Question

If $x$ is real, then the expression
$\frac{x^2+34x-71}{x^2+2x-7}$

• A. lies between $4$ and $7$
• B. lies between $5$ and $9$
• C. has no value between $4$ and $7$
• D. has no value between $5$ and $9$

Answer 1

Answer: (D)

Let $\frac{x^2+34x-71}{x^2+2x-7}=m$
$\therefore x^2+34x-71=m{x}^2+2mx-7m$
$\Rightarrow x^2(1-m)+(34-2m)x+7m-71=0$
Since $x$ is real
$\therefore$ Disc $\ge$ 0
$\therefore (34-2m{)}^2-4(1-m)(7m-71)\ge 0$
$(17-m{)}^2-(1-m)(7m-71)\ge 0$
$\Rightarrow 289+m^2-34m-[7m-71-7m^2+71m]\ge 0$
$\Rightarrow 8m^2-112m+360\ge 0$
$\Rightarrow m^2-14m+45\ge 0$
$\Rightarrow (m-9)(m-5)\ge 0$
$\Rightarrow m>9,m>5$ or $m<9,m<5$.
$\Rightarrow m>9$ or $m<5$
$\therefore m$ has no value between $5$ and $9$.