Question

Range of $f\left(x\right)=\frac{R\left(x\right)}{(x^2-3x+2)}$ is

• A. $[-2,2]$
• B. $(-\infty ,-2-\sqrt{3})\cup [-2+\sqrt{3},\infty ]$
• C. $(-\infty ,-7,-4\sqrt{3}]\cup [-7+4\sqrt{3},\infty )$
• D. none of these

Answer 1

Answer: (C)

$(-\infty ,-7,-4\sqrt{3}]\cup [-7+4\sqrt{3},\infty )$

Let unknown polynomial be $P\left(x\right)$. Let $Q\left(x\right)$ and $R\left(x\right)$ be the quotient and remainder, respectively, when it is divided by $(x-3)(x-4)$. Then,
$P\left(x\right)=(x-3)(x-4)Q\left(x\right)+R\left(x\right)$
Then, we have
$R\left(x\right)=ax+b$
$\Rightarrow P\left(x\right)=(x-3)(x-4)Q\left(x\right)+ax+b$
Given that $P\left(3\right)=2$ and $P\left(4\right)=1$. Hence,
$3a+b=2$ and $4a+b=1$
$\Rightarrow a=-1$ and $b=5$
$\Rightarrow R\left(x\right)=5-x$
$f\left(x\right)=y=\frac{-x+5}{x^2-3x+2}$
$\Rightarrow y{x}^2+(1-3y)x+2y-5=0$
Now, $x$ is real, then
$D\ge 0$
$\Rightarrow (1-3y{)}^2-4y(2y-5)\ge 0$
or $y^2+14y+1\ge 0$
or $y\in (-\infty ,\frac{-14-\sqrt{192}}{2}]\cup [\frac{-14+\sqrt{192}}{2},\infty )$
$(-\infty ,-7-4\sqrt{3}]\cup [-7+4\sqrt{3},\infty )$