A) \[\frac{7}{6}\]
B) \[\frac{8}{7}\]
C) \[\frac{9}{8}\]
D) \[\frac{10}{9}\]
Correct Answer: C
Solution :
\[(a-b){{x}^{2}}+ax+1=0\left\langle _{2\alpha }^{\alpha } \right.\] \[3\alpha \,=\frac{-a}{a-b};2\alpha =\frac{1}{a-b}\] \[\therefore \,\frac{2{{a}^{2}}}{{{(a-b)}^{2}}9}=\frac{1}{a-b}\,\Rightarrow \,2{{a}^{2}}=9(a-b)\] \[\Rightarrow \,2{{a}^{2}}-9a+9b=0\] Also,\[a\,\in \,R,\] \[\therefore \,\Delta \,\,\underline{>}\,0\Rightarrow \,81-72b\,\underline{>}\,0\Rightarrow \,b\,\underline{<}\,\frac{9}{8}.\]
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