A) -1
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
\[x=\sqrt{2+x}\] Þ \[{{x}^{2}}-x-2=0\] Þ \[(x-2)(x+1)=0\] Þ \[x=2,-1\] But\[\sqrt{2+\sqrt{2+.....}}\ne -1\], so it is equal to 2.
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