Question

Range of $f\left(x\right)=\frac{x^2+34x-71}{x^2+2x-7}$ is

• A. $[5,9]$
• B. $(5,9]$
• C. $(-\infty ,5]\cup [9,\infty )$
• D. $(-\infty ,5)\cup (9,\infty )$

Answer 1

The correct option is C

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

Explanation for the correct option:

To find the range of $f\left(x\right)$:

Considering the given function as

$\frac{x^2+34x-71}{x^2+2x-7}=y$ On simplifying we get

$$\begin{gathered} \Rightarrow x^2+34x-71=y{x}^2+2xy-7y \\ \Rightarrow x^2(1-y)+x(34-2y)-71+7y=0 \end{gathered}$$

this is a quadratic equation and for this to have real-valued solutions its discriminant should be greater than or equal to $0$

discriminant of a quadratic equation $a{x}^2+bx+c$ is $b^2-4ac$

$$\begin{gathered} {(34-2y)}^2-4(1-y)\left(\right(7y-71)\ge 0 \\ \Rightarrow (1156+4y^2-136y)-4(7y-71-7y^2+71y)\ge 0 \\ \Rightarrow 32y^2-448y+1440\ge 0 \\ \Rightarrow y^2-14y+45\ge 0 \\ \Rightarrow (y-9\left)\right(y-5)\ge 0 \end{gathered}$$

by wavy curve method: we see that

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

Hence, option (C) is the correct answer