Question

The solution of $2^x + 2^{|x|} \geq 2\sqrt{2}$ is

• A. $(-\infty, log_2(\sqrt{2} + 1))$
• B. $(0, \infty)$
• C. $(\frac{1}{2}, log_2(\sqrt{2} - 1))$
• D. $(-\infty, log_2(\sqrt{2} - 1)] \cup [\frac{1}{2}, \infty)$

Answer 1

Answer: (D)

$(-\infty, log_2(\sqrt{2} - 1)] \cup [\frac{1}{2}, \infty)$

Answer 2

1. Case $x \geq 0$: $|x|=x \implies 2^x + 2^x \geq 2\sqrt{2} \implies 2 \cdot 2^x \geq 2\sqrt{2} \implies 2^{x+1} \geq 2^{3/2} \implies x+1 \geq 3/2 \implies x \geq 1/2$. Solution: $[1/2, \infty)$.
2. Case $x < 0$: $|x|=-x \implies 2^x + 2^{-x} \geq 2\sqrt{2}$. Let $y=2^x$, so $0<y<1$. Inequality becomes $y+1/y \geq 2\sqrt{2} \implies y^2 - 2\sqrt{2}y + 1 \geq 0$. Roots are $\sqrt{2}\pm 1$. So $y \leq \sqrt{2}-1$ or $y \geq \sqrt{2}+1$. With $0<y<1$, we get $0<y \leq \sqrt{2}-1$. Thus $0 < 2^x \leq \sqrt{2}-1 \implies x \leq \log_2(\sqrt{2}-1)$. Solution: $(-\infty, \log_2(\sqrt{2}-1)]$.
3. Union of solutions: $(-\infty, \log_2(\sqrt{2}-1)] \cup [1/2, \infty)$.