Maths

x² – x – 4 = 0

  $P_n={\alpha }^n-{\beta }^n$                    

$?=\frac{P_{15}{P}_{16}-P_{14}{P}_{16}-P_{15}^2+P_{14}{P}_{15}}{P_{13}{P}_{14}}$

$=\frac{\left(P_{15}-P_{14}\right)\left(P_{16}-P_{15}\right)}{P_{13}{P}_{14}}$

$=\frac{4P_{13}\cdot 4P_{14}}{P_{13}\cdot P_{14}}=16$

P_n = aⁿ - bⁿ   

$={\alpha }^{n-1}\cdot \alpha -{\beta }^{n-1}\cdot \beta$

$={\alpha }^{n-1}\left({\alpha }^2-4\right)-{\beta }^{n-1}\left(\beta -4\right)$

$P_n={\alpha }^{n+1}-{\beta }^{n+1}-4\left({\alpha }^{n-1}-{\beta }^{n-1}\right)$

$P_n=P_{n+1}-4P_{n-1}\Rightarrow P_{n+1}-P_n=4P_{n-1}$                  

np + npq = 82.5 Þ np (1 + q) = 82.5

$np\cdot npq=1350\Rightarrow {\left(np\right)}^{2}q=1350$

n = ?

$\Rightarrow \frac{{\left(1+q\right)}^2}{q}=\frac{82.5*82.5}{1350}$

$q^2+2q+1=\frac{121}{24}q$

-> 24q² – 73q + 24 = 0

q $=\frac{73\pm \sqrt[]{73^2-4*576}}{48}$  

$=\frac{73\pm 55}{48}=\frac{128}{48},\frac{18}{48}=\frac{8}{3},\frac{3}{8}$

$p=\frac{5}{8},q=\frac{3}{8}$

                $n\cdot \frac{5}{8}\cdot \frac{11}{8}=\frac{165}{2}$

n = 96

Data contradiction.

Common tangents

$T_1:y=mx\pm \sqrt[]{4m^2+9}$      

$T_2:y=mx\pm \sqrt[]{42m^2-13}$    

So, 4m² + 9 = 42 m² – 13 Þ m =   $\pm 2$

  $\Rightarrow c=\pm 5$              

So tangent  $y=\pm 2x\pm 5$

y = 2x + 5 does not pass through 4^th quadrant compare this tangent with T = 0 to get pt. of intersection

$\frac{x{x}_2}{42}-\frac{y{y}_2}{143}=1\Rightarrow x_2=\frac{-84}{5}$

$f\left(x\right)=\frac{2}{\sqrt[]{3}}{\displaystyle {\int }_0^{\sqrt[]{3}}f\left(\frac{{\lambda }^{2}x}{3}\right)d\lambda }$

Substitute $\frac{{\lambda }^{2}x}{3}=t$ $\frac{{}^{}}{}$

$f\left(x\right)=\frac{1}{\sqrt[]{x}}{\displaystyle {\int }_0^x\frac{f\left(t\right)}{\sqrt[]{t}}dt}$

differentiate using leibneitz rule

$\sqrt[]{x}.f\text{'}\left(x\right)+f\left(x\right).\frac{1}{2\sqrt[]{x}}=\frac{f\left(x\right)}{\sqrt[]{x}}$

f (1) = $\sqrt[]{3}\Rightarrow f\left(x\right)=\sqrt[]{3x}$

Use [x + n] = n + [x], where $n\in z$  f (x) = [x] – 10 will be minimum for $x\in [0,10]$  break the limits as G.I.F. is discontinuous at integral points.

${\displaystyle \underset{0}{\overset{10}{\int }}f\left(x\right)dx=-10-9-....-1}$    

$=\frac{-10.11}{2}$

${\displaystyle \underset{0}{\overset{10}{\int }}\left|f\left(x\right)\right|dx=10+9+....+1}$

Find a, α

$\left({\alpha }^2+{\alpha }^2-3\right)+{\left({\alpha }^2-{\alpha }^2-1\right)}^5=0$                

$\alpha =\sqrt[]{2}$      

zx differentiate the curve

$2x+2y.y^1+5{\left(x^2-y^2-1\right)}^4\left(2x-2y\right)=0$ ……. (i)

differentiate equation (i)

$\left(\alpha ,\alpha \right)\Rightarrow y\text{'}\text{'}=\frac{-23}{4\sqrt[]{2}}$          

$V=\frac{1}{3}\pi r^{2}h$

$V=\frac{\pi }{8}{h}^{3}ta{n}^2\alpha$

$\frac{dv}{dt}=\pi h^{2}ta{n}^2\alpha .\frac{dh}{dt}$

CSA = $\pi r\mathcal{l}$ $$

$\frac{d\left(CSA\right)}{dt}=\frac{15}{16}\pi 2h.\frac{dh}{dt}$

$\frac{15}{8}*\frac{2}{3}=5m^2/hrs$

$tn=\frac{2\left(2^3+4^3+6^3+....+{\left(2n\right)}^3\right)-\left(1^3-2^3+3^3+.....{\left(2n\right)}^3\right)}{n\left(4n+3\right)}$

^{$\frac{2*2^3*{\left(\frac{n*\left(n+1\right)}{2}\right)}^2-{\left(\frac{2n*\left(2n+1\right)}{2}\right)}^2}{n\left(4n+3\right)}$}

^{tn = n}

^{$now{\displaystyle \sum _{n=1}^{15}{T}_n}=\frac{15.16}{2}$}

Kindly consider the following figure