\( A_{2}=\{n \in \mathbb{N}: 3+2|4 n-3|>13\} \) each set, give TWO elements belonging
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To solve the inequality \( 3 + 2|4n - 3| > 13 \), we start by simplifying it. Subtracting 3 from both sides gives: \[ 2|4n - 3| > 10 \] Dividing by 2: \[ |4n - 3| > 5 \] This absolute inequality can be split into two cases: 1. \( 4n - 3 > 5 \) 2. \( 4n - 3 < -5 \) **Solving the first case:** \[ 4n - 3 > 5 \implies 4n > 8 \implies n > 2 \] Thus, \( n \) could be 3, 4, 5, ... **Solving the second case:** \[ 4n - 3 < -5 \implies 4n < -2 \implies n < -0.5 \] Since \( n \) must be a natural number (\( \mathbb{N} \)), this case does not yield any valid solutions. Hence, the only valid solutions are \( n \in \mathbb{N} \) such that \( n > 2 \). Therefore, two elements in the set are \( 3 \) and \( 4 \).
