Library / China Mathematical Competition (Jiangxi) /1 of 9

China Mathematical Competition (Jiangxi)

Algebra Difficulty 3.9 AMC 10/12 Find the answer China

Let $k$ be a real number such that the inequality $\sqrt{x-3} + \sqrt{6-x} \ge k$ has a solution. The maximum value of $k$ is:

Pick one

Solution

Set $y = \sqrt{x-3} + \sqrt{6-x}$ , $3 \le x \le 6$ .
Then
$\begin{aligned} y^2 &= (x-3) + (6-x) + 2\sqrt{(x-3)(6-x)} \\ &\le 2[(x-3) + (6-x)] = 6. \end{aligned}$
So $0 < y \le \sqrt{6}$ , and the maximum value of $k$ is $\sqrt{6}$ . Answer: D.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.