Range of the function f(x) = frac{x^2 + x + 2 | vedclass

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(C) Let $y = \frac{x^2 + x + 2}{x^2 + x + 1}$.We can rewrite the function as $y = \frac{(x^2 + x + 1) + 1}{x^2 + x + 1} = 1 + \frac{1}{x^2 + x + 1}$.T

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$(1, 7/3]$

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(C) Let $y = \frac{x^2 + x + 2}{x^2 + x + 1}$.We can rewrite the function as $y = \frac{(x^2 + x + 1) + 1}{x^2 + x + 1} = 1 + \frac{1}{x^2 + x + 1}$.To find the range,we need to find the range of the expression $g(x) = x^2 + x + 1$.Completing the square for $g(x)$,we get $g(x) = (x + 1/2)^2 + 3/4$.The minimum value of $g(x)$ is $3/4$ (at $x = -1/2$) and the maximum value is $\infty$.Thus,the range of $g(x)$ is $[3/4, \infty)$.Now,the range of $\frac{1}{g(x)}$ is $(0, 1/(3/4)] = (0, 4/3]$.Adding $1$ to this range,the range of $f(x)$ is $(1, 1 + 4/3] = (1, 7/3]$.

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

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