A) 8 and -5
B) -5 and 9
C) 0 and 9
D) 5 and 9
Correct Answer: D
Solution :
Let\[y=\frac{{{x}^{2}}+34x-71}{{{x}^{2}}+2x-7}\] \[\Rightarrow \]\[{{x}^{2}}(y-1)+2(y-17)x+(71-7y)=0\] For real values of\[x\], its discriminant,\[D>0\] \[\therefore \] \[4{{(y-17)}^{2}}-4(y-1)(71-7y)\ge 0\] \[\Rightarrow \] \[({{y}^{2}}-34y+289)\] \[\Rightarrow \] \[{{y}^{2}}-14y+45\ge 0\] \[-(71y-7{{y}^{2}}-71+7y)\ge 0\] \[\Rightarrow \] \[{{y}^{2}}-14y+45\ge 0\] \[\Rightarrow \] \[(y-5)(y-9)\ge 0\] \[\Rightarrow \] \[y\le 5\]and\[y\ge 9\] Hence, \[y\]does not lie between 5 and 9
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