Quadratic Equations
Range of rational function = all reals; discriminant condition
MJMT_Full_Test_11
Grade 12
Question:
For $x\in\mathbb{R}$, the expression $\dfrac{x^2+2x+c}{x^2+4x+3c}$ can take all real values if $c\in$
$(1,2)$
$[-1,1]$
$(0,1)$
$(-1,0)$
Step-by-Step Solution
Key Concept: Set expression $=y$; rearrange to $(y-1)x^2+2(2y-1)x+c(3y-1)=0$. For all $y$ to be achievable, discriminant $\geq0$ for all $y$: leads to $c(c-1)\leq0$, i.e., $0\leq c\leq1$, combined with $4-3c>0$.
$c\in(0,1)$.
Correct Answer: 3