Maths

R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12), . (20, 20)}

n (R₁) = 66

R₂ = {a is integral multiple of b}

So n (R₁ – R₂) = 66 – 20 = 46

as R₁ Ç R₂ = { (a, a) : a Î s} = { (1, 1), (2, 2), ., (20, 20)}

RHL $\underset{x\to 0^{+}}{lim}\frac{co{s}^{-1}\left(1-x^2\right)si{n}^{-1}\left(1-x\right)}{x-x^3}$

$\underset{x\to 0^{+}}{lim}\frac{\pi }{2}\cdot \frac{co{s}^{-1}\left(1-x^2\right)}{x}$

$\frac{\pi }{2}\underset{x\to 0^{+}}{lim}\frac{-1}{\sqrt[]{{\left(1-\left(1-x^2\right)\right)}^2}}\left(-2x\right)$

$=\frac{\pi }{2}\underset{x\to 2^{+}}{lim}\frac{2x}{\sqrt[]{2x^2-x^4}}=\pi \underset{x\to 0^{+}}{lim}\frac{x}{x\sqrt[]{2-x^2}}$

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

LHL $\underset{x\to 0^{+}}{lim}\frac{co{s}^{-1}\left(1-{\left(1+x\right)}^2\right)si{n}^{-1}\left(1-\left(1+x\right)\right)}{1\cdot \left(1-{\left(1+x\right)}^2\right)}$

$=\underset{x\to 0^{-}}{lim}\frac{co{s}^{-1}\left(-x^2-2x\right),si{n}^{-1}\left(-x\right)}{-x^2-2x}$            

$=\frac{\pi }{2}\underset{x\to 0^{-}}{lim}\frac{-si{n}^{-1}x}{-x\left(x+2\right)}=\frac{\pi }{2}*\frac{1}{2}=\frac{\pi }{2}$            

$tanB*tanC=\frac{\sqrt[]{x}}{\sqrt[]{x^2+x+1}}*\frac{1}{\sqrt[]{x\left(x^2+x+1\right)}}$

$=\frac{1}{x^2+x+1}=ta{n}^{2}A$            

tan² A = tan B tan C

It is only possible when A = B = C at x = 1

A = 30°, B = 30°, C = 30° $\left[tanA=tanB=tanC=\frac{1}{\sqrt[]{3}}\right]$                                 

$A+B=\frac{\pi }{2}-C$

$gof\left(x\right)=\{\begin{array}{cc}\hfill f\left(x\right),\hfill & \hfill f\left(x\right)<0\hfill \\ \hfill f\left(x\right),\hfill & \hfill f\left(x\right)>0\hfill \end{array}$

$=\{\begin{array}{lll}{e}^{lnx}=x\hfill & \left(0,1\right)\hfill & e^{-x}\hfill \\ \left(-\infty ,0\right)\hfill & lnx\hfill & \left(1,\infty \right)\hfill \end{array}$

Therefore, gof (x) is many one and into

$\sqrt[]{3}-\sqrt[]{2}=\frac{\left(\sqrt[]{3}+\sqrt[]{2}\right)\left(\sqrt[]{3}-\sqrt[]{2}\right)}{\left(\sqrt[]{3}+\sqrt[]{2}\right)}=\frac{1}{\sqrt[]{3}+\sqrt[]{2}}$

Let $\sqrt[]{3}+\sqrt[]{2}=t$  

$t^x+\frac{1}{t^x}=\frac{10}{3}$

Let  $t^x=y$ $y+\frac{1}{y}=\frac{10}{3}$           

y = 3 or   $\frac{1}{3}$

${\left(\sqrt[]{3}+\sqrt[]{2}\right)}^x=3$ or $\frac{1}{3}$  

$xlog\left(\sqrt[]{3}+\sqrt[]{2}\right)=ln3$ or –ln3

$x=\frac{ln3}{ln\left(\sqrt[]{3}+\sqrt[]{3}\right)}$   or $\frac{-ln3}{\sqrt[]{3}+\sqrt[]{2}}$  

two real values of x

$f\left(x\right)=\{\begin{array}{cc}\hfill e^{-x},\hfill & \hfill x<0\hfill \\ \hfill lnx,\hfill & \hfill x>0\hfill \end{array}$

