Question

If $b^2 - 4ac = 0$, $a > 0$, then the domain of the function $y = log(ax^3 + (a + b)x^2 + (b + c)x + c)$ is

• A. R-\left\\{-\frac{b}{2a}\right\\}
• B. R-\left\\{\left\\{-\frac{b}{2a}\right\\}\cup\left\\{x : x \ge-1\right\\}\right\\}
• C. R-\\{\left\\{-\frac{b}{2a}\right\\}\cap (\infty, -1]\\}
• D. None of these

Answer 1

Answer: (C)

R-\\{\left\\{-\frac{b}{2a}\right\\}\cap (\infty, -1]\\}

Answer 2

We have $y=log\left(ax^{3}+bx^{2}+cx+ax^{2}+bx+c\right)$ $= log\left(x\left(ax^{2} + bx + c\right) + ax^{2} + bx + c\right)$ $= log\left(\left(x +1)( ax^{2} + bx + c\right)\right)$ $=log \left[\left(x+1\right)a\left(\left(x+\frac{b}{2a}\right)^{2}+\frac{c}{a}-\frac{b^{2}}{4a^{2}}\right)\right]$ $=log\left[\left(x+1\right)a\left(\left(x+\frac{b}{2a}\right)^{2}-\frac{b^{2}-4ac}{4a^{2}}\right)\right]$ $=log\left[a\left(x+1\right)\left(x+\frac{b}{2a}\right)^{2}\right]$ $\therefore y$ is defined if $x >-1$ and $x \ne-\frac{b}{2a}$ $\therefore y$ is defined if x \notin \left\\{-\frac{b}{2a}\right\\} \cap (-\infty, -1] $\therefore$ Domain $=R\\{\\{-\frac{b}{2a}\\}\cap(-\infty, -1]\\}$