Inequality & Modulus: Level 3

Solution (B): This nested absolute value unwraps in two stages. Case 1: \(|x| – 5 = 3 \implies |x| = 8 \implies x = 8\) or \(x = -8\). Case 2: \(|x| – 5 = -3 \implies |x| = 2 \implies x = 2\) or \(x = -2\). The four possible values are {8, -8, 2, -2}. The sum is \(8 + (-8) + 2 + (-2) = 0\).

Solution (A): 1. Solve the first inequality: \(15 – 3n > 6 \implies 9 > 3n \implies 3 > n\), or \(n < 3\). 2. Solve the second inequality: \(\frac{5n}{4} > -5 \implies 5n > -20 \implies n > -4\). 3. We need integers \(n\) such that \(-4 < n < 3\). 4. The set of integers is \(\{-3, -2, -1, 0, 1, 2\}\). 5. The product of all these integers is 0, because 0 is in the set.

Solution (B): Test a value in the range, such as \(x = -0.5\). (A) \(x = -0.5\) (B) \(x^2 = (-0.5)^2 = 0.25\) (C) \(x^3 = (-0.5)^3 = -0.125\) (D) \(\frac{1}{x} = \frac{1}{-0.5} = -2\) (E) \(x-1 = -0.5 – 1 = -1.5\) The only positive value is \(0.25\), which is \(x^2\).

Solution (D): The “triangle inequality” states \(|x+y| \le |x| + |y|\). The equality holds when \(x\) and \(y\) have the same sign. Our problem is \(|a + (-b)| = |a| + |-b|\). (Since \(|-b| = |b|\)). This equality holds only when \(a\) and \(-b\) have the same sign. Case 1: \(a > 0\) and \(-b > 0 \implies b < 0\). (Opposite signs) Case 2: \(a < 0\) and \(-b < 0 \implies b > 0\). (Opposite signs) In both cases, \(a\) and \(b\) must have opposite signs, so their product \(ab\) must be negative.

Solution (E): 1. \(ab > 0\) means \(a\) and \(b\) have the same sign (both positive or both negative). 2. \(a+b < 0\) means their sum is negative. This is only possible if they are both negative. 3. Therefore, \(a < 0\) (I is true) and \(b < 0\) (II is true). 4. \(a/b \implies \text{Negative} / \text{Negative} = \text{Positive}\). So \(a/b > 0\) (III is true). All three statements must be true.

Solution (D): \(x\) can be \(\{2, -2\}\). \(y\) can be \(\{5, -5\}\). We need to find the smallest value of \((x-y)^2\), which means we need the smallest absolute value of \((x-y)\). 1. \(2 – 5 = -3\) 2. \(2 – (-5) = 7\) 3. \(-2 – 5 = -7\) 4. \(-2 – (-5) = 3\) The possible values for \((x-y)\) are \(\{-3, 7, -7, 3\}\). The smallest absolute value is 3. The smallest value of \((x-y)^2\) is \(3^2 = 9\).

Solution (D): We are given \(a\) is a positive fraction and \(b\) is greater than 1. (A) \(a+b\): (Positive) + (Positive > 1) must be > 1. (Must be true). (B) \(b/a\): (Number > 1) / (Fraction < 1). The result is > 1. (Must be true). (C) \(ab < 1\): e.g., \(a=0.1, b=2 \implies ab = 0.2 < 1\). (Can be true). (D) \(a - b > 0\): This means \(a > b\). But we are given \(a < 1\) and \(b > 1\), so \(a < b\). This statement CANNOT be true. (E) \(a/b < 1\): (Positive fraction) / (Number > 1). This is always < 1. (Must be true).

Solution (A): \(a\) can be \(\{3, -3\}\). \(b\) can be \(\{5, -5\}\). We need to find all possible values of \(a – b\): 1. \(3 – 5 = -2\) 2. \(3 – (-5) = 8\) 3. \(-3 – 5 = -8\) 4. \(-3 – (-5) = 2\) The set of possible values is \(\{-8, -2, 2, 8\}\). The largest value is 8. The smallest value is -8. The product is \(8 \times (-8) = -64\).

Solution (E): We evaluate each term. 1. Since \(x < 0\), \(|x| = -x\). So, \( \frac{|x|}{x} = \frac{-x}{x} = -1 \). 2. Since \(x\) is not 0, \(x^0 = 1\). 3. The sum is \( (-1) + 1 = 0 \).

Solution (C): To find the range of \(x-y\), we find the smallest and largest possible values. 1. Smallest \(x-y\): Use the smallest \(x\) (1) and subtract the largest \(y\) (3). Smallest = \(1 – 3 = -2\). 2. Largest \(x-y\): Use the largest \(x\) (2) and subtract the smallest \(y\) (2). Largest = \(2 – 2 = 0\). The range is \(-2 < x-y < 0\).