Range of function f(x)=x2+x+2x2+x+1x∈R is

Let y=x2+x+2x2+x+1=1+1x2+x+1=1+1(x+1/2)2+3/4>1⇒ 1y−1=x2+x+1=x+122+34≥34Thus, y>1 and y−1≤4/3⇒y≤7/3∴ 1<y≤7/3

Range of function f(x)=x2+x+2x2+x+1x∈R is

Let y=x2+x+2x2+x+1=1+1x2+x+1=1+1(x+1/2)2+3/4>1⇒ 1y−1=x2+x+1=x+122+34≥34Thus, y>1 and y−1≤4/3⇒y≤7/3∴ 1<y≤7/3

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Range of function $f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1}$
$x \in R$ is

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(1, 7/3]

(1, 3/2)

(1, 7/5]

$(1, \infty)$

answer is C.

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Detailed Solution

Let $y = \frac{x^{2} + x + 2}{x^{2} + x + 1} = 1 + \frac{1}{x^{2} + x + 1}$

$\begin{aligned} = 1 + \frac{1}{(x + 1 / 2)^{2} + 3 / 4} > 1 \\ \Rightarrow \frac{1}{y - 1} = x^{2} + x + 1 = {(x + \frac{1}{2})}^{2} + \frac{3}{4} \geq \frac{3}{4} \end{aligned}$

Thus, $y > 1$ and $y - 1 \leq 4 / 3 \Rightarrow y \leq 7 / 3$

$∴ 1 < y \leq 7 / 3$

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Range of function $f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1}$
$x \in R$ is