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)