Given equation is,
\(x\,+\sqrt{x}\,+\sqrt{x+2}\,+\sqrt{x^2+2x}\) = 3
\(\Rightarrow\) \(x\,+\sqrt{x}\,+\sqrt{x+2}\,+\sqrt{x^2+2x}-3=0\)
Let f(x) = \(x\,+\sqrt{x}\,+\sqrt{x+2}\,+\sqrt{x^2\,+\,2x}-3\)
Therefore, the domain of function f(x) is [0,∞).
f(o) = - 3 < 0
f(1) = 1 + 1 + \(\sqrt{3}\,+\sqrt{3}\)
= 2 + \(2\sqrt{3}\) > 0
Hence, at least one solution lies between 0 and 1.
f1(x) = 1 + \(\frac{1}{2\sqrt{x}}\) + \(\frac{1}{2\sqrt{x\,+\,2}}\) + \(\frac{2(x\,+\,1)}{2\sqrt{x^2\,+\,2x}}\)> 0 (\(\because\) \(x\in\) [0,∞)]
Hence, f(x) is increasing in its domain
therefore, only one positive solution of given equation exist which lies between 0 and 1.
Hence, option (C) is correct