Class 11th

$\left|z-\frac{1}{z}\right|=2$

${\left|z\right|}_{max}=?$

$\left|z-\frac{1}{2}\right|\ge \left|\left|z\right|-\frac{1}{\left|z\right|}\right|$

$2\ge \left|r-\frac{1}{r}\right|$

0 $\le r^2+2r-1\&r^2-2r-1\le 0$

$r=\frac{-2\pm \sqrt[]{8}}{2}$ $r=\frac{2\pm \sqrt[]{8}}{2}$

$=-1\pm \sqrt[]{2}$ $=1\pm \sqrt[]{2}$

r $\ge \sqrt[]{2}-1\&0\le r\le 1+\sqrt[]{2}$

$\sqrt[]{2}-1\le r\le \sqrt[]{2}+1$

L₁ : 3x – 4y + 12 = 0

L₂ : 8x + 6y + 11 = 0

$\left(\alpha ,\beta \right)$ lies on that angle which contain origin

$\therefore$ Equation of angle bisector of that angle which contain origin is

$\frac{3x-4y+12}{5}=\frac{8x+6y+11}{10}$

$\Rightarrow 2x+14y-13=0$

$\left(\alpha ,\beta \right)$ lies on it

$\Rightarrow 2\alpha +14\beta -13=0$ …… (i)

$\Rightarrow 3\alpha -4\beta +7=0$ ……. (ii)

Solving (i) & (ii)

$\alpha =-\frac{23}{25}\&\beta =\frac{53}{50}$

$\therefore \alpha +\beta =\frac{7}{50}$

$\Rightarrow 100\left(\alpha +\beta \right)=14$

(a) Polluted water has low value of dissolved oxygen, but high value of B.O.D., because chemical and organic matter uses dissolved oxygen to decompose.

(b) Eutrophication is result of excessive growth of weed in water bodies, which consumes dissolved oxygen, thus decreases oxygen.

Cu is least reactive metal and it lies below hydrogen in reactivity series.

${\left(N{H}_4\right)}_{2}C{r}_2{O}_7\underset{}{\overset{\Delta }{\to }}\boxed{N_2}+C{r}_2{O}_3+4H_{2}O$

$KMn{O}_4+HC\mathcal{l}\to MnC{\mathcal{l}}_2+KC\mathcal{l}+\boxed{C{\mathcal{l}}_2}+H_{2}O$

$A\mathcal{l}+NaOH+H_{2}O\to Na\left[A\mathcal{l}{\left(OH\right)}_4\right]+\boxed{H_2}$

$NaN{O}_3\to NaN{O}_2+\boxed{O_2}$

$V_{rms}=\sqrt[]{\frac{3RT}{M}}As.v_1=v_2$

$V_{rms}\alpha \sqrt[]{T}$       ${\eta }_2=4n_1$

$T_1=T_2\Rightarrow V_{rms-1}=V_{rms-2}$

$P_1=\frac{n_{1}RT}{v}{P}_z=\frac{n_{2}RT}{v}$

$\frac{P_1}{P_2}=\frac{n_1}{n_2}=\frac{1}{4}$

$l=\underset{n\to x}{lim}\frac{1}{2^n}\left(\frac{1}{\sqrt[]{1-\frac{1}{2^n}}}+\frac{1}{\sqrt[]{1-\frac{2}{2^n}}}+\frac{1}{\sqrt[]{1-\frac{3}{2^n}}}+.....+\frac{1}{\sqrt[]{1-\frac{2^n-1}{2^n}}}\right)$

Let 2n = t and if n $\infty$ then t $\infty$

$l=\underset{n\to x}{lim}\frac{1}{t}\left({\displaystyle \sum _{r=1}^{t=1}\frac{1}{\sqrt[]{1+\frac{r}{t}}}}\right)$

$={\left[2x^{\frac{1}{2}}\right]}_0^1=2$

(a) low solubility of LiF in H₂O due to high lattice energy.

(b) $O_2^{-}$  has unpaired e^- thus paramagnetic.

(c) Na in liq NH₃ forms electronated ammonia thus conducting in nature.

(d) $d_K<d_{Na}$ .

$\underset{x\to \frac{x}{4}}{lim}\frac{8\sqrt[]{2}-{\left(cosx+sinx\right)}^7}{\sqrt[]{2}-\sqrt[]{2}sin2x}\left(\frac{0}{0}form\right)$

$=\underset{x\to \frac{x}{4}}{lim}\frac{-7{\left(cosx+sinx\right)}^6\left(-sinx+cosx\right)}{-2\sqrt[]{2}cos2x}$ $\underset{x\to \frac{\pi }{4}}{Lt}\frac{{\left(cosx+sinx\right)}^5\left(co{s}^{2}x-si{n}^{2}x\right)}{2\sqrt[]{2}cos2x}=\frac{7{\left(\sqrt[]{2}\right)}^5}{2\sqrt[]{2}}=14$

In clark's method Ca (OH)₂ is used for softening hard water.

$Ca{\left(HC{O}_3\right)}_2+Ca{\left(OH\right)}_2\to 2CaC{O}_3+2H_{2}O$

$Mg{\left(HC{O}_3\right)}_2+2Ca{\left(OH\right)}_2\to 2CaC{O}_3+Mg{\left(OH\right)}_{2}2H_{2}O$    

Thus CaCO₃ and Mg (OH)₂ are formed in reaction.