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