Class 11th

$\underset{\left(acid\right)}{H_{2}S}+\underset{Base}{H_{2}O}?H_3{O}^{+}+H{S}^{-}$

$\underset{\left(acid\right)}{H_{2}O}+\underset{\left(Base\right)}{N{H}_3}?N{H}_{4}OH$

With H₂S water acts like base and with NH₃ it acts like acid.

Let P (h, k) be the mid point of the chord x² – y² = 4

$\therefore$ its equation is xh – yk = h² – k²

$Or,y=\left(\frac{h}{x}\right)x+\frac{k^2-h^2}{k}$ if this line is tangent to y² = 8x then $\frac{k^2-h^2}{k}$ = $\frac{2}{h/k}=\frac{2k}{h}$

$h\left(k^2-h^2\right)=2k^2$              

$\therefore$ Required locus is 2y² = x (y² – x²)

$\Rightarrow x^3=y^2\left(x-2\right)$

  $ta{n}^{-1}2=\theta$

->tan q = 2

$sin\left(\theta -\alpha \right)=\frac{1}{\sqrt[]{5}}$

->4 – 2 tanq = 1 + 2 tan a tan a = $\frac{3}{4}$  

$\alpha =ta{n}^{-1}\frac{3}{4}$          

In electro-refining, impure metal (blister copper) is used as an anode while precious metal like Au, Pt gets deposited as anode mud.

$C_2=a_2+b_2=\left(a_1-3\right)+2b_1=12$

$a_1+2b_1=15.............\left(i\right)$

$a_1+4b_1=19............\left(ii\right)$              

Solving (i) & (ii), b₁ = 2, a₁ = 11

$=5\left[22+9\left(-3\right)\right]+2\left(\frac{2^{10}-1}{2-1}\right)$

$=2^{11}-27=2^6*2^5-27=2021$               

$\frac{3+7+x+y}{4}=5$              

$\Rightarrow x+y=10.........\left(i\right)$

$\frac{1}{4}\left(9+49+x^2+y^2\right)-25=10$              

x² + y² = 82 .(ii)

$\therefore \overline{x}=\frac{\left(3+2x\right)+\left(7+2y\right)+\left(x+y\right)+\left(x-y\right)}{4}$              

$=\frac{10+4x+2y}{4}$              

Solving (i) & (ii), x = 9, y = 1

$\therefore \overline{x}=\frac{48}{4}=12$          

$\frac{{\left(2i\right)}^n}{{\left(1-i\right)}^{n-2}}$

$=\frac{{\left(2i\right)}^n{\left(1+i\right)}^{n-2}}{{\left(1+1\right)}^{n-2}}$            

$=-4{\left(-1+i\right)}^{n-2}$              

  $now{\left(-1+i\right)}^2=-2i$

$\therefore -4{\left(-1+i\right)}^{n-2}\in \left(+\right)ve$  integer for n – 2 = 4

->n = 6

$\overrightarrow{a}=\widehat{i}+2\widehat{j}+\widehat{k}$                

$\overrightarrow{b}=2\widehat{i}+4\widehat{j}-5\widehat{k}\overrightarrow{c}=-\lambda \widehat{i}+2\widehat{j}+3\widehat{k}$              

$\overrightarrow{b}+\overrightarrow{c}=\left(2-\lambda \right)\widehat{i}+6\widehat{j}-2\widehat{k}$

${\left(12-\lambda \right)}^2={\left(2-\lambda \right)}^2+40$

$\lambda =5$