Maths

Given curves $\frac{x^2}{9}+\frac{y^4}{4}=1.........\left(i\right)$

&     $x^2+y^2=\frac{31}{4}...........\left(ii\right)$      

Equation of any tangent to (i) be y = mx +  $\sqrt[]{9m^2+4}............\left(iii\right)$

For common tangent (iii) also should be tangent to (ii) so by condition of common tangency

$9m^2+4=\frac{31}{4}\left(1+m^2\right)$

OR 36m² + 16 = 31 + 31m²

=>m² = 3

${\displaystyle \underset{0}{\overset{\pi }{\int }}\left|sin2x\right|dx}$

$=2{\displaystyle \underset{0}{\overset{\pi /2}{\int }}sin2xdx}$ =2 ${\left[\frac{-cos2x}{2}\right]}_0^{\pi /2}$ = $2\left(\frac{1}{2}-\left(-\frac{1}{2}\right)\right)=2$

$y^2=a\left(x+\frac{\sqrt[]{a}}{2}\right),a>0$

2yy1 = a

$y^2=2y{y}_1\left(x+\frac{\sqrt[]{2y{y}_1}}{2}\right)$

$y=2y_{1}x+y_1\sqrt[]{2y{y}_1}$

${\left(y-2y_{1}x\right)}^2=y_1^2.2y{y}_1=2y{y}_1^3$

$\therefore order=1,degree=3$

Hence, degree – order = 3 – 1 = 2

$30{}^{30}{C}_0+29{}^{30}{C}_1+.....+2{}^{30}{C}_{28}+1{.}^{30}{C}_{29}=n.2^m$

$\therefore n=15,m=0$

$\therefore n+m=15+30=45$

$\left(x-1\right)\left(x^2-x+1\right)=0$

$x=1,x=\frac{1\pm \sqrt[]{3}i}{2}=\frac{1}{2}+\frac{\sqrt[]{3}}{2}i,\frac{1}{2}-\frac{\sqrt[]{3}}{2}i=e^{i\frac{\pi }{3}},e^{-i\frac{\pi }{3}}$

Sum of 162th power of roots = $1+e^{i54\pi }+e^{-i54\pi }=1+1+1=3$

$3sinx+4cosx=k+1,cos\alpha =\frac{3}{5},sin\alpha =\frac{4}{5}$

$5sin\left(x+\alpha \right)=k+1,$

$\therefore -5\le k+1\le 5\Rightarrow -6\le k\le 4$

$\therefore$ total number of integral values of k is 11.

Let $A\left(-2,-21,29\right),B\left(-1,-16,23\right),P\left(\lambda ,2,1\right),Q\left(4,-2,2\right)$

Given $\overrightarrow{AB}\perp \overrightarrow{PQ}$

$\therefore \overrightarrow{AB}.\overrightarrow{PQ}=0$

$\left(\widehat{i}+5\widehat{j}-6\widehat{k}\right).\left(\left(4-\lambda \right)\widehat{i}-4\widehat{j}+\widehat{k}\right)=0$

$4-\lambda -20-6=0$

$\lambda =-22$

$\therefore {\left(\frac{\lambda }{11}\right)}^2+\left(\frac{-4\lambda }{11}\right)-4=4+8-4=8$

$e^{siny}cosy\frac{dy}{dx}+e^{siny}cosx=cosx$

Put esin y = t

$e^{siny}cosy\frac{dy}{dx}=\frac{dt}{dx}$

$\therefore \frac{dt}{dx}+tcosx=cosx$

$I.F=e^{{\displaystyle \int cosxdx}}=e^{sinx}$

$\therefore t{e}^{sinx}={\displaystyle \int e^{sinx}cosxdx}$

Put sin x = u, cos xdx = du

$\therefore putx=0,y\left(0\right)=0,1=1+c\Rightarrow c=0$

$\therefore 1+0+0+0=1$

 $lo{g}_4\left(x-1\right)=lo{g}_2\left(x-3\right)$

$\Rightarrow \frac{1}{2}lo{g}_2\left(x-1\right)=lo{g}_2\left(x-3\right)$

$\Rightarrow lo{g}_2\left(x-1\right)=lo{g}_2{\left(x-3\right)}^2$

$\Rightarrow x-1=x^2-6x+9$

$\Rightarrow x=2,5$

also x – 1 > 0 and x – 3 > 0

x > 1 & x > 3

Hence, x = 5 possible. Only one solution.

$\sqrt[]{3}co{s}^{2}x=\left(\sqrt[]{3}-1\right)cosx+1$

$\left(\sqrt[]{3}cosx+1\right)\left(cosx-1\right)=0$

$\therefore cosx=-\frac{1}{\sqrt[]{3}}\left(rejected\right)$

Hence, cos x = 1 x = 0 one solution