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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Functions

BITSAT 2009 · Mathematics · Functions

The range of the function (f(x)= (x^2-x+1)/(x^2+x+1) ) where ( x ∈...

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

Options

  1. A((-\infty, 3])
  2. B((-\infty, \infty))
  3. C([3, \infty))
  4. D(\left[\frac{1}{3},3\right])

Correct answer

D. (\left[\frac{1}{3},3\right])

Step-by-step solution

$\begin{aligned}& \text{Let } y=\frac{x^2-x+1}{x^2+x+1}\\ & \Rightarrow \mathrm{x}^2(\mathrm{y}-1)+\mathrm{x}(\mathrm{y}+1)+(\mathrm{y}-1)=0 \\ & \Rightarrow x=\frac{-(y+1) \pm \sqrt{(y+1)^2-4(y-1)^2}}{2(y-1)}\\ & =\frac{-(y+1) \pm \sqrt{-3 y²+10 y-3}}{2(y-1)} \text{ is real iff }\\ & \mathrm{y}-1 ≠ 0 \Rightarrow \mathrm{y} \neq1 \\ & \text{ If } y=1 \text{ then original equation gives } \mathrm{x}=0 \text{, }\\ & \text{ so taking } \mathrm{y}=1 \\ & \text{ Also }3 y²-10 y+3 \leq 0 \\ & \Rightarrow(3 y-1)(y-3) \leq 0 \\ & \Rightarrow \mathrm{y} \in\left[\frac{1}{3},3\right] \quad \therefore \text{ Range is }\left[\frac{1}{3},3\right] \\ & \end{aligned}$)
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