A) equals 1
B) Lies between 0 and 1
C) lies between I and 2
D) is greater than 2
Correct Answer: C
Solution :
Let \[x=\sqrt{1+\sqrt{1+\sqrt{1+.....}}}\] \[(a+b)+c=a+(b+c)\] Squaring on both sides \[(a\times b)\times c=a\times (b\times c)\] \[a+0=a=0+a\] Solving, we get \[a\times 1=a=1\times a\] But x cannot be negative, so \[a+(-a)=0=(-a)+a\] which lies between 1 and 2.
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