Q37 / 110 · VITEEE 2022 · Mathematics
Range of the function \(f(x)=\frac{x^2+x+2}{x^2+x+1} ; x \in R\) is
- A\((1, \infty)\)
- B\((1,11 / 7]\)
- Correct answerC\((1,7 / 3]\)
- D\((1,7 / 5]\)
Solution
We have
\(\begin{aligned}
f(x) & =\frac{x^2+x+2}{x^2+x+1}=\frac{\left(x^2+x+1\right)+1}{x^2+x+1} \\
& =1+\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}
\end{aligned}\)
We can see here that as \(x \rightarrow \infty, f(x) \rightarrow 1\) which is the \(\min\) value of \(f(x)\). i.e. \(f_{\min }=1\). Also \(f(x)\) is max when \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\) is min which is so when \(x=-1 / 2\)
i.e. when \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=\frac{3}{4}\).
\(\therefore f_{\max }=1+\frac{1}{3 / 4}=7 / 3 \quad \therefore R_f=(1,7 / 3]\)
Hence, the correct answer is option: 'C'.