Answer:(a)(ii) Since \(\mathrm{f}(x)\)is of degree \({6}\), there is a total of \({6}\) roots. As there is only one real root for \(x\ge 0\), i.e. \(x=\beta \), \text{ and } \(\mathrm{f}(x)=\mathrm{f}(-x)\), thus \(x=-\beta \) is the other real root. With only \({2}\) real roots, the equation \(\mathrm{f}(x)=0\) has \({4}\) non-real roots. (a)(iii) The roots in modulus-argument form are: \({{z}_{1}}=r{{\mathrm{e}}^{i\theta }}\), \({{z}_{2}}=r{{\mathrm{e}}^{-i\theta }}\), \({{z}_{3}}=-r{{\mathrm{e}}^{i\theta }}\), \({{{z}_{4}}=-r{{\mathrm{e}}^{-\mathrm{i}\theta }}=r{{\mathrm{e}}^{\mathrm{i}(\pi -\theta )}}}\), \({{z}_{5}}=\beta {{\mathrm{e}}^{\mathrm{i}\pi }}\), \({{{z}_{6}}=\beta {{\mathrm{e}}^{\mathrm{i}\theta }}}\) are the remaining roots of \(\mathrm{f}(x)=0\). (a)(iv) \(A={{\beta }^{2}}\), \(B=2\), \(C=-2\) (b) \(\frac{1}{2}\)