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Algebra Difficulty 3.8 AMC 10/12 Find the answer China

Suppose f(x)=x3+log2(x+x2+1)f(x) = x^3 + \log_2(x + \sqrt{x^2+1}). For any a,bRa, b \in \mathbb{R}, to satisfy f(a)+f(b)0f(a) + f(b) \ge 0, the condition a+b0a + b \ge 0 is:

Pick one

Solution

Obviously f(x)=x3+log2(x+x2+1)f(x) = x^3 + \log_2(x + \sqrt{x^2+1}) is an odd function and is monotonically increasing. So, if a+b0a+b \ge 0, i.e. aba \ge -b, we get f(a)f(b)f(a) \ge f(-b), f(a)f(b)f(a) \ge -f(b), and that means f(a)+f(b)0f(a)+f(b) \ge 0.

On the other hand, if f(a)+f(b)0f(a) + f(b) \ge 0, then f(a)f(b)=f(b)f(a) \ge -f(b) = f(-b). So aba \ge -b, a+b0a+b \ge 0. Answer: A.

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