Maths

$\frac{a^{a^x}+a^{1-a^x}}{2}\ge \sqrt[]{a^{a^x}.a^{1-a^x}}$ $\left(AM\ge GM\right)$

$\Rightarrow a^{a^x}+a^{1-a^x}\ge 2\sqrt[]{a}$

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

P {a person chosen from the group has chest disorder}

$=\frac{160}{400}*0.35+\frac{100}{400}*0.20+\frac{140}{400}*0.10$

$P\left(\frac{Personissmoker\&non-vegetarian}{Personhaschestdisorder}\right)$

$=\frac{\frac{160}{400}*0.35}{\frac{160}{400}*0.35+\frac{100}{400}*0.20+\frac{140}{400}*0.10}$

$=\frac{40*0.7}{40*0.7+25*0.4+35*0.2}=\frac{23}{23+10+7}=\frac{28}{45}$

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

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

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

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l}a=4c=1\\ a=2c=2\\ a=1c=4\end{array}\}3ways$  

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability =  $\frac{5}{216}$   

Let sinθ= t ${\displaystyle \int \frac{sin\theta \left(2sin\theta .cos\theta \right)\left(si{n}^6\theta +si{n}^4\theta +si{n}^2\theta \right)\sqrt[]{2si{n}^4\theta +3si{n}^2\theta +6}}{2si{n}^2\theta }}$ dθ

sinθ = t

cos θ. dθ = dt

${\displaystyle \int \frac{u^{1/2}}{12}}du=\frac{u^{3/2}}{18}+C=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C=\frac{{\left(2si{n}^6\theta +3si{n}^4\theta +6si{n}^2\theta \right)}^{3/2}}{18}+C$

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)

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