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AP EAMCET · Maths · Application of Derivatives

If \(x\) is real and \(\alpha, \beta\) are maximum and minimum values of \(\frac{x^2-x+1}{x^2+x+1}\) respectively then \(\alpha+\beta=\)

  1. A \(\frac{10}{3}\)
  2. B \(\frac{8}{3}\)
  3. C \(\frac{4}{3}\)
  4. D \(\frac{2}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{10}{3}\)

Step-by-step Solution

Detailed explanation

\(y=\frac{x^2-x+1}{x^2+x+1}\) \(\begin{aligned} & \Rightarrow \frac{d y}{d x}=\frac{(2 x-1)\left(x^2+x+1\right)-(2 x+1)\left(x^2-x+1\right)}{\left(x^2+x+1\right)^2}=0 \\ & \Rightarrow \frac{2 x^2-2}{\left(x^2+x+1\right)^2}=0 \Rightarrow x= \pm 1 \end{aligned}\)…