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 $x < 0$.

Case 1: $x \geq 0$ If $x \geq 0$, then $|x| = x$. The inequality becomes: $2^x + 2^x \geq 2\sqrt{2}$ $2 \cdot 2^x \geq 2\sqrt{2}$ $2^{x+1} \geq 2^{3/2}$ Comparing exponents: $x+1 \geq \frac{3}{2} \implies x \geq \frac{1}{2}$. The solution for this case is $x \in [\frac{1}{2}, \infty)$.

Case 2: $x < 0$ If $x < 0$, then $|x| = -x$. The inequality becomes: $2^x + 2^{-x} \geq 2\sqrt{2}$ Let $y = 2^x$. Since $x < 0$, $0 < y < 1$. The inequality is $y + \frac{1}{y} \geq 2\sqrt{2}$. Multiplying by $y$: $y^2 + 1 \geq 2\sqrt{2}y \implies y^2 - 2\sqrt{2}y + 1 \geq 0$. The roots of $y^2 - 2\sqrt{2}y + 1 = 0$ are $y = \sqrt{2} \pm 1$. So, $y \leq \sqrt{2} - 1$ or $y \geq \sqrt{2} + 1$.

Substituting back $y = 2^x$: $2^x \leq \sqrt{2} - 1$ or $2^x \geq \sqrt{2} + 1$.

For $2^x \geq \sqrt{2} + 1$, we get $x \geq \log_2(\sqrt{2} + 1)$. This contradicts $x < 0$. For $2^x \leq \sqrt{2} - 1$, we get $x \leq \log_2(\sqrt{2} - 1)$. This is consistent with $x < 0$ as $\log_2(\sqrt{2} - 1) < 0$. The solution for this case is $x \in (-\infty, \log_2(\sqrt{2} - 1)]$.

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