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The range of $y=\large\frac{x^2-x+1}{x^2+x+1}$ if $x$ is real

$\begin{array}{1 1}(A)\;\big[-\large\frac{1}{3},\normalsize 3\big]&(B)\;\big[\large\frac{1}{3},\normalsize 3\big]\\(C)\;\big[ 3,-\large\frac{1}{3}\big]&(D)\;[-3,3]\end{array} $

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$y=\large\frac{x^2-x+1}{x^2+x+1}$
$\Rightarrow (y-1)x^2+(y+1)x+(y-1)=0$
Since x is real
$\Rightarrow D \geq 0$
$\Rightarrow (y+1)^2-4(y-1)^2\geq 0$
$\Rightarrow (y-3)^2(3y-1)\leq 0$
$\Rightarrow y \in \big[\large\frac{1}{3}$$,3\big]$
Hence (B) is the correct answer.
answered Apr 14, 2014 by sreemathi.v
 

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