Question

. Determine the range of (x ^ 2 + x + 1)/(x ^ 2 - x + 1)​

Answer 1

SolutioN :

$\tt \rightarrow \: f(x) = \dfrac{ {x}^{2} + x + 1 }{ {x}^{2} - x + 1 }$

$\tt \rightarrow \: \dfrac{ Quadratic}{ Quadratic } \: \: form.$

$\tt \rightarrow \: \dfrac{ {x}^{2} + 2x - x+ 1 }{ {x}^{2} - x + 1 }$

$\tt \rightarrow \: \dfrac{ {x}^{2} - x+ 1 }{ {x}^{2} - x + 1 } + \frac{2x}{ {x}^{2} - x + 1 }$

$\tt \rightarrow \: 1 + \frac{2x}{ {x}^{2} - x + 1 }$

$\tt \rightarrow \: 1 + \frac{2}{ x- 1 + \frac{1}{x} }$

$\tt \rightarrow \: 1 + \frac{2}{ x + \frac{1}{x} - 1 }$

Now, Apply AM and GM.

$\tt \rightarrow \: AM \geq GM$

$\tt \rightarrow \: \frac{x + \dfrac{1}{x} }{2} \geq \sqrt{x .\dfrac{1}{x} }$

$\tt \rightarrow \: x + \dfrac{1}{x} \geq 2.$

Now,

$\tt \rightarrow \: 1 + \frac{2}{ [-2 , 2 ]- 1 }$

$\tt \rightarrow \: 1 + \frac{2}{ [-3 , 1 ] }$

$\tt \rightarrow \: 1 + [- \frac{2}{3} , 2]$

$\tt \rightarrow \: [\frac{1}{3} , 3]$

Therefore, the range of function f(x) is [ 1/3 , 3 ]

Answer 2

Answer:

$\sf \: Let \: y = \: \big \: \frac{x {}^{2} - x + 1 }{x {}^{2} + x + 1}$

$\sf \: \big ⇒yx {}^{2} + yx + y \: = \: x {}^{2} - x + 1$

$\sf⇒ ( x - 1)x {}^{2} + (y + 1)x + (y - 1) = 0$

$\sf \: If \:  x \:∈ \: R, \: then$

$\sf \: Discriminent \: \geqslant 0$

$\sf⇒(y + 1) {}^{2} - 4(y - 1) {}^{2} \geqslant 0$

$\sf⇒ \: 3y {}^{2} - 10y - 3 \geqslant 0$

$\sf⇒ \: 3y {}^{2} - 10y + 3 \leqslant 0$

$\sf⇒ \: (3y - 1) \: (y - 3) \leqslant 3$

$\sf \: Therefore \: range = [\frac{1}{3}, 3]$