A) \[\left( 2-\sqrt{2},+2+\sqrt{2} \right)\]
B) \[\left( 2-\sqrt{3},+2+\sqrt{3} \right)\]
C) \[\left( -\sqrt{2},\sqrt{2} \right)\]
D) \[\left( -\sqrt{3},\sqrt{3} \right)\]
Correct Answer: B
Solution :
[b] \[{{A}_{1}}=\frac{3a+b}{4},{{A}_{2}}\frac{a+b}{2},{{A}_{3}}=\frac{a+3b}{4}\] For equation \[{{A}_{1}}{{x}^{2}}+2{{A}_{2}}x+{{A}_{3}}=0,\] \[D=4({{A}_{2}}^{2}-{{A}_{1}}{{A}_{3}})\] \[=4\left( {{\left( \frac{a+b}{2} \right)}^{2}}-\frac{(b+3a)}{4}\frac{(a+3b)}{4} \right)\] \[=\frac{4\left( {{a}^{2}}+{{b}^{2}}-4ab \right)}{16}<0\] \[\Rightarrow \,\,\,\,\,\,\,{{\left( \frac{a}{b} \right)}^{2}}-4\left( \frac{a}{b} \right)+1<0\] \[\Rightarrow \,\,\,\,2-\sqrt{3}<\frac{a}{b}<2+\sqrt{3}\]You need to login to perform this action.
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