Class 11th

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 $\Rightarrow \left|\frac{1-\frac{1}{2}+1}{\sqrt[]{2}}\right|=r$   

$r=\frac{3}{2\sqrt[]{2}}$       

2x + y = 1

$m_1=\frac{-2}{1}=-2and{m}_1{m}_2=-1$          

$-2.m_2=-1$           

$m_2=\frac{1}{2}$           

y² = 6x

y² = 4 (3/2) x

$y=mx+\frac{a}{m}$         

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

$y=\frac{x}{2}+3$           

$A\to \left(B\to A\right)$

$\sim A\vee \left(B\to A\right)$

$\sim A\vee \left(\sim B\vee A\right)$

$\left(\sim A\vee B\right)\vee \left(\sim A\vee A\right)=tA\to \left(A\cup B\right)$

 $cosec\left(2co{t}^{-1}\left(5\right)+co{s}^{-1}\left(\frac{4}{5}\right)\right)$

Let cot^-1 (5) = and cos^-1 $\left(\frac{4}{5}\right)=\alpha$

cot = 5 cos = $\frac{4}{5}$

$=cosec\left(2\theta +\alpha \right)=\frac{1}{sin\left(2\theta +\alpha \right)}$

$as\{\begin{array}{l}sin2\theta =\frac{2tan\theta }{1+ta{n}^2\theta }=\frac{2\left(\frac{1}{5}\right)}{1+\frac{1}{25}}=\frac{5}{13}\\ cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=\frac{1-\frac{1}{25}}{1+\frac{1}{25}}=\frac{12}{13}\end{array}$

Let A = {a, b, c}, B = {1, 2, 3, 4, 5} n (A * B) = 15

x = number of one-one functions from A to B.

${=}^5{C}_3.3!=60$

y = number of one-one functions for A to (A * B)

${=}^{15}{C}_3.3!=15*14*13=2730$

$\frac{Y}{x}=\frac{2730}{60}\Rightarrow 2y=91x$