A) \[\frac{4}{9}and\,\,1\]
B) \[\frac{2}{9}and\,\,4\]
C) \[1\,\,and\,\,4\]
D) \[4\,\,and\,\,\frac{40}{3}\]
Correct Answer: A
Solution :
(a): Let \[\frac{\left( x-1 \right)\left( x+3 \right)}{\left( x-2 \right)\left( x+4 \right)}=y\] then, \[{{x}^{2}}+2x-3=y\left( {{x}^{2}}+2x-8 \right)\] or, \[{{x}^{2}}\left( 1-y \right)+2x\left( 1-y \right)+\left( 8y-3 \right)=0\] For x to be real, its Discriminant\[D\ge 0\], or, \[{{\left[ 2\left( 1-y \right) \right]}^{2}}-4\left( 1-y \right)\left( 8y-3 \right)\ge 0\] \[\Rightarrow \left( 1-y \right)\left( 4-9y \right)\ge 0\Rightarrow \left( 1-y \right)\left( \frac{4}{9}-y \right)\ge 0\] It shows that \[\left( 1-y \right)\] and \[\left( \frac{4}{9}-y \right)\] should be of the same sign. We get three cases; for y to be on the real number line, namely, \[y\in \left( -\infty ,\frac{4}{9} \right)\]; \[y\in \left( \frac{4}{9},1 \right)\] and \[y\in \left( 1,\infty \right)\] Case I: If \[y<\frac{4}{9}\], then,\[\left( 1-y \right)\left( \frac{4}{9}-y \right)>0\] Case II: If \[\frac{4}{9}<y<1\], then, \[\left( 1-y \right)\left( \frac{4}{9}-y \right)<0\] Case III: If \[y>1\] then, \[\left( 1-y \right)\left( \frac{4}{9}-y \right)>0\] Hence, for x to be real, y, (i.e., the value of the given expression) cannot lie between\[\frac{4}{9}\] and 1.
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