Question

The given expression $\frac{x^2+34x-71}{x^2+2x-7}$ lies in the range

• A. $y\in (-\infty ,9]\cup [5,\infty )$
• B. $y\in (-\infty ,4]\cup [5,\infty )$
• C. $y\in (-\infty ,5]\cup [9,\infty )$
• D. None of these

Answer 1

Answer: (C)

$y\in (-\infty ,5]\cup [9,\infty )$

Let $\frac{x^2+34x-71}{x^2+2x-7}=y$
$x^2(y-1)+(2y-34)x+71-7y=0$
For real values of $x$, discriminant $\ge 0$
$\therefore (2y-34{)}^2-4(y-1\left)\right(71-7y)\ge 0$
$8y^2-112y+360\ge 0$
$y^2-14y+45>0$
$(y-9)(y-5)\ge 0$
$yϵ(-\infty ,5]\cup [9,\infty ])$
Hence, $y$ can never lie between $5$ and $9$.