Question
Range of the function $$f\left( x \right) = \frac{{{x^2} + x + 2}}{{{x^2} + x + 1}};x \in R$$ is
A.
$$\left( {1,\infty } \right)$$
B.
$$\left( {1,\frac{{11}}{7}} \right]$$
C.
$$\left( {1,\frac{7}{3}} \right]$$
D.
$$\left( {1,\frac{7}{5}} \right]$$
Answer :
$$\left( {1,\frac{7}{3}} \right]$$
Solution :
We have $$f\left( x \right) = \frac{{{x^2} + x + 2}}{{{x^2} + x + 1}} = \frac{{\left( {{x^2} + x + 1} \right) + 1}}{{{x^2} + x + 1}}$$
$$ = 1 + \frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{3}{4}}}$$
We can see here that as $$x \to \infty ,f\left( x \right) \to 1$$ which is the min value of $$f\left( x \right)\,{\text{.i}}{\text{.e}}{\text{.}}\,{f_{\min }} = 1.$$ Also $$f\left( x \right)$$ is max when $${\left( {x + \frac{1}{2}} \right)^2} + \frac{3}{4}$$ is min which is so when $$x = - \frac{1}{2}$$
$$\eqalign{
& {\text{i}}{\text{.e when}}\,{\left( {x + \frac{1}{2}} \right)^2} + \frac{3}{4} = \frac{3}{4} \cr
& \therefore {f_{\max }} = 1 + \frac{3}{4} = \frac{7}{3} \cr
& \therefore {R_f} = \left( {1,\frac{7}{3}} \right] \cr} $$