Class 11th

$t_n=\frac{1}{2^n.3^{11-n}}={\left(\frac{1}{3}\right)}^{11}.{\left(\frac{3}{2}\right)}^n$

$S_{10}={\left(\frac{1}{3}\right)}^{11}.\frac{\left(\frac{3}{2}\right).\left({\left(\frac{3}{2}\right)}^{10}-1\right)}{\frac{3}{2}-1}$

$=\frac{1}{3^{10}}.\frac{3^{10}-2^{10}}{2^{10}}$

$=1{+}^{10}{C}_1.2^1{+}^{10}{C}_2.2^2+....{+}^{10}{C}_9.2^9$

k = 21 + 6m ® R = 3

= A + B (say)

(1 + x)¹⁵ =

differentiating 15 (1 + x)¹⁴ =   $\sum r{.}^{15}{C}_r{x}^{r-1}$

put  = x = -1

$\Rightarrow 0={-}^{15}{C}_1+2{.}^{15}{C}_2-.......=A$      

  $B{=}^{14}{C}_{13}=2^{13}$            

$\therefore A+B=2^{13}-14$           

$e^{\left(co{s}^{2}x+co{s}^{4}x+co{s}^{6}x+.....\infty \right)}lo{g}_{e}2$  

$=e^{\frac{co{s}^{2}x}{1-co{s}^{2}x}lo{g}_{e}2}=e^{co{t}^{2}xlo{g}_{e}2}=2^{co{t}^{2}x}$          

t² – 9t + 8 = 0

(t – 8) (t – 1) = 0

$t=2^{co{t}^{2}x}=8=2^3$

$\Rightarrow co{t}^{2}x=3=co{t}^2\frac{\pi }{6}$

$\frac{2sinx}{sinx+\sqrt[]{3}cosx}=\frac{2*\frac{1}{2}}{\frac{1}{2}+\sqrt[]{3}*\frac{\sqrt[]{3}}{2}}=\frac{1}{2}$            

x² + y² = 36         ……(i)

y² = 9x      .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A={\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{36-x^2}-3\sqrt[]{x}\right)dx}$  

$={\left[\frac{x\sqrt[]{36-x^2}}{2}+18si{n}^{-1}\frac{x}{6}\right]}_0^3-3.{\frac{x^{3/2}}{3/2}]}_0^3$   

  $=3\pi -\frac{3\sqrt[]{3}}{2}$           

$\therefore$ Required area = $=\frac{1}{2}\pi {\left(6\right)}^2+2\left(3\pi -\frac{3\sqrt[]{3}}{2}\right)=\left(24\pi -3\sqrt[]{3}\right)sq.unit$

BC = CN = x = 75

$cot\theta =\frac{3h}{75}$            

$tan\theta =\frac{h}{75}$          

$\frac{3h^2}{75*75}=1\Rightarrow h=\frac{75}{\sqrt[]{3}}=25\sqrt[]{3}$

p + q = 2

p⁴ + q⁴ = 272

  ${\left(p^2+q^2\right)}^2-2p^2{q}^2=272$          

${\left[{\left(p+q\right)}^2-2pq\right]}^2-2p^2{q}^2=272$           

Let pq = t -> (4 – 2t)² – 2t² = 272

2t² – 16 t – 256 = 0

->t = pq = 16

Required equation x² – 2x + 16 = 0

A  (A -> B) -> B is a tautology

Number of mol of A reacted = 1*10? = 10?
Number of molecule of A = 10? * 6 *10²³
Quantum efficiency = (6 *10¹? ) / (1.2 *10¹? ) = 0.50

According to question,

  $\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$         

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$           

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$  

Obviously B (2, 2) satisfying condition (i)