

->2lm = 0

->lm = 0

l = 0 or m = 0

->m = n Þ l = n

if we take direction consine of line

$0,\frac{1}{\sqrt[]{2}},\frac{1}{\sqrt[]{2}}and\frac{1}{\sqrt[]{2}},0,\frac{1}{\sqrt[]{2}}$  

cos α =   $\frac{1}{2}$

$si{n}^4\alpha +co{s}^4\alpha ={\left(\frac{\sqrt[]{3}}{2}\right)}^4={\left(\frac{1}{2}\right)}^4=\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=\frac{5}{8}$           

xyz = 24

24 = 2³ * 3

Let's distribute 2, 3 among 3 variables. No. of positive integral solution =

No. of ways to distribute = 

$\therefore In\Delta AQP$         

$tan30°=\frac{PQ}{AQ}$ $\frac{1}{\sqrt[]{3}}=\frac{h}{x+y}$        

x + y = $\sqrt[]{3}h.....\left(i\right)$  

$\therefore ln\Delta BQP$          

$tan45°=\frac{h}{y}$          

h = y .(ii)

(i) & (ii) x + y = $\sqrt[]{3}y$  

$\Rightarrow x=\left(\sqrt[]{3}-1\right)y.......\left(iii\right)$     

Let the speed be S

$\frac{x}{S}=20$

x = 20.S

from (iii)

$\frac{y}{S}=10\left(\sqrt[]{3}+1\right)$          

sin 2qθ+ tan 2θ> 0

$\frac{2tan\theta }{1+ta{n}^2\theta }+\frac{2tan\theta }{1-ta{n}^2\theta }>0$           

Let tan q = x

  $\frac{2x}{1+x^2}+\frac{2x}{1-x^2}>0$          

$tan\theta <-1or0<tan\theta <1$          

  $\theta \in \left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$          

(2 – i) z = (2 + i) $\overline{z}$ , put z = x + iy

  $y=\frac{x}{2}........\left(i\right)$

(ii) $\left(2+i\right)z+\left(i-2\right)\overline{z}-4i=0$  

x + 2y = 2

(iii) $iz+\overline{z}+1+i=0$  

Equation of tangent x – y + 1 = 0

Solving (i) and (ii)

$x=1.y=\frac{1}{2}\Rightarrow centre\left(1,\frac{1}{2}\right)$

Perpendicular distance of point $\left(1,\frac{1}{2}\right)$ from x – y + 1 = 0 is p = r $\Rightarrow \left|\frac{1-\frac{1}{2}+1}{\sqrt[]{2}}\right|=r$   

$r=\frac{3}{2\sqrt[]{2}}$