A) \[x=1\]
B) \[0<x<1\]
C) x is infinite
D) \[x=2\]
Correct Answer: D
Solution :
\[x=\sqrt{2+\sqrt{2+\sqrt{2+....}}}\] ...(i) Squaring both sides of (1), we get \[{{x}^{2}}=2+\sqrt{2+\sqrt{2+\sqrt{2+....}}}\] \[\Rightarrow \] \[{{x}^{2}}=2+x\] \[\Rightarrow \] \[{{x}^{2}}-x-2=0\] \[\Rightarrow \] \[x=\frac{1\,\pm \,\sqrt{1+18}}{2}=\frac{1\pm \sqrt{9}}{2}\] Since x cannot be negative, therefore, neglect \[\frac{1-\sqrt{9}}{2}\] Thus, \[x=2\]
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