niteshwaghray wrote:
If \(z^2=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\), where the given expression extends till infinity, which of the following statements must be true ?
I. Two values are possible for z
II. 4 - z^2 = 2
III. z^8 = 16
A. I only
B. II only
C. III only
D. I, II and III
E. None of the above
\(z^2=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\);
\(z^2=\sqrt{2+({\sqrt{2+\sqrt{2+\sqrt{2+...}})}}}\), as the expression under square root extends infinitely then expression in brackets would equal to z^2 itself so we can rewrite given expression as \(z^2=\sqrt{2+z^2}\). Square both sides \(z^4=2+z^2\) --> \(z=\sqrt{2}\) or \(z=-\sqrt{2}\).
We have two possible values, so I is true. Also, both of them satisfy II and III.
Answer: D.