If b² – 4ac = 0 and a > 0, then the domain of the function f(x) = log (ax³ + (2a + b) x² + (2b + c) x + 2c) is

$\begin{array}{c}\mathrm{f}\left(\mathrm{x}\right)=\log \left({\mathrm{ax}}^3+(2\mathrm{a}+\mathrm{b}){\mathrm{x}}^2+(2\mathrm{b}+\mathrm{c})\mathrm{x}+2\mathrm{c}\right)\\ =\log \left(\left({\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\right)(\mathrm{x}+2)\right)\\ =\log \left({\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\right)+\log (\mathrm{x}+2)\end{array}$

Since a > 0 and D = 0,

$\begin{array}{l}\therefore {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\ge 0\mathrm{\forall }\mathrm{x}\in \mathrm{R}\\ \therefore \log \left({\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\right)\text{ is defined if }{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\ne 0\\ \Rightarrow \mathrm{a}\left({\mathrm{x}}^2+\frac{\mathrm{b}}{\mathrm{a}}\mathrm{x}+\frac{\mathrm{c}}{\mathrm{a}}\right)\ne 0\\ \Rightarrow \mathrm{a}\left[{\left(\mathrm{x}+\frac{\mathrm{b}}{2\mathrm{a}}\right)}^2-\left(\frac{{\mathrm{b}}^2-4\mathrm{ac}}{4{\mathrm{a}}^2}\right)\right]\ne 0\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{a}{\left(\mathrm{x}+\frac{\mathrm{b}}{2\mathrm{a}}\right)}^2\ne 0\left[\because {\mathrm{b}}^2-4\mathrm{ac}=0\right]\\ \Rightarrow \mathrm{x}\ne -\frac{\mathrm{b}}{2\mathrm{a}}\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\text{ is defined for }\mathrm{x}\ne -\frac{\mathrm{b}}{2\mathrm{a}}\text{ and }\mathrm{x}+2>0\\ \therefore \text{ Domain of }\mathrm{f}=(-2,\mathrm{\infty })-\left\{-\frac{\mathrm{b}}{2\mathrm{a}}\right\}.\end{array}$