For x∈R, the range of x2+4x+1x2+x+1 is

Let y=x2+4x+1x2+x+1⇒ y−1x2+y−4x+y−1=0Now x∈R⇒Δ≥0⇒−2≤y≤2⇒ Range is −2,2

For x∈R, the range of x2+4x+1x2+x+1 is

Let y=x2+4x+1x2+x+1⇒ y−1x2+y−4x+y−1=0Now x∈R⇒Δ≥0⇒−2≤y≤2⇒ Range is −2,2

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For $x \in R,$ the range of $\frac{x^{2} + 4 x + 1}{x^{2} + x + 1}$ is

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[3, 5]

$[- \infty, - 3]$

$ϕ$

[-2, 2]

answer is D.

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Detailed Solution

Let $y = \frac{x^{2} + 4 x + 1}{x^{2} + x + 1} \Rightarrow (y - 1) x^{2} + (y - 4) x + (y - 1) = 0$

Now $x \in R \Rightarrow Δ \geq 0 \Rightarrow - 2 \leq y \leq 2 \Rightarrow R a n g e i s [- 2, 2]$

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For x∈R, the range of x2+4x+1x2+x+1 is

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For $x \in R,$ the range of $\frac{x^{2} + 4 x + 1}{x^{2} + x + 1}$ is