A function f is defined as $f(x) = \ln(5-x)$, $x < 5$.

1. Express $x$ in terms of $y$ and set up the integral for the volume of revolution about the y-axis.

Given the curve $y = \ln(5-x)$:
$e^y = 5 - x$
$x = 5 - e^y$

The region is bounded by the x-axis ($y=0$) and the y-axis. At the y-axis, $x=0$, which gives the upper limit:
$y = \ln(5 - 0) = \ln 5$

The volume $V$ generated when rotated completely about the y-axis is given by:
$V = \pi \int_{y_1}^{y_2} x^2 \, dy$
$V = \pi \int_{0}^{\ln 5} (5 - e^y)^2 \, dy$

2. Evaluate the definite integral using a graphing calculator.

Using a graphing calculator to evaluate the integral:
$V \approx 38.44019$
$V = 38.44 \text{ (correct to 2 d.p.)}$

1. Analyse the equation for $x \ge 0$ and $x < 0$ to find the valid range

For $x \ge 0$, $|x| = x$, so $f(|x|) = f(x)$.
The equation simplifies to $f(x) = |f(x)|$, which is true if and only if $f(x) \ge 0$.
$\ln(5-x) \ge 0 \implies 5-x \ge 1 \implies x \le 4$.
Combining with $x \ge 0$, we get $0 \le x \le 4$.

For $x < 0$, $|x| = -x$, so $f(|x|) = \ln(5+x)$.
Since $x < 0$, $5-x > 5 \implies \ln(5-x) > \ln(5) > 0$. Thus, $|f(x)| = f(x) = \ln(5-x)$.
The equation becomes $\ln(5+x) = \ln(5-x) \implies 5+x = 5-x \implies 2x = 0 \implies x = 0$.
Since this does not satisfy $x < 0$, there are no solutions for $x < 0$.

Therefore, the set of values is $0 \le x \le 4$.