The mass, \(x\) grams, of a certain substance present in a chemical reaction at time \(t\) minutes satisfies the differential equation \(\frac{\mathrm{d}x}{\mathrm{d}t}=k\left( 1+x-{{x}^{2}} \right)\), where \(0\le x\le \frac{1}{2}\) and \(k\) is a constant. It is given that \(x=\frac{1}{2}\) and \(\frac{\mathrm{d}x}{\mathrm{d}t}=-\frac{1}{4}\) when \(t=0\).
Show that \(k=-\frac{1}{5}\).[1]
By first expressing \(1+x-{{x}^{2}}\) in completed square from, find \(t\) in terms of \(x\).[5]
Hence find
the exact time taken for the mass of the substance present in the chemical reaction to become half of its initial value,[1]
the time taken for there to be none of the substance present in the chemical reaction, giving your answer correct to \(3\) decimal places.[1]
Express the solution of the differential equation in the form \(x=\mathrm{f}\left( t \right)\) and sketch the part of the curve with this equation which is relevant in this context.[5]