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AP EAMCET · Maths · Functions

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

  1. A \((-\infty, 1) \cup(1, \infty)\)
  2. B \((-\infty,-1] \cup[1, \infty)\)
  3. C \((-\infty,-2] \cup[3, \infty)\)
  4. D \((-\infty,-1] \cup[3, \infty)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \((-\infty,-1] \cup[3, \infty)\)

Step-by-step Solution

Detailed explanation

\(y=\frac{x^2+x+1}{x}\) \(\begin{aligned} & y x=x^2+x+1 \\ & x^2+(1-y) x+1=0\end{aligned}\) \(\therefore \quad f(x)\) is real valued function i.e., \(x \in \mathrm{R}\)…
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