Class 12th

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

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

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

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

Sphalerite is ore of zinc consists of ZnS during its concentration group 1 cyanides are used as depressants like NaCN

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

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

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

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

$\Rightarrow x=2,5$

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

Information missing. The question was droppedby NTA.

 $y=\left|\left|x-1\right|-2\right|$

Area bounded region = $\frac{1}{2}*4*2=4$

R = { (P, Q) |P and Q are at the same distance from the origin}.

at $\left(1,-1\right),x^2+y^2={\left(1\right)}^2+{\left(-1\right)}^2=1+1=2$

$P_1:3x+15+21z=9$           

P₂ : x – 3y – z = 5

P₃ : 2x + 10y + 14z = 5

Ratio of the direction cosines of P₁ and P₂

$\frac{3}{2}=\frac{15}{10}=\frac{21}{14}$

Hence, P₁ and P₃ are parallel.

 $\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)=\left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{a}-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}=0-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}\left(as\overrightarrow{a}.\overrightarrow{b}=0given\right)$

$\overrightarrow{a}*\left(\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)\right)=-{\left|\overrightarrow{a}\right|}^2\overrightarrow{a}*\overrightarrow{b}$

$\overrightarrow{a}*\left(\overrightarrow{a}*\left(\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)\right)\right)=-{\left|\overrightarrow{a}\right|}^2\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)=-{\left|\overrightarrow{a}\right|}^2\left(-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}\right)={\left|\overrightarrow{a}\right|}^4\overrightarrow{b}$

Let n be the number of times.

$p=\frac{1}{2},q=\frac{1}{2}$

According to question,

${\Rightarrow }^n{C}_7{=}^n{C}_9\Rightarrow n-7=9\Rightarrow n=16$

$p\left(x=2\right){=}^{16}{C}_2{\left(\frac{1}{2}\right)}^2.{\left(\frac{1}{2}\right)}^{14}=\frac{15}{2^{13}}$