Maths

 $\Delta =\left|\begin{array}{ccc}\hfill 8\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 1\hfill \\ \hfill \lambda \hfill & \hfill -3\hfill & \hfill 0\hfill \end{array}\right|=12-3\lambda$

So for $\lambda$ = 4, it is having infinitely many solutions. ${\Delta }_x=\left|\begin{array}{ccc}\hfill -2\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \mu \hfill & \hfill -3\hfill & \hfill 0\hfill \end{array}\right|$ = 6 $3\mu =0\Rightarrow -6-3\mu =0$

For $\mu =-2$ distance of $\left(4,-2,-\frac{1}{2}\right)$ from 8x + y + 4z + 2= 0 $\left|\frac{32-2-2+2}{\sqrt[]{64+1+16}}\right|=\frac{10}{3}$ units

 $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt[]{1-x^2}},$ $I.F.e^{{\displaystyle \int \frac{xdx}{x^2-1}}}=\sqrt[]{\left|x^2-1\right|}=\sqrt[]{1-x^2}$ $\left(?x\in \left(-1,1\right)\right)$

Solution of differential equation is $y\sqrt[]{1-x^2}={\displaystyle \int \left(x^4+2x\right)dx=\frac{x^5}{5}+x^2+c}$

Curve is passing through origin, c = 0 $y=\frac{x^5+5x^2}{5\sqrt[]{1-x^2}}$

$\therefore {\displaystyle \underset{-\frac{\sqrt[]{3}}{2}}{\overset{\frac{\sqrt[]{3}}{2}}{\int }}\frac{x^5+5x^2}{5\sqrt[]{1-x^2}}dx=\frac{\pi }{3}-\frac{\sqrt[]{3}}{4}}$

 $\frac{z_2-0}{\left(1+2i\right)-0}=\left|\frac{OB}{OA}\right|e^{\frac{i\pi }{4}}\Rightarrow z_2=\left(1+2i\right)\left(1+i\right)=-1+3i\therefore arg{z}_2=\pi -ta{n}^{-1}3and\left|z_2\right|=\sqrt[]{10}$

$z_1-2z_2=3-4i\therefore arg\left(z_1-2z_2\right)=-ta{n}^{-1}\frac{4}{3}\Rightarrow \left|z_1-2z_2\right|=\left|2+4i+1-3i\right|=\sqrt[]{10}$

 $l={\displaystyle \underset{0}{\overset{20\pi }{\int }}{\left(\left|sinx\right|+\left|cosx\right|\right)}^{2}dx=20{\displaystyle \underset{0}{\overset{\pi }{\int }}\left(1+\left|sin2x\right|\right)dx=40{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+\left|sin2x\right|\right)dx=40{\left(x-\frac{cos2x}{2}\right)}_0^{\frac{\pi }{2}}}}}$

$=40\left(\frac{\pi }{2}+\frac{1}{2}+\frac{1}{2}\right)$ = 20 (p + 2)

f (3x)- f (x) = x

Replace $x\to \frac{x}{3}\Rightarrow f\left(x\right)-f\left(\frac{x}{3}\right)=\frac{x}{3}$

Again replace $x\to \frac{x}{3}\Rightarrow f\left(\frac{x}{3}\right)-f\left(\frac{x}{3^2}\right)-f\left(\frac{x}{3^2}\right)=\frac{x}{3^2}$

$\Rightarrow f\left(3x\right)-f\left(0\right)=\frac{3x}{2}puttingx=\frac{8}{3}\Rightarrow f\left(8\right)-f\left(0\right)=4\therefore f\left(0\right)=3$

Also putting x = $\frac{14}{3}$ in f (3x) – 3 = $\frac{3x}{2}\Rightarrow$ F (14) – 3 = 7 f (14) = 10

 $Letf\left(x\right)=lo{g}_{cosx}cosecx=\frac{logcosecx}{logcosx}$

$f\text{'}\left(x\right)=\frac{logcosx.sinx\left(-cosecxcotx-\left(logcosecx\right)\frac{1}{cosx}.\left(sinx\right)\right)}{{\left(logcosx\right)}^2}$

$Atx=\frac{\pi }{4}f\text{'}\left(\frac{\pi }{4}\right)=\frac{-log\left(\frac{1}{\sqrt[]{2}}\right)+log\sqrt[]{2}}{{\left(log\frac{1}{\sqrt[]{2}}\right)}^2}=\frac{2}{log\sqrt[]{2}}\therefore atx=\frac{\pi }{4},lo{g}_e\left(2f\text{'}\left(x\right)\right)=4$

 $\beta =\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)},\alpha \in R\underset{x\to 0}{lim}\frac{\frac{\alpha }{3}-\left(\frac{e^{3x}-1}{3x}\right)}{\alpha x\left(\frac{e^{3x}-1}{3x}\right)}$

$=\underset{x\to 0}{lim}\frac{1-\frac{\left(1+3x+\frac{9x^2}{2}+..........-1\right)}{3x}}{1+3x+\frac{9x^2}{2}+........-1}=-\frac{1}{2}\therefore \alpha +\beta =\frac{5}{2}$

$f_a\left(x\right)=ta{n}^{-1}2x-3ax+7\Rightarrow f_a\text{'}\left(x\right)=\frac{2}{1+4x^2}-3a\Rightarrow f_a\text{'}\left(x\right)\ge 0\Rightarrow 3a\le \frac{2}{1+4x^2}$

$a_{max}=\frac{2}{3}\left(\frac{1}{1+4*\frac{{\pi }^2}{36}}\right)=\frac{6}{9+{\pi }^2}=\overline{a}\therefore f_a\left(\frac{\pi }{8}\right)=ta{n}^{-1}\frac{2\pi }{8}-3\frac{\pi }{8}\frac{6}{9+{\pi }^2}+7=8-\frac{9\pi }{4\left(9+{\pi }^2\right)}$

Let equation of normal to x² = y at Q (t, t²) is x + 2ty = t + 2t³

It passes through the point (1, -1) so, 2t³ + 3t – 1 = 0

Let f(t) = 2t³ + 3t – 1 f   $\left(\frac{1}{4}\right)f\left(\frac{1}{3}\right)<0\Rightarrow t\in \left(\frac{1}{4},\frac{1}{3}\right)$

Let P(1 – sin q, -1 + cos q) $\therefore$  slope of normal = slope of CP $-\frac{1}{2t}=\frac{cos\theta }{-sin\theta }\Rightarrow 2t$ Þ = tan q according to question $x=1-sin\theta =1-\frac{2t}{\sqrt[]{1+4t^2}}=g\left(t\right)\therefore g\left(t\right)=1-\frac{2t}{\sqrt[]{1+4t^2}}$ ,  

Þ g'(t) < 0 g(t) is decreasing function in  $t\in \left(\frac{1}{4},\frac{1}{3}\right)\Rightarrow g\left(t\right)\in \left(0.440,0.485\right)\in \left(\frac{1}{4},\frac{1}{2}\right)$

$A\text{'}BA=\left[111\right]\left[\begin{array}{ccc}\hfill 9^2\hfill & \hfill -10^2\hfill & \hfill 11^2\hfill \\ \hfill 12^2\hfill & \hfill 13^2\hfill & \hfill -14^2\hfill \\ \hfill -15^2\hfill & \hfill 16^2\hfill & \hfill 17^2\hfill \end{array}\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=$

$\left[9^2+12^2-15^2-10^2+13^2+16^{2}11^2-14^2+17^2\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$

$=\left[9^2+12^2-15^2-10^2+13^2+16^2+11^2-14^2+17^2\right]=\left[539\right]$