A) \[\sqrt{2}\]
B) 2
C) \[2\sqrt{2}\]
D) \[2+\sqrt{2}\]
Correct Answer: B
Solution :
[b] Let \[x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+}}}}...\infty .\] Then, \[x=\sqrt{2+x}\] \[\therefore \] \[2+x={{x}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}-x-2=0\] \[\Rightarrow \] \[{{x}^{2}}-2x+x-2=0\] \[\Rightarrow \] \[x\,\,(x-2)+1\,\,(x-2)=0\] \[\Rightarrow \] \[(x-2)(x+1)=0\] \[\Rightarrow \] \[x=2\] \[[\therefore x\ne -1]\] |
You need to login to perform this action.
You will be redirected in
3 sec