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

The inequality $2^x + 2^{|x|} \geq 2\sqrt{2}$ is solved by considering two cases: $x \geq 0$ and $x < 0$. For $x \geq 0$, $|x| = x$, so the inequality becomes $2^x + 2^x \geq 2\sqrt{2}$, which simplifies to $2 \cdot 2^x \geq 2\sqrt{2}$, or $2^{x+1} \geq 2^{3/2}$. Since the base is greater than 1, we have $x+1 \geq \frac{3}{2}$, leading to $x \geq \frac{1}{2}$. The solution for this case is $[\frac{1}{2}, \infty)$.

For $x < 0$, $|x| = -x$, so the inequality is $2^x + 2^{-x} \geq 2\sqrt{2}$. Let $y = 2^x$. Since $x < 0$, we have $0 < y < 1$. The inequality becomes $y + \frac{1}{y} \geq 2\sqrt{2}$. Multiplying by $y$ (which is positive), we get $y^2 + 1 \geq 2\sqrt{2}y$, or $y^2 - 2\sqrt{2}y + 1 \geq 0$. The roots of $y^2 - 2\sqrt{2}y + 1 = 0$ are $y = \sqrt{2} \pm 1$. Since the quadratic opens upwards, the inequality holds for $y \leq \sqrt{2} - 1$ or $y \geq \sqrt{2} + 1$. Combining with the condition $0 < y < 1$, we must have $0 < y \leq \sqrt{2} - 1$. Substituting back $y = 2^x$, we get $2^x \leq \sqrt{2} - 1$. Taking $\log_2$ on both sides, we have $x \leq \log_2(\sqrt{2} - 1)$. The solution for this case is $(-\infty, \log_2(\sqrt{2} - 1)]$.

The overall solution is the union of the solutions from both cases: $(-\infty, \log_2(\sqrt{2} - 1)] \cup [\frac{1}{2}, \infty)$.