Maths

$2sin\frac{\pi }{8}sin\frac{2\pi }{8}tan\frac{3\pi }{8}sin\left(\pi -\frac{5\pi }{8}\right)sin\left(\pi -\frac{6\pi }{8}\right)sin\left(\pi -\frac{7\pi }{8}\right)$

$={\left(\frac{1}{\sqrt[]{2}}\right)}^2.2si{n}^2\frac{\pi }{8}si{n}^2\frac{3\pi }{8}$

^{$=\frac{1}{4}.{\left(2sin\frac{\pi }{8}sin\frac{3\pi }{8}\right)}^2$}

$=\frac{1}{4}.{\left(2sin\frac{\pi }{8}sin\frac{3\pi }{8}\right)}^2$

${\displaystyle \underset{0}{\overset{5}{\int }}\frac{x+\left[x\right]}{e^{x-\left[x\right]}}}dx$

$={\displaystyle \sum _{r=1}^5{\displaystyle \underset{r-1}{\overset{r}{\int }}\frac{\left(x+1-r\right)}{e^{x+1-r}}}dx}$

Put x + 1 – r = t ->dx = dt

$=5{\left[-t{e}^{-t}-e^{-t}\right]}_0^1$              

$=5\left[-1e^{-1}-e^{-1}+e^0\right]=5\left(1-\frac{2}{e}\right)$

$\Rightarrow \alpha =-10\beta =5\Rightarrow {\left(\alpha +\beta \right)}^2=25$

$\Rightarrow \frac{1+x}{x}\ge 1and\frac{1+x}{x}\le -1$

$\Rightarrow \frac{1}{x}\ge 0\frac{1+2x}{x}\le 0$

Domain $\left[-\frac{1}{2},0\right)\cup \left(0,\infty \right)$

$=\left[-\frac{1}{2},\infty \right)-\left\{0\right\}$  

$f\left(x\right)=\left\{x\right\},g\left(x\right)=1?\left\{x\right\}$

$f\left(x\right)=e^{x^{2}ln\left(\frac{2}{x}\right)}=e^{x^2\left(ln2-lnx\right)}$

$f\text{'}\left(x\right)=e^{x^2\left(ln2-lnx\right)}\left[2x\left(ln2-lnx\right)-x\right]$            

$={\left(\frac{2}{x}\right)}^{x^2}x\left[2\left(ln2-lnx\right)-1\right]$          

$f\text{'}\left(x\right)=0\Rightarrow 2\left(ln2-lnx\right)-1=0$   

2ln x = $ln\frac{4}{e}\Rightarrow x^2=\frac{4}{e}$  

Let P (h, k) be the mid point of the chord x² – y² = 4

$\therefore$ its equation is xh – yk = h² – k²

$Or,y=\left(\frac{h}{x}\right)x+\frac{k^2-h^2}{k}$ if this line is tangent to y² = 8x then $\frac{k^2-h^2}{k}$ = $\frac{2}{h/k}=\frac{2k}{h}$

$h\left(k^2-h^2\right)=2k^2$              

$\therefore$ Required locus is 2y² = x (y² – x²)

$\Rightarrow x^3=y^2\left(x-2\right)$

System of equations can be written as

  $\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill \lambda \hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}5\\ \mu \\ 1\end{array}\right)$             

R₃ – R₂, R₂ – R₁

For unique solution $\lambda \ne 5,\mu \in R.P=\frac{5}{6}$  

For no solution $\lambda =5,\mu \ne 3q=\frac{1}{6}*\frac{5}{6}=\frac{5}{36}$  

  $ta{n}^{-1}2=\theta$

->tan q = 2

$sin\left(\theta -\alpha \right)=\frac{1}{\sqrt[]{5}}$

->4 – 2 tanq = 1 + 2 tan a tan a = $\frac{3}{4}$  

$\alpha =ta{n}^{-1}\frac{3}{4}$          

$P\left(\frac{x\ge 5}{x>2}\right)=\frac{P\left(x\ge 5\cap x>2\right)}{P\left(x>2\right)}=\frac{P\left(x\ge 5\right)}{P\left(x>2\right)}$

$=\frac{1-P\left(x\le 4\right)}{1-P\left(x\le 2\right)}$

$=\frac{1-\left[{\left(\frac{5}{6}\right)}^3*\frac{1}{6}+{\left(\frac{5}{6}\right)}^2*\frac{1}{6}+\left(\frac{5}{6}\right)\frac{1}{6}+\frac{1}{6}\right]}{1-\left[\frac{5}{6}*\frac{1}{6}+\frac{1}{6}\right]}$

$={\left(\frac{5}{6}\right)}^4*{\left(\frac{6}{5}\right)}^2={\left(\frac{5}{6}\right)}^2=\frac{25}{36}$

$C_2=a_2+b_2=\left(a_1-3\right)+2b_1=12$

$a_1+2b_1=15.............\left(i\right)$

$a_1+4b_1=19............\left(ii\right)$              

Solving (i) & (ii), b₁ = 2, a₁ = 11

$=5\left[22+9\left(-3\right)\right]+2\left(\frac{2^{10}-1}{2-1}\right)$

$=2^{11}-27=2^6*2^5-27=2021$