$g\left(x\right)=\{\begin{array}{cc}\hfill e^x,\hfill & \hfill x<0\hfill \\ \hfill x,\hfill & \hfill x>0\hfill \end{array}$

|r₁ – r₂| < c₁c₂ < r₁ + r₂

-> $\left|2-\sqrt[]{4{\lambda }^2-9}\right|<\left|2\lambda \right|<2+\sqrt[]{4{\lambda }^2-9}$

$\left|2\lambda \right|-2<\sqrt[]{4{\lambda }^2-9}$    

$4{\lambda }^2+4-8\left|\lambda \right|<4{\lambda }^2-9$

  $\lambda >\frac{13}{8},\lambda <-\frac{13}{8}$           

$\sqrt[]{4{\lambda }^2-9}>0$

$\lambda >\frac{3}{2},\lambda <-\frac{3}{2}$

$\lambda \in \left(-\infty ,-\frac{13}{8}\right)\cup \left(\frac{13}{8},\infty \right)$           

Now,

$\left|2-\sqrt[]{4{\lambda }^2-9}\right|<\left|2\lambda \right|$            

$4+4{\lambda }^2-9-4\sqrt[]{4{\lambda }^2-9}<4{\lambda }^2$  

$4\sqrt[]{4{\lambda }^2-9}>-5\Rightarrow \lambda \in R$  

$\lambda \in \left(-\infty ,-\frac{13}{8}\right)\cup \left(\frac{13}{8},\infty \right)$  

P (2W and 2B) = P (2B, 6W) * P (2W and 2B)

+ P (3B, 5W) * P (2W and 2B)

+ P (4B, 4W) * P (2W and 2B)

+ P (5B, 3W) * P (2W and 2B)

+ P (6B, 2W) * P (2W and 2B)

(15 + 30 + 36 + 30 + 15)

$=\frac{36}{126}$

$=\frac{18}{63}$

$=\frac{6}{21}$

$=\frac{2}{7}$

x² – y² cosec²q = 5 $\frac{x^2}{1}-\frac{y^2}{si{n}^2\theta }=5$                        

x² cosec²q + y² = 5  $\frac{x^2}{si{n}^2\theta }+\frac{y^2}{1}=5$        

$e_H=\sqrt[]{7}{e}_e$                  

and $e_H=\sqrt[]{1+\frac{si{n}^2\theta }{1}}$  

-> $\sqrt[]{1+si{n}^2\theta }=\sqrt[]{7}\sqrt[]{1-si{n}^2\theta }$

1 + sin²q = 7 – 7 sin²q

->8sin²q = 6

-> $sin\theta =\sqrt[]{\frac{3}{4}}=\frac{\sqrt[]{3}}{2}$  

-> $\theta =\frac{\pi }{3}$  

5f(x) + 4f $\left(\frac{1}{x}\right)$  = x² – 4           .(1)

Replace x by  $\frac{1}{x}$

5f  $\left(\frac{1}{x}\right)$  + 4f(x) = $\frac{1}{x^2}$  – 4   .(2)

5 * equation (1) – 4 * equation (2)

$9f\left(x\right)=5x^{2-}\frac{4}{x^2}-4$            

$y=9f\left(x\right)\cdot x^2=\frac{5x^4-4-4x^2}{x^2}{x}^2$            

y = 5x⁴ – 4 – 4x²

y = 20x³ – 8x > 0

4x(5x² – 2) > 0

    $x\in \left(-\sqrt[]{\frac{2}{5}},0\right)\cup \left(\sqrt[]{\frac{2}{5}},\infty \right)$

Total ways to partition 5 into 4 parts are:

5 0 

4 1 0  $\frac{5!}{4!}=5$

3 2 0 $\frac{5!}{3!\cdot 2!}=10$

3 1 0 $\frac{5!}{2!2!2!}=15$

2 1 $\frac{5!}{2!*3!}=10$

51 Total way