(a) To find \(gf(x)\), substitute \(g(x) = ax + 1\) into \(f(x)\):

\(gf(x) = g(f(x)) = g\left(x + \frac{1}{x}\right) = a\left(x + \frac{1}{x}\right) + 1\).

(b) Given \(gf(2) = 11\), substitute \(x = 2\) into \(gf(x)\):

\(gf(2) = a\left(2 + \frac{1}{2}\right) + 1 = 11\).

\(a\left(2.5\right) + 1 = 11\)

\(2.5a + 1 = 11\)

\(2.5a = 10\)

\(a = 4\).

(c) The function \(f(x) = x + \frac{1}{x}\) is not one-one because it has a minimum point at \(x = 1\), meaning it is not strictly increasing or decreasing. Therefore, \(f\) does not have an inverse.

(d) To find \(g^{-1}(x)\), solve \(y = ax + 1\) for \(x\):

\(y - 1 = ax\)

\(x = \frac{y - 1}{a}\)

For \(a = 5\), \(g^{-1}(x) = \frac{x - 1}{5}\).

Substitute \(f(x) = x + \frac{1}{x}\) into \(g^{-1}(x)\):

\(g^{-1}f(x) = \frac{x + \frac{1}{x} - 1}{5}\).

(e) The composite function \(fg\) cannot be formed because the domain of \(f\) (\(x > 0\)) does not include the whole range of \(g\), which is all real numbers.