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(D) The function is $f(x) = \log \{ax^3 + (a+b)x^2 + (b+c)x + c\}$.Factorizing the expression inside the logarithm:$ax^3 + ax^2 + bx^2 + bx + cx + c =

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$R - (\{-\frac{b}{2a}\} \cup (-\infty, -1])$

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(D) The function is $f(x) = \log \{ax^3 + (a+b)x^2 + (b+c)x + c\}$.Factorizing the expression inside the logarithm:$ax^3 + ax^2 + bx^2 + bx + cx + c = ax^2(x+1) + bx(x+1) + c(x+1) = (ax^2 + bx + c)(x+1)$.Since $b^2 = 4ac$,the quadratic $ax^2 + bx + c$ can be written as $a(x + \frac{b}{2a})^2$.Thus,$f(x) = \log \{a(x + \frac{b}{2a})^2(x+1)\}$.For $f(x)$ to be defined,the argument of the logarithm must be strictly positive:$a(x + \frac{b}{2a})^2(x+1) > 0$.Given $a > 0$,the term $a(x + \frac{b}{2a})^2$ is always $\geq 0$. It is $0$ when $x = -\frac{b}{2a}$.Therefore,we require $x+1 > 0$ and $x 
eq -\frac{b}{2a}$.This implies $x > -1$ and $x 
eq -\frac{b}{2a}$.The domain $D$ is $(-1, \infty) - \{-\frac{b}{2a}\}$.This is equivalent to $R - (\{-\frac{b}{2a}\} \cup (-\infty, -1])$.Thus,the correct option is $D$.

Given that $a, b$ and $c$ are real numbers such that $b^2 = 4ac$ and $a > 0$. The maximal possible set $D \subseteq R$ on which the function $f: D \rightarrow R$ given by $f(x) = \log \{ax^3 + (a+b)x^2 + (b+c)x + c\}$ is defined,is

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