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China Mathematical Competition

Algebra Difficulty 3.8 AMC 10/12 Find the answer China

Suppose $f(x) = x^3 + \log_2(x + \sqrt{x^2+1})$ . For any $a, b \in \mathbb{R}$ , to satisfy $f(a) + f(b) \ge 0$ , the condition $a + b \ge 0$ is:

Pick one

Solution

Obviously $f(x) = x^3 + \log_2(x + \sqrt{x^2+1})$ is an odd function and is monotonically increasing. So, if $a+b \ge 0$ , i.e. $a \ge -b$ , we get $f(a) \ge f(-b)$ , $f(a) \ge -f(b)$ , and that means $f(a)+f(b) \ge 0$ .

On the other hand, if $f(a) + f(b) \ge 0$ , then $f(a) \ge -f(b) = f(-b)$ . So $a \ge -b$ , $a+b \ge 0$ . Answer: A.

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