For all real values of x,the minimum value of | vedclass

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(B) Let $f(x) = \frac{x^2-x+1}{x^2+x+1}$.We can rewrite the expression as:$f(x) = \frac{(x^2+x+1) - 2x}{x^2+x+1} = 1 - \frac{2x}{x^2+x+1}$.To minimize

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Let $f(x) = \frac{x^2-x+1}{x^2+x+1}$.We can rewrite the expression as:$f(x) = \frac{(x^2+x+1) - 2x}{x^2+x+1} = 1 - \frac{2x}{x^2+x+1}$.To minimize $f(x)$,we need to maximize the term $\frac{2x}{x^2+x+1}$.For $x 
eq 0$,we have $\frac{2x}{x^2+x+1} = \frac{2}{x + \frac{1}{x} + 1}$.Since $x + \frac{1}{x} \geq 2$ for $x > 0$ and $x + \frac{1}{x} \leq -2$ for $x Let $y = \frac{x^2-x+1}{x^2+x+1} \implies y(x^2+x+1) = x^2-x+1 \implies (y-1)x^2 + (y+1)x + (y-1) = 0$.For $x$ to be real,the discriminant $D \geq 0$:$(y+1)^2 - 4(y-1)^2 \geq 0 \implies (y+1-2y+2)(y+1+2y-2) \geq 0 \implies (3-y)(3y-1) \geq 0$.This implies $\frac{1}{3} \leq y \leq 3$.Thus,the minimum value is $\frac{1}{3}$.

For all real values of $x$,the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

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