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Mathematics Question on Relations and functions

The range of the function f(x)=x2−x+1x2+x+1f\left(x\right)=\frac{x^{2}-x+1}{x^{2}+x+1} where x∈Rx \in R

A

(−∞,3](-\infty, 3]

B

(−∞,∞)\left(-\infty, \infty\right)

C

[3,∞)[3, \infty)

D

[13,3]\left[\frac{1}{3}, 3\right]

Answer

[13,3]\left[\frac{1}{3}, 3\right]

Explanation

Solution

We have, y=x2−x+1x2+x+1y=\frac{x^{2}-x+1}{x^{2}+x+1} ⇒y x2+yx+y−x2+x−1=0\Rightarrow y\,x^{2}+yx+y-x^{2}+x-1=0 ⇒x2(y−1)+x(y+1)+(y−1)=0\Rightarrow x^{2}\left(y-1\right)+x\left(y+1\right)+\left(y-1\right)=0 ∴x=−(y+1)±(y+1)2−4(y−1)22(y−1)\therefore x=\frac{-\left(y+1\right) \pm\sqrt{\left(y+1\right)^{2}-4\left(y-1\right)^{2}}}{2\left(y-1\right)} =−(y+1)±−3y2+10y−32(y−1)=\frac{-\left(y+1\right) \pm\sqrt{-3y^{2}+10y-3}}{2\left(y-1\right)} If y=1y = 1 then original equation gives x=0x = 0. Also, −3y2+10y−3≥0-3y^{2}+10y-3 \ge 0 ⇒3y2−10y+3≤0\Rightarrow 3y^{2}-10y+3 \le 0 ⇒3y2−9y−y+3≤0\Rightarrow 3y^{2}-9y-y+3 \le 0 ⇒(3y−1)(y−3)≤0\Rightarrow \left(3y-1\right)\left(y-3\right) \le 0 ⇒y∈[13,3]\Rightarrow y \in\left[\frac{1}{3}, 3\right] ∴\therefore Range is [13,3]\left[\frac{1}{3}, 3\right].