Maths

Given G.P's 2, 2², 2³, …60 term and 4, 4², 4³, … of 60

Now G.M. = ${\left(2\right)}^{\frac{225}{8}}\Rightarrow {\left(2,2^2,2^3,....\right)}^{\frac{1}{60+n}}={\left(2\right)}^{\frac{225}{8}}\Rightarrow n=\frac{57}{8},20son=20$

$\therefore {\displaystyle \sum _{k=1}^{n}k(n-k)}20*\frac{20*21}{2}-\frac{20*21*41}{6}=1330$

 $\frac{x^2}{{\left(\sqrt[]{\frac{5}{2}}\right)}^2}+\frac{y^2}{{\left(\sqrt[]{\frac{5}{3}}\right)}^2}=1$

Equation of tangent having slope m is

$y=mx\pm \sqrt[]{\frac{5}{3}{m}^2+\frac{5}{3}},$ which passes through (1, 3) and we get m₁ + m₂ = -4 and m₁m₂ = $\frac{44}{9}$

$\therefore$ Acute angle between the tangents is α  = tan^-1 $\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=ta{n}^{-1}\left(\frac{24}{7\sqrt[]{5}}\right)$

Since a is a odd natural number then $\left|{\displaystyle \underset{1}{\overset{3}{\int }}{y}^{a}dy}\right|=\frac{364}{3}\Rightarrow \left|{\left(\frac{y^{a+1}}{a+1}\right)}_1^3\right|=\frac{364}{3}\Rightarrow \frac{3^{a+1}}{a+1}=\frac{364}{3}$

a = 5

| (A + I) (adj A + I)| = 4 |A adj A + A + Adj A + I| = 4 | (A)I + A + adj A + I|= 4|A| = 1

|A + adj A| = 4

$A=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\Rightarrow adjA=\left[\begin{array}{cc}\hfill a\hfill & \hfill -b\hfill \\ \hfill -c\hfill & \hfill d\hfill \end{array}\right]\Rightarrow \left|\begin{array}{cc}\hfill \left(a+d\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \left(a+d\right)\hfill \end{array}\right|=4\Rightarrow a+d=\pm 2$

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

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

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

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

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