All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
When $x=\frac{1}{2}, \frac{dx}{dt}=-\frac{1}{4}$,

$-\frac{1}{4}=k(1+0.5-0.25)$

$k=-0.2=-\frac{1}{5}$

(ii)
$1+x-x^{2} = 1-[(x-\frac{1}{2})^{2} - \frac{1}{4}] = -(x-\frac{1}{2})^{2} + \frac{5}{4}$

$\frac{dx}{dt} = -\frac{1}{5}(1+x+x^{2})$

$\frac{1}{1+x+x^{2}}(\frac{dx}{dt}) = -\frac{1}{5}$

$\int \frac{1}{\frac{5}{4}-(x-\frac{1}{2})^{2}} dx = \int -\frac{1}{5} dt$

$\frac{1}{\sqrt{5}} \mathrm{ln} \frac {\frac{\sqrt{5}}{2}+x-0.5}{\frac{\sqrt{5}}{2}-(x-0.5)} = -\frac{1}{5}t + C$

When $t=0, x=0.5, C = 0$.

$\frac{1}{\sqrt{5}} \mathrm{ln} \frac {\sqrt{5}+2x-1}{\sqrt{5}-2x+1} = -\frac{1}{5}t$

$\frac{-5}{\sqrt{5}} \mathrm{ln} \frac {\sqrt{5}+2x-1}{\sqrt{5}-2x+1} = t$

$t = \frac{5}{\sqrt{5}} \mathrm{ln} \frac {\sqrt{5}-2x+1}{\sqrt{5}+2x-1}$

$t = {\sqrt{5}} \mathrm{ln} \frac {\sqrt{5}-2x+1}{\sqrt{5}+2x-1}$

(iii)
(a) When $x=0.25$, $t= {\sqrt{5}} \mathrm{ln} \frac {\sqrt{5}-0.5+1}{\sqrt{5}+0.5-1} = {\sqrt{5}} \mathrm{ln} \frac {2\sqrt{5}+1}{2\sqrt{5}-1}$

(b) When $x=0$, $t=2.152\mathrm{min}$ to 3 d.p.

(iv)
$t = {\sqrt{5}} \mathrm{ln} \frac {\sqrt{5}-2x+1}{\sqrt{5}+2x-1}$

$\frac{t}{\sqrt{5}} = \mathrm{ln} \frac {\sqrt{5}-(2x-1)}{\sqrt{5}+(2x-1)}$

Let $2x-1=y$

$\frac{t}{\sqrt{5}} = \mathrm{ln} \frac {\sqrt{5}-y}{\sqrt{5}+y}$

$e^{\frac{t}{\sqrt{5}}} = \frac {\sqrt{5}-y}{\sqrt{5}+y}$

$(\sqrt{5}+y)e^{\frac{t}{\sqrt{5}}} = \sqrt{5}-y$

$\sqrt{5} e^{\frac{t}{\sqrt{5}}} + y e^{\frac{t}{\sqrt{5}}} = \sqrt{5}-y$

$y + y e^{\frac{t}{\sqrt{5}}} = \sqrt{5}-\sqrt{5} e^{\frac{t}{\sqrt{5}}}$

$y(1 + e^{\frac{t}{\sqrt{5}}}) = \sqrt{5}-\sqrt{5} e^{\frac{t}{\sqrt{5}}}$

$2x-1= \frac{\sqrt{5}-\sqrt{5} e^{\frac{t}{\sqrt{5}}}}{(1 + e^{\frac{t}{\sqrt{5}}})}$

$x = \frac{1}{2}[1 + \frac{\sqrt{5}-\sqrt{5} e^{\frac{t}{\sqrt{5}}}}{(1 + e^{\frac{t}{\sqrt{5}}})}]$

Plotting into Graphic Calculator,

Some students will overlook to make $t$ in terms of $x$ and are careless when it comes to completing the square. For (iiia), some students did not identify that the initial mass was 0.5g and half of it refers to 0.25g. As for (iv), it is certainly very tedious, but there are a lot of marks, just do it carefully. Many students overlook the given range for $x$ values too, and do label the graph when $t = 0$ and $t = 2.152$.