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If $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ are positive quantities, then the condition for the difference between the arithmetic mean and the geometric mean to be greater than 1 is
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Verified Answer
The correct answer is:
$\left|\sqrt{\mathrm{x}_{1}}+\sqrt{\mathrm{x}_{2}}\right|>\sqrt{2}$
Arithmetic mean of $\mathrm{x}_{1}, \mathrm{x}_{2}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}$
Geometric mean of $\mathrm{x}_{1}, \mathrm{x}_{2}=\sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}$
Given, $\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}-\sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}>1$
$\Rightarrow \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}>\sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}+1$
$\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}>2 \sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}+2$
$\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}-2 \sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}>2$
$\Rightarrow\left(\sqrt{\mathrm{x}_{1}}\right)^{2}+\left(\sqrt{\mathrm{x}_{2}}\right)^{2}-2 \sqrt{\mathrm{x}_{1}} \sqrt{\mathrm{x}_{2}}>2$
$\Rightarrow\left(\sqrt{\mathrm{x}_{1}}-\sqrt{\mathrm{x}_{2}}\right)^{2}>2$
$\Rightarrow\left|\sqrt{\mathrm{x}_{1}}-\sqrt{\mathrm{x}_{2}}\right|>\sqrt{2}$
Correct option (c).
Geometric mean of $\mathrm{x}_{1}, \mathrm{x}_{2}=\sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}$
Given, $\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}-\sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}>1$
$\Rightarrow \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}>\sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}+1$
$\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}>2 \sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}+2$
$\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}-2 \sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}>2$
$\Rightarrow\left(\sqrt{\mathrm{x}_{1}}\right)^{2}+\left(\sqrt{\mathrm{x}_{2}}\right)^{2}-2 \sqrt{\mathrm{x}_{1}} \sqrt{\mathrm{x}_{2}}>2$
$\Rightarrow\left(\sqrt{\mathrm{x}_{1}}-\sqrt{\mathrm{x}_{2}}\right)^{2}>2$
$\Rightarrow\left|\sqrt{\mathrm{x}_{1}}-\sqrt{\mathrm{x}_{2}}\right|>\sqrt{2}$
Correct option (c).
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