Question Number 102292 by Boykisss last updated on 08/Jul/20
Answered by Rio Michael last updated on 09/Jul/20
![(√((1−x)/(1 + kx))) = (1−x)^(1/(2.)) (1 + kx)^(−(1/2)) = [1 + (1/2)(−x)+ (((1/2)(−(1/2)))/(2!))(−x)^2 + ...][1 + (−(1/2))(kx)+ ((−(1/2)(−(3/2)))/2)(kx)^2 ] = [1−(1/2)x −(1/8)x^2 +...][1−((kx)/2) + ((3k^2 x^2 )/8)+...] = 1 −((kx)/2) + ((3k^2 x^2 )/8) − (x/2) + ((kx^2 )/4)−(1/8)x^2 neglecting higher powers of x. = 1 + (−(k/2)−(1/2))x + (((3k^2 )/8) + (k/4)−(1/8))x^2 +... ⇒ −(k/2)−(1/2) =−2 ⇒ ((k+ 1)/2) = 2 ⇔ k= 3 (ii)(a) if all have equal chances then number of ways =^(10) C_4 ways (b) if two are married and will not attend seperately, then they must attend together⇒ we need two more friends =^8 C_2 (c) two are not in terms say A and B are not in terms ⇒ if A attends then B doesnot: therefore^8 C_3 + ⇒ if Battends thenA doesnot: therefore^8 C_3](Q102488.png)
$$\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}\:+\:{kx}}}\:=\:\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}.}} \left(\mathrm{1}\:+\:{kx}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left[\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(−{x}\right)+\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}!}\left(−{x}\right)^{\mathrm{2}} \:+\:...\right]\left[\mathrm{1}\:+\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({kx}\right)+\:\frac{−\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}}\left({kx}\right)^{\mathrm{2}} \right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{x}\:−\frac{\mathrm{1}}{\mathrm{8}}{x}^{\mathrm{2}} +...\right]\left[\mathrm{1}−\frac{{kx}}{\mathrm{2}}\:+\:\frac{\mathrm{3}{k}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{8}}+...\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}\:−\frac{{kx}}{\mathrm{2}}\:+\:\frac{\mathrm{3}{k}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{8}}\:−\:\frac{{x}}{\mathrm{2}}\:+\:\frac{{kx}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}{x}^{\mathrm{2}} \:\:\:\mathrm{neglecting}\:\mathrm{higher}\:\mathrm{powers}\:\mathrm{of}\:{x}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}\:+\:\left(−\frac{{k}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right){x}\:+\:\left(\frac{\mathrm{3}{k}^{\mathrm{2}} }{\mathrm{8}}\:+\:\frac{{k}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}\right){x}^{\mathrm{2}} +... \\ $$$$\Rightarrow\:−\frac{{k}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:=−\mathrm{2}\:\Rightarrow\:\:\:\frac{{k}+\:\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{2}\:\:\Leftrightarrow\:{k}=\:\mathrm{3} \\ $$$$\left(\mathrm{ii}\right)\left(\mathrm{a}\right)\:\mathrm{if}\:\mathrm{all}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{chances}\:\mathrm{then}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:=\:^{\mathrm{10}} {C}_{\mathrm{4}} \:\mathrm{ways} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{if}\:\mathrm{two}\:\mathrm{are}\:\mathrm{married}\:\mathrm{and}\:\mathrm{will}\:\mathrm{not}\:\mathrm{attend}\:\mathrm{seperately},\:\mathrm{then}\:\mathrm{they}\:\mathrm{must} \\ $$$$\mathrm{attend}\:\mathrm{together}\Rightarrow\:\mathrm{we}\:\mathrm{need}\:\mathrm{two}\:\mathrm{more}\:\mathrm{friends}\:=^{\mathrm{8}} {C}_{\mathrm{2}} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{two}\:\mathrm{are}\:\mathrm{not}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{say}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are}\:\mathrm{not}\:\mathrm{in}\:\mathrm{terms} \\ $$$$\Rightarrow\:\mathrm{if}\:\:{A}\:\mathrm{attends}\:\mathrm{then}\:\mathrm{B}\:\mathrm{doesnot}:\:\mathrm{therefore}\:^{\mathrm{8}} {C}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$$\Rightarrow\:\mathrm{if}\:{B}\mathrm{attends}\:\mathrm{then}{A}\:\mathrm{doesnot}:\:\mathrm{therefore}\:^{\mathrm{8}} {C}_{\mathrm{3}} \\ $$