A) \[a=2,b=32\]
B) \[a=3,b=12\]
C) \[a=4,b=16\]
D) \[a=12,b=3\]
Correct Answer: A
Solution :
Let \[\beta =\alpha r,=\alpha {{r}^{2}},\delta =\alpha {{r}^{3}},r>1\] For equation \[{{x}^{2}}-3x+a=0\] \[\alpha +\beta =3\] \[\Rightarrow \] \[\alpha +\alpha r=3\] \[\Rightarrow \] \[\alpha (1+r)=3\] ...(i) and \[\alpha .\beta =a\] \[\Rightarrow \] \[a=\alpha .\alpha r={{\alpha }^{2}}r\] ..(ii) For equation\[{{x}^{2}}-12x+b=0\] \[\gamma +\delta =12\] \[\Rightarrow \] \[\alpha {{r}^{2}}+\alpha {{r}^{3}}=12\] \[\Rightarrow \] \[\alpha {{r}^{2}}(1+r)=12\] ??(iii) and \[\gamma \delta =b\] \[\Rightarrow \] \[b=\alpha {{r}^{2}}.\alpha {{r}^{3}}={{\alpha }^{2}}{{r}^{5}}\] ...(iv) From Eqs. (i) and (iii), \[{{r}^{2}}.3=12\] \[\Rightarrow \] \[{{r}^{2}}=4\] \[\Rightarrow \] \[r=2\] Then, \[\alpha (1+2)=3\] \[\Rightarrow \] \[\alpha =1\] On putting the values of \[\alpha \]and r in Eqs. (ii) and (iv), we get \[a={{\alpha }^{2}}.r={{(1)}^{2}}\times 2=2\] and \[b={{(1)}^{2}}{{(2)}^{5}}=32\]
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