Q41 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics

The range of the real valued function $\mathrm{f}(\mathrm{x})=\frac{x^2+x+1}{x}$ is

  • A
    $(-\infty, 1) \cup(1, \infty)$
  • B
    $(-\infty,-1] \cup[1, \infty)$
  • C
    $(-\infty,-2] \cup[3, \infty)$
  • Correct answer
    D
    $(-\infty,-1] \cup[3, \infty)$

Solution

$y=\frac{x^2+x+1}{x}$ $\begin{aligned} & y x=x^2+x+1 \\ & x^2+(1-y) x+1=0\end{aligned}$ $\therefore \quad f(x)$ is real valued function i.e., $x \in \mathrm{R}$ $\begin{aligned} & \therefore \quad \mathrm{D} \geq 0 \\ & (1-y)^2-4(1)(1) \geq 0 \\ & 1+y^2-2 y-4 \geq 0 \\ & y^2-2 y-3 \geq 0 \\ & y^2-3 y+y-3 \geq 0 \\ & y(y-3)+1(y-3) \geq 0 \\ & (y+1)(y-3) \geq 0 \\ & \therefore \quad y \in(-\infty,-1] \cup[3, \infty) \\ & \therefore \quad \text { Range of } f(x) \text { is }(-\infty,-1] \cup[3, \infty)\end{aligned}$

Hence, the correct answer is option: 'D'.

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