A) 5 and 4
B) 5 and ?4
C) ? 5 and 4
D) None of these
Correct Answer: C
Solution :
\[\frac{{{x}^{2}}+14x+9}{{{x}^{2}}+2x+3}=y\] Þ \[{{x}^{2}}+14x+9={{x}^{2}}y+2xy+3y\] Þ \[{{x}^{2}}(y-1)+2x(y-7)+(3y-9)=0\] Since x is real, \\[4{{(y-7)}^{2}}-4(3y-9)(y-1)>0\] Þ \[4({{y}^{2}}+49-14y)-4(3{{y}^{2}}+9-12y)>0\] Þ \[4{{y}^{2}}+196-56y-12{{y}^{2}}-36+48y>0\] Þ \[8{{y}^{2}}+8y-160<0\] Þ \[{{y}^{2}}+y-20<0\] Þ \[(y+5)(y-4)<0\]; \ y lies between ? 5 and 4.
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