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  4. If b2 - 4ac = 0, a > 0, then the domain of the function y = log(ax3 + (a + b)x2 + (b + c)x + c) is

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Q. 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

Relations and Functions

Solution:

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 $=R\{\{-\frac{b}{2a}\}\cap(-\infty, -1]\}$