InterviewSolution
Saved Bookmarks
| 1. |
Let `S(x)=1+x-x^(2)-x^(3)+x^(4)+x^(5)-x^(6)-x^(7)+"........+"oo`, where `0ltxlt1`. If `S(x)=(sqrt(2)+1)/(2)`, then the value of `(x+1)^(2)` is |
|
Answer» `S(x)=1+x-x^(2)-x^(3)+x^(4)+x^(5)-x^(6)-x^(7)+"....."+oo` where `x in(0,1)` `S(x)=(1+x)-x^(2)(1+x)+x^(4)+(1+x)-x^(6)(1+x)+"....."+oo` `impliesS(x)=(1+x)[1-x^(2)+x^(4)-x^(6)+"....."+oo]` `impliesS(x)=(1+x)((1)/(1+x^(2)))" " [:.S_(oo)=(a)/(1-r) " for "GP]` According to the question, `S(x)=(sqrt(2)+1)/(2)` So, `(1+x)/(1+x^(2))=(sqrt(2)+1)/(2)` `implies=2+2x=(sqrt(2)+1)x^(2)+sqrt(2)+1` `implies(sqrt(2)+1)x^(2)-2x-2+sqrt(2)+1=0` `implies(sqrt(2)+1)x^(2)-2x+sqrt(2)-1=0` `implies(sqrt(2)+1)x^(2)-2x+(1)/(sqrt(2)+1)=0` `implies[(sqrt(2)+1)x]^(2)-2(sqrt(2)+1)x+1=0` `implies[(sqrt(2)+1)x-1]^(2)=0` `implies x=(1)/(sqrt(2)+1)" " ["repeated "]` So, ` x=sqrt(2)-1` `:. (x+1)^(2)=2`. |
|