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
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
Updated On: Jul 6, 2022
- $R-\left\{-\frac{b}{2a}\right\}$
- $R-\left\{\left\{-\frac{b}{2a}\right\}\cup\left\{x : x \ge-1\right\}\right\}$
- $R-\{\left\{-\frac{b}{2a}\right\}\cap (\infty, -1]\} $
- None of these
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The Correct Option is C
Solution and Explanation
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 $=R\{\{-\frac{b}{2a}\}\cap(-\infty, -1]\}$
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Notes on Relations and functions
Concepts Used:
Relations and functions
A relation R from a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.
Representation of Relation and Function
Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, represent this function in different forms.
- Set-builder form - {(x, y): f(x) = y2, x ∈ A, y ∈ B}
- Roster form - {(1, 1), (2, 4), (3, 9)}
- Arrow Representation
