The problem defines a function $g(x) = \frac{1}{2} - \frac{3x}{1-x}$. We need to find: (i) $g(2)$ (ii) the value of $x$ such that $g^{-1}(1) = x$, which means we need to find $x$ such that $g(x) = 1$.

1. Problem Description

The problem defines a function $g(x) = \frac{1}{2} - \frac{3x}{1-x}$. We need to find:

(i) $g(2)$

(ii) the value of $x$ such that $g^{-1}(1) = x$, which means we need to find $x$ such that $g(x) = 1$.

2. Solution Steps

(i) To find $g(2)$, we substitute $x = 2$ into the expression for $g(x)$:

$g(2) = \frac{1}{2} - \frac{3(2)}{1-2} = \frac{1}{2} - \frac{6}{-1} = \frac{1}{2} + 6 = \frac{1}{2} + \frac{12}{2} = \frac{13}{2}$.

(ii) To find the image of 1 under $g^{-1}$, we need to find the value of $x$ such that $g(x) = 1$.

So, we set $g(x) = 1$ and solve for $x$:

$\frac{1}{2} - \frac{3x}{1-x} = 1$

$-\frac{3x}{1-x} = 1 - \frac{1}{2} = \frac{1}{2}$

$-6x = 1-x$

$-5x = 1$

$x = -\frac{1}{5}$

3. Final Answer

(i) $g(2) = \frac{13}{2}$

(ii) The image of 1 under $g^{-1}$ is $-\frac{1}{5}$.