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If $x$ is real,$\large\frac{x^2+3yx-71}{x^2+2x-7}$ cannot lie between

$\begin{array}{1 1}(A)\;(-9,-5)&(B)\;(-9,5)\\(C)\;(5,9)&(D)\;(-5,9)\end{array} $

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$y=\large\frac{x^2+3yx-71}{x^2+2x-7}$
$\Rightarrow (y-1)x^2+2(y-17)x-(7y-71)=0$
$x$ is real
$\therefore D \geq 0$
$\Rightarrow 4(y-17)^2+4(y-1)(7y-71)\geq 0$
$\Rightarrow (y-5)(y-9)\geq 0$
$\Rightarrow y\notin (5,9)$
Hence (C) is the correct answer.
answered Apr 14, 2014 by sreemathi.v
 

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