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Determine the range of
(x² + x + 1)/(x² - x + 1)​

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Answer 1

Answer:

Let y =

= x2 + x + 1x2 - x + 1

= x2 + x + 1x2 - x + 1⇒yx2 + yx+y=x2-x+1 ⇒ (x1)x2 + (y + 1)x+ (y − 1) = 0

= x2 + x + 1x2 - x + 1⇒yx2 + yx+y=x2-x+1 ⇒ (x1)x2 + (y + 1)x+ (y − 1) = 0If x € R, then

= x2 + x + 1x2 - x + 1⇒yx2 + yx+y=x2-x+1 ⇒ (x1)x2 + (y + 1)x+ (y − 1) = 0If x € R, theninant >= 0

⇒ (y + 1)2 - 4(y − 1)2 ≥ 0

⇒-3y2 +10y - 320

⇒-3y2 +10y - 320⇒ 3y2 - 10y +3≤0

⇒-3y2 +10y - 320⇒ 3y2 - 10y +3≤0⇒ (3y - 1)(y - 3) ≤ 08

⇒-3y2 +10y - 320⇒ 3y2 - 10y +3≤0⇒ (3y - 1)(y - 3) ≤ 08⇒ 31 ≤ y ≤ 3

⇒-3y2 +10y - 320⇒ 3y2 - 10y +3≤0⇒ (3y - 1)(y - 3) ≤ 08⇒ 31 ≤ y ≤ 3=Range = [31, 3]

Answer 2

Answer:

Let y = (x2 + x + 1)/(x2 - x + 1)

⇒ x2y – xy + y = x2 + x + 1

⇒ x2y – xy + y – x2 – x – 1 = 0

⇒ x2(y – 1) – x(y + 1) + (y – 1) = 0

x is real ⇒ b2 – 4ac ≥ 0 ⇒ (y + 1)2 – 4(y – 1)2 ≥ 0

⇒ (y + 1)2 – (2y – 2)2 ≥ 0

⇒ (y + 1 + 2y – 2) (y + 1 – 2y + 2) ≥ 0

⇒ (3y – 1) (–y + 3) ≥ 0

⇒ –(3y – 1) (y – 3) ≥ 0

a = coeff. of y2 = –3 < 0.,

But The expression ≥ 0 ⇒ y lies between 1/3 and 3

∴ The range of (x2 + x + 1)/(x2 - x + 1) is [1/3, 3]