Maths

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

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

(t + 1)dx = (2x + (t + 1)³)dt

$\frac{dx}{dt}-\frac{2x}{t+1}={\left(t+1\right)}^2$

I.F. $=e^{{\displaystyle \int -\frac{2}{t+1}dt}}=\frac{1}{{\left(t+1\right)}^2}$  

Solution is

$\frac{x}{{\left(t+1\right)}^2}={\displaystyle \int 1dt}$  

x = (t + c) (t + 1)²

$?$ x (0) = 2 then c = 2

x = (t + 2) (t + 1)²

 x (1) = 12

${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{8\sqrt[]{2}cosx}{\left(1+e^{sinx}\right)\left(1+si{n}^{4}x\right)}}$ dx

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left\{\left(\frac{8\sqrt[]{2}cosx}{\left(1+e^{sinx}\right)\left(1+si{n}^{4}x\right)}+\frac{8\sqrt[]{2}cosx}{\left(1+e^{-sinx}\right)\left(1+si{n}^{4}x\right)}\right)\right\}dx}$            

$=8\sqrt[]{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{cosx}{1+si{n}^{4}x}dx}$            

Let sin x = t

$I=8\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^4}}$            

$=4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}}dt}$       

$=4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{t^2}\right)dt}{{\left(t-\frac{1}{t}\right)}^2+2}}-4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1-\frac{1}{t^2}\right)dt}{{\left(t+\frac{1}{t}\right)}^2-2}}$            

$=4\sqrt[]{2}\cdot \frac{1}{\sqrt[]{2}}{\left(ta{n}^{-1}\frac{t-\frac{1}{t}}{\sqrt[]{2}}\right)}_0^1-4\sqrt[]{2}\cdot \frac{1}{2\sqrt[]{2}}{\left[log\left|\frac{t+\frac{1}{t}-\sqrt[]{2}}{t+\frac{1}{t}+\sqrt[]{2}}\right|\right]}_0^1$         

$=2\pi -2log\left|\frac{2-\sqrt[]{2}}{2+\sqrt[]{2}}\right|$        

$=2\pi +2log\left(3+2\sqrt[]{2}\right)$           

a = b = 2

3, 7, 11, 15, 19, 23, 27, . 403 = AP₁

2, 5, 8, 11, 14, 17, 20, 23, . 401 = AP₂

so common terms A.P.

11, 23, 35, ., 395

->395 = 11 + (n – 1) 12

->395 – 11 = 12 (n – 1)

$\frac{384}{12}=n-1$    

32 = n – 1

n = 33

Sum =  $\frac{33}{2}$ [2*11+ (32)12]

=  $\frac{33}{2}$ [22 + 384]

= 6699

|A| = 3

|B| = 1

->|C| = |ABA^T| = |A|B|A⁷| = |A|²|B|

= 9

->|X| = |A|C|²|A^T|

= 3 * 9² * 3 = 9 * 9² = 729