A) (3, 12)
B) (12, 3)
C) (2, 32)
D) (4, 16)
Correct Answer: C
Solution :
Since \[\alpha ,\ \beta ,\ \gamma ,\ \delta \] form an increasing G.P., so \[\alpha \delta =\beta \gamma \]where\[\alpha <\beta <\gamma <\delta \]. On solving \[{{x}^{2}}-3x+a=0\], we get \[x=\frac{1}{2}(3\pm \sqrt{9-4a})\]. Also \[\alpha <\beta \]. Hence \[\alpha =\frac{1}{2}(3-\sqrt{9-4a}),\ \beta =\frac{1}{2}(3+\sqrt{9-4a})\] Similarly from \[{{x}^{2}}-12x+b=0\], we get \[\gamma =\frac{1}{2}(12-\sqrt{144-4b}),\ \delta =\frac{1}{2}(12+\sqrt{144-4b})\] Substituting these values of \[\alpha ,\ \beta ,\ \gamma ,\ \delta \] in \[\alpha \delta =\beta \gamma \] and simplifying, we get\[(a,\ b)=(2,\ 32)\]. Trick: Check the alternates; only (c) satisfies the condition.
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