Correct Option: CStep-by-step Solution:1. Simplify each term of the expression:Consider the first term: \\(\\frac{x^2+2x+2}{x+1}\\).We can rewrite the numerator as \\(x^2+2x+1+1 = (x+1)^2+1\\).So, \\(\\frac{(x+1)^2+1}{x+1} = \frac{(x+1)^2}{x+1} + \frac{1}{x+1} = (x+1) + \frac{1}{x+1}\\).Similarly, the second term simplifies to \\((y+1) + \frac{1}{y+1}\\).Thus, the expression \(E\) becomes \(E = \left(x+1 + \frac{1}{x+1}\right) + \left(y+1 + \frac{1}{y+1}\right)\).2. Introduce substitution for simplification:Let \(a = x+1\) and \(b = y+1\).Since \(x, y > 0\), it implies \(a > 1\) and \(b > 1\).The expression transforms to \(E = a + \frac{1}{a} + b + \frac{1}{b}\).3. Utilize the given constraint:We are given \(x+y=1\).Adding 2 to both sides: \\(x+1+y+1 = 1+2\\).So, \(a+b = 3\).4. Minimize the transformed expression:We need to minimize \(E = a + \frac{1}{a} + b + \frac{1}{b}\) subject to \(a+b=3\) and \(a,b > 1\).Consider the function \(f(t) = t + \frac{1}{t}\). For \(t>1\), this function is convex (its second derivative \(f''(t) = \frac{2}{t^3}\) is positive). For a sum of convex functions with a linear constraint, the minimum occurs when the variables are equal.Therefore, the minimum value of \(E\) occurs when \(a=b\).Given \(a+b=3\), if \(a=b\), then \(2a=3 \implies a = \frac{3}{2}\). Thus, \(a = \frac{3}{2}\) and \(b = \frac{3}{2}\).5. Calculate the minimum value:Substitute \(a = \frac{3}{2}\) and \(b = \frac{3}{2}\) back into the expression for \(E\):\(E_{min} = \left(\frac{3}{2} + \frac{1}{\frac{3}{2}}\right) + \left(\frac{3}{2} + \frac{1}{\frac{3}{2}}\right)\)\(E_{min} = \left(\frac{3}{2} + \frac{2}{3}\right) + \left(\frac{3}{2} + \frac{2}{3}\right)\)\(E_{min} = 2 \times \left(\frac{3}{2} + \frac{2}{3}\right)\)\(E_{min} = 2 \times \left(\frac{9+4}{6}\right)\)\(E_{min} = 2 \times \frac{13}{6}\)\(E_{min} = \frac{13}{3}\).The final answer is \\(\\frac{13}{3}\\).