Maths

18 = 3² * 2

For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even.

As we know no. of the form                          9 k -> 1000

9 k + 1 -> 1000

9 k + 2 -> 1000

9 k + 3 -> 1000  -> Total no. = 2000

9 k + 4 -> 1000

9 k + 5 -> 1000

9 k + 6 ->1000

9 k + 7 -> 1000

9 k + 8 -> 1000

In which half will be even & half be odd so Required no. = 1000

Given sequence is -16, 8, -4, 2, .

are in G.P. with first term a = -16 & common ratio r =  $\frac{-1}{2}$

Now $t_p=a{r}^{P-1}=-16{\left(-\frac{1}{2}\right)}^{p-1}\&t_q=a{r}^{q-1}=-16{\left(-\frac{1}{2}\right)}^{q-1}$

So A.M. = -8   $\left[{\left(-\frac{1}{2}\right)}^{p-1}+{\left(\frac{-1}{2}\right)}^{q-1}\right]=\alpha \left(let\right)$

Given equation is 4x² – 9x + 5 = 0 gives $x=1,\frac{5}{4}$

From roots we get possible value of b = 1 so

$16{\left(-\frac{1}{2}\right)}^{\frac{P+q-2}{2}}=1OR{\left(-\frac{1}{2}\right)}^{\frac{p+q-2}{2}}=\frac{1}{16}-{\left(-\frac{1}{2}\right)}^4$

$\Rightarrow \frac{p+q-2}{2}=4\Rightarrow p+q=10$

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

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

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