Question

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

• A. Necessary and sufficient
• B. Necessary but not sufficient
• C. Not necessary but sufficient
• D. Neither necessary nor sufficient

Answer 1

Answer: (A)

Necessary and sufficient

Answer 2

We have

$$f(x)=x^3+\log_2\Bigl(x+\sqrt{x^2+1}\Bigr).$$

Observation: Notice that

$$\log_2\Bigl(x+\sqrt{x^2+1}\Bigr)$$

is an odd function because

$$\log_2\Bigl(-x+\sqrt{x^2+1}\Bigr) + \log_2\Bigl(x+\sqrt{x^2+1}\Bigr)=\log_2\Bigl[(x+\sqrt{x^2+1})(-x+\sqrt{x^2+1})\Bigr]=\log_2(1)=0.$$

Also, $x^3$ is odd. Hence,

$$f(-x)=-f(x).$$

Thus, $f$ is odd.

Now, for any $a,b\in\Bbb R$,

$$f(a)+f(b)\ge0 \quad\Longleftrightarrow\quad f(b)\ge -f(a)=f(-a).$$

Since $f$ is strictly increasing (its derivative is always positive), the inequality $f(b)\ge f(-a)$ is equivalent to

$$b\ge -a\quad\Longleftrightarrow\quad a+b\ge0.$$

Thus the condition $a+b\ge0$ is both necessary and sufficient to guarantee $f(a)+f(b)\ge0$.