Q267 / 270 · NDA 2017 (Phase 2) · Mathematics
If $x_1$ and $x_2$ are positive quantities, then the condition for the difference between the arithmetic mean and the geometric mean to be greater than 1 is
- A$\mathrm{x}_1+\mathrm{x}_2 \gt 2 \sqrt{\mathrm{x}_1 \mathrm{x}_2}$
- B$\sqrt{\mathrm{x}_1}+\sqrt{\mathrm{x}_2} \gt \sqrt{2}$
- Correct answerC$\left|\sqrt{\mathrm{x}_1}-\sqrt{\mathrm{x}_2}\right| \gt \sqrt{2}$
- D$\mathrm{x}_1+\mathrm{x}_2 \lt 2\left(\sqrt{\mathrm{x}_1 \mathrm{x}_2}+1\right)$
Solution
Arithmetic mean of $x_1, x_2=\frac{x_1+x_2}{2}$
Geometric mean of $\mathrm{x}_1, \mathrm{x}_2=\sqrt{\mathrm{x}_1 \mathrm{x}_2}$
Given, $\frac{\mathrm{x}_1+\mathrm{x}_2}{2}-\sqrt{\mathrm{x}_1 \mathrm{x}_2} \gt 1$
$\Rightarrow \frac{\mathrm{x}_1+\mathrm{x}_2}{2} \gt \sqrt{\mathrm{x}_1 \mathrm{x}_2}+1$
$\begin{aligned} & \Rightarrow \mathrm{x}_1+\mathrm{x}_2 \gt 2 \sqrt{\mathrm{x}_1 \mathrm{x}_2}+2 \\ & \Rightarrow \mathrm{x}_1+\mathrm{x}_2-2 \sqrt{\mathrm{x}_1 \mathrm{x}_2} \gt 2 \\ & \Rightarrow\left(\sqrt{\mathrm{x}_1}\right)^2+\left(\sqrt{\mathrm{x}_2}\right)^2-2 \sqrt{\mathrm{x}_1} \sqrt{\mathrm{x}_2} \gt 2\end{aligned}$
$\begin{aligned} & \Rightarrow\left(\sqrt{\mathrm{x}_1}-\sqrt{\mathrm{x}_2}\right)^2 \gt 2 \\ & \Rightarrow\left|\sqrt{\mathrm{x}_1}-\sqrt{\mathrm{x}_2}\right| \gt \sqrt{2}\end{aligned}$
Hence, the correct answer is option: 'C'.