AP EAMCET · Maths · Basic of Mathematics
The set \(\left\{x \in \mathbb{R}: 16\left(2^X\right)>16^{\frac{-1}{x}}\right\}=\)
- A \(\{\mathrm{x} \in \mathbb{R}: \mathrm{x}>0\}\)
- B \(\{\mathrm{x} \in \mathbb{R}: \mathrm{x} < 0\}\)
- C \(\mathbb{R}\)
- D \(\{x \in \mathbb{R}: x>2\}\)
Answer & Solution
Correct Answer
(A) \(\{\mathrm{x} \in \mathbb{R}: \mathrm{x}>0\}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text {} 16\left(2^x\right)>16^{-\frac{1}{x}} \Rightarrow 2^{x+4}>2^{-4 / x} \\ & \Rightarrow x+4>-4 / x \Rightarrow x^2+4 x+4>0 \\ & \Rightarrow(x+2)^2>0 \end{aligned}\) which is true for all \(x \in R\)
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