AP EAMCET · Maths · Functions
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 \((-\infty,-2] \cup[3, \infty)\)
- D \((-\infty,-1] \cup[3, \infty)\)
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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