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Question: If x is real, then greatest and least values of \(\frac{x^{2} - x + 1}{x^{2} + x + 1}\) are...

If x is real, then greatest and least values of x2−x+1x2+x+1\frac{x^{2} - x + 1}{x^{2} + x + 1} are

A

3, –1/2

B

3, 1/3

C

– 3, –1/3

D

None of these

Answer

3, 1/3

Explanation

Solution

Let y=x2−x+1x2+x+1y = \frac{x^{2} - x + 1}{x^{2} + x + 1}

x2(y−1)+(y+1)x+(y−1)=0x^{2}(y - 1) + (y + 1)x + (y - 1) = 0

∵ x is real, therefore b2−4ac≥0b^{2} - 4ac \geq 0

⇒ (y+1)2−4(y−1)(y−1)≥0(y + 1)^{2} - 4(y - 1)(y - 1) \geq 0 ⇒ 3y2−10y+3≤03y^{2} - 10y + 3 \leq 0

⇒ (3y−1)(y−3)≤0(3y - 1)(y - 3) \leq 0 ⇒ (y−13)(y−3)≤0\left( y - \frac{1}{3} \right)(y - 3) \leq 0 ⇒ 13≤y≤3\frac{1}{3} \leq y \leq 3

Thus greatest and least values of expression are 3, 1/3 respectively.