Question

Consider an unknown polynomial which when divided by $(x-3)$ and $(x-4)$ leaves remainders $2$ and $1$ respectively. Let $R\left(x\right)$ be the remainder when this polynomial is divided by $(x-3)(x-4)$.

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)

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

Given $R\left(x\right)$ is the remainder when unkown polynomial is divided by $(x-3)(x-4)$.
Clearly, degree of $R\left(x\right)$ will be less than 2 (degree of divisor).
So, let $R\left(x\right)=ax+b$
Given $R\left(3\right)=2$
$\Rightarrow 3a+b=2$ ....(1)
Also, $R\left(4\right)=1$
$\Rightarrow 4a+b=1$ ....(2)
Solving (1) and (2), we get
$a=-1,b=5$
So $R\left(x\right)=-x+5$
Now, let $y=\frac{-x+5}{x^2-3x+2}$
$\Rightarrow y{x}^2+x(-3y+1)+(2y-5)=0$
$\Rightarrow y^2+14y+1\ge 0$ (Since $x$ is real, $D\ge 0$)
$\Rightarrow (y+7+4\sqrt{3})(y+7-4\sqrt{3})\ge 0$
$\Rightarrow y\in (-\infty ,-7-4\sqrt{3}]\cup [-7+4\sqrt{3},\infty )$