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Freundlich adsorption isotherm

  $\frac{x}{m}=k{p}^{1/n}$          

$log\frac{x}{m}=\frac{1}{n}logp+logk$           

Comparing with y = mx + C

so, slope of given graph is  $\frac{1}{n}$  whose value varies from 0 to 1.

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

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

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

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

$\lambda =-22$

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

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Siderite – FeCO₃ (ore of iron)

Calamine – ZnCO₃ (ore of zinc)

Malachite – CuCO₃.Cu (OH)₂ (ore of copper)

Cryolite – Na₃AlF₆ (ore of aluminium)

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

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

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Information missing. The question was droppedby NTA.

 $y=\left|\left|x-1\right|-2\right|$

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