Question

Solution of \( 2^x + 2^{|x|} \geq 2\sqrt{2} \) is:

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

Hint:

When solving inequalities involving absolute values, consider the different cases for \( x \geq 0 \) and \( x<0 \), and apply logarithms where appropriate to simplify the expressions.

Answer 1

The Correct Option is D

We are given the inequality: \[ 2^x + 2^{|x|} \geq 2\sqrt{2}. \] Case 1: \( x \geq 0 \) In this case, \( |x| = x \), so the inequality becomes: \[ 2^x + 2^x \geq 2\sqrt{2}, \] which simplifies to: \[ 2 \cdot 2^x \geq 2\sqrt{2}. \] Dividing both sides by 2: \[ 2^x \geq \sqrt{2}. \] Taking the logarithm (base 2) of both sides: \[ x \geq \log_2(\sqrt{2}). \] Since \( \log_2(\sqrt{2}) = \frac{1}{2} \), we get: \[ x \geq \frac{1}{2}. \] Thus, the solution for \( x \geq 0 \) is \( [\frac{1}{2}, \infty) \). Case 2: \( x<0 \) In this case, \( |x| = -x \), so the inequality becomes: \[ 2^x + 2^{-x} \geq 2\sqrt{2}. \] Multiply both sides by \( 2^x \): \[ 1 + 2^{2x} \geq 2^{x+1}. \] Rearranging terms: \[ 2^{2x} - 2^{x+1} + 1 \geq 0. \] Let \( y = 2^x \), so the inequality becomes: \[ y^2 - 2y + 1 \geq 0. \] This factors as: \[ (y - 1)^2 \geq 0. \] Since the square of any real number is non-negative, this inequality is always true. 
Thus, there are no further restrictions on \( x \) for \( x<0 \). Thus, the solution for \( x<0 \) is \( (-\infty, \log_2(\sqrt{2} + 1)) \). 
Conclusion: The solution to the inequality is: \[ (-\infty, \log_2(\sqrt{2} + 1)) \cup \left[\frac{1}{2}, \infty\right). \]