Q50 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics
If $x$ is real and $\alpha, \beta$ are maximum and minimum values of $\frac{x^2-x+1}{x^2+x+1}$ respectively then $\alpha+\beta=$
- Correct answerA$\frac{10}{3}$
- B$\frac{8}{3}$
- C$\frac{4}{3}$
- D$\frac{2}{3}$
Solution
$y=\frac{x^2-x+1}{x^2+x+1}$
$\begin{aligned} & \Rightarrow \frac{d y}{d x}=\frac{(2 x-1)\left(x^2+x+1\right)-(2 x+1)\left(x^2-x+1\right)}{\left(x^2+x+1\right)^2}=0 \\ & \Rightarrow \frac{2 x^2-2}{\left(x^2+x+1\right)^2}=0 \Rightarrow x= \pm 1 \end{aligned}$
$\operatorname{Maximum}$ value $(\alpha)=f(-1)=\frac{1+1+1}{1-1+1}=3$ Minimum value $(\beta)=f(1)=\frac{1}{3} \Rightarrow \alpha+\beta=\frac{10}{3}$.
$\begin{aligned} & \Rightarrow \frac{d y}{d x}=\frac{(2 x-1)\left(x^2+x+1\right)-(2 x+1)\left(x^2-x+1\right)}{\left(x^2+x+1\right)^2}=0 \\ & \Rightarrow \frac{2 x^2-2}{\left(x^2+x+1\right)^2}=0 \Rightarrow x= \pm 1 \end{aligned}$
$\operatorname{Maximum}$ value $(\alpha)=f(-1)=\frac{1+1+1}{1-1+1}=3$ Minimum value $(\beta)=f(1)=\frac{1}{3} \Rightarrow \alpha+\beta=\frac{10}{3}$.
Hence, the correct answer is option: 'A'.