First slide

Theory of equations

Question

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

Moderate

A

$(- \infty, \log_{2} (\sqrt{2} + 1)$
B

$[\log_{2} (\sqrt{2} + 1), \infty)$
C

$(\frac{1}{2}, \log_{2} (\sqrt{2} - 1))$
D

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

Solution

We have,
$\begin{array}{ll} 2^{x} + 2^{x} \geq 2 \sqrt{2} & (x \geq 0) \\ \Rightarrow 2^{x} \geq \sqrt{2} \Rightarrow x \geq \frac{1}{2} \\ and 2^{x} + 2^{- x} \geq 2 \sqrt{2} & (x < 0) \end{array}$
$\begin{array}{lc} \Rightarrow & t + \frac{1}{t} \geq 2 \sqrt{2} \\ \Rightarrow & t^{2} - 2 \sqrt{2} t + 1 \geq 0 \\ \Rightarrow & [t - (\sqrt{2} - 1)] [t - (\sqrt{2} + 1)] \geq 0 \\ \Rightarrow & t \leq \sqrt{2} - 1 or t \geq \sqrt{2} + 1 \end{array}$
but t>0
$\begin{array}{lc} \Rightarrow 0 < 2^{x} \leq \sqrt{2} - 1 or 2^{x} \geq \sqrt{2} + 1 \\ \Rightarrow - \infty < x \leq \log_{2} (\sqrt{2} - 1) \\ or x \geq \log_{2} (\sqrt{2} + 1) \end{array}$
(but not acceptable as x < 0)
$∴ x \in (- \infty, \log_{2} (\sqrt{2} - 1)] \cup [\frac{1}{2}, \infty)$

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