If x is real and alpha, beta are maximum and | vedclass

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(A) Let $y = \frac{x^2-x+1}{x^2+x+1}$.To find the extrema,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{(2x-1)(x^2+x+1) - (2x+1)(x^2

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$\frac{10}{3}$

Vedclass · Answer

(A) Let $y = \frac{x^2-x+1}{x^2+x+1}$.To find the extrema,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{(2x-1)(x^2+x+1) - (2x+1)(x^2-x+1)}{(x^2+x+1)^2} = 0$.Simplifying the numerator:$(2x^3 + 2x^2 + 2x - x^2 - x - 1) - (2x^3 - 2x^2 + 2x + x^2 - x + 1) = 0$.$(2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1) = 0$.$2x^2 - 2 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.For $x = -1$,$y = \frac{(-1)^2 - (-1) + 1}{(-1)^2 + (-1) + 1} = \frac{1+1+1}{1-1+1} = \frac{3}{1} = 3$.For $x = 1$,$y = \frac{1^2 - 1 + 1}{1^2 + 1 + 1} = \frac{1}{3}$.Thus,the maximum value $\alpha = 3$ and the minimum value $\beta = \frac{1}{3}$.Therefore,$\alpha + \beta = 3 + \frac{1}{3} = \frac{10}{3}$.

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

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