Question

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

We have $y=log\left(a{x}^3+b{x}^2+cx+a{x}^2+bx+c\right)$
$=log\left(x\left(a{x}^2+bx+c\right)+a{x}^2+bx+c\right)$
$=log\left(\left(x+1)(a{x}^2+bx+c\right)\right)$
$=log\left[\left(x+1\right)a\left({\left(x+\frac{b}{2a}\right)}^2+\frac{c}{a}-\frac{b^2}{4a^2}\right)\right]$
$=log\left[\left(x+1\right)a\left({\left(x+\frac{b}{2a}\right)}^2-\frac{b^2-4ac}{4a^2}\right)\right]$
$=log\left[a\left(x+1\right){\left(x+\frac{b}{2a}\right)}^2\right]$
$\therefore y$ 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 $=R\{\{-\frac{b}{2a}\}\cap (-\infty ,-1]\}$