This question tests properties of absolute values that hold for all real numbers. The absolute value |x| represents the distance from x to zero, which is always non-negative. For any real number x, we have |x| ≥ 0. When x ≥ 0, |x| = x, so |x| ≥ x with equality. When x < 0, |x| = -x > 0 > x, so |x| > x. Therefore, |x| ≥ x is always true. The incorrect option |x| ≤ x would require |x| = x for all x, which fails when x is negative since then |x| = -x > x.

This question tests the geometric interpretation of absolute value equations. The equation |x-2| = |y-2| means that x and y are equidistant from 2 on the number line. Since we're told x < 2 < y, x must be to the left of 2 and y to the right of 2. For these points to be equidistant from 2, we need 2-x = y-2, which gives us x + y = 4. This makes geometric sense: if x is d units to the left of 2 and y is d units to the right of 2, then x = 2-d and y = 2+d, so x + y = 4. A common error would be to think x and y are equal, but they're actually reflections of each other across the point 2.

This question tests the relationship between absolute values and ordering. The absolute value |a| represents the distance from a to zero, regardless of sign. Given that |a| < |b| and b < 0, we know b is negative and its distance from zero is greater than a's distance from zero. Since |b| = -b (because b is negative), we have |a| < -b. This means a could be any value between b and -b, which includes both positive and negative values. For example, if b = -10, then |b| = 10, and a could be any value with |a| < 10, such as a = 5 or a = -7. The key insight is that knowing |a| < |b| doesn't determine the sign of a.

This question tests absolute value relationships and ordering. The absolute value |a| represents the distance from a to zero, and similarly for |b|. Given a < 0 < b, we know a is negative and b is positive. Since |a| > |b|, the distance from a to zero exceeds the distance from b to zero. This means the negative value a has greater magnitude than the positive value b, so a + b < 0. For example, if a = -5 and b = 3, then |a| = 5 > |b| = 3, and a + b = -2 < 0.

This question tests understanding of absolute value inequalities. The inequality |t-4| ≥ 3 means the distance from t to 4 is at least 3 units. This occurs when t ≤ 1 or t ≥ 7, representing points that are 3 or more units away from 4 on the number line. Among the given choices, we need to check which values satisfy this condition: |2-4| = 2 < 3 (no), |6-4| = 2 < 3 (no), |5-4| = 1 < 3 (no), |1-4| = 3 ≥ 3 (yes), |4-4| = 0 < 3 (no). Only t = 1 satisfies the inequality. A common mistake is to think values close to 4 satisfy the inequality, when actually we need values far from 4.

This question tests comparing distances using absolute value inequalities. The inequality |x-5| < |x-1| means x is closer to 5 than to 1 on the number line. To find when this occurs, we can square both sides (since both are non-negative): (x-5)² < (x-1)². Expanding gives x²-10x+25 < x²-2x+1, which simplifies to -10x+25 < -2x+1, then -8x < -24, so x > 3. Geometrically, this makes sense: the point x = 3 is equidistant from 1 and 5, and any point to the right of 3 is closer to 5. A common error is to try case analysis without recognizing the geometric interpretation of being closer to one point than another.

This question tests finding maximum values given a range constraint. With -2 < x < 1, we need to evaluate each expression across this interval. For x itself, the range is (-2, 1) with maximum approaching 1. For |x|, the maximum is 2 when x approaches -2. For -x, the range is (-1, 2) with maximum approaching 2. For |x+1|, since x+1 ranges from (-1, 2), the maximum is 2. For |x-1|, since x-1 ranges from (-3, 0), we have |x-1| ranging from 0 to 3, with maximum 3 when x approaches -2. Therefore, |x-1| achieves the greatest value of 3. A common error is to evaluate only at the endpoints without considering how absolute value affects negative inputs.

This question tests the relationship between absolute value and ordering. The condition |p| = |q| means p and q have the same distance from zero. Combined with p < q, there are only two possibilities: either both are positive (impossible since then p < q would contradict |p| = |q|), or p is negative and q is positive with p = -q. Since p < q and |p| = |q|, we must have p < 0 < q with p = -q. This means p and q are opposites, confirming that p = -q. A common mistake is to think p and q could both be negative, but if they were both negative with equal absolute values, they would be equal, contradicting p < q.

This question tests the relationship between absolute value and ordering. The absolute value |a| represents the distance from a to zero, and similarly for |b|. Given |a| = |b|, both numbers are equidistant from zero. Combined with a < b, this means a must be negative and b must be positive, with a = -b. For example, if a = -3 and b = 3, then |a| = |b| = 3 and a < b. This confirms that a = -b is the true statement.

This question tests relationships between absolute values and signs of numbers. The absolute value |x| represents the distance from x to zero, regardless of sign. Given |x| < |y| and x < 0 < y, we know x is negative and y is positive. Since |x| < |y|, the distance from x to zero is less than the distance from y to zero. This means -x < y (since |x| = -x when x < 0). Adding x to both sides gives 0 < x+y, so x+y > 0. The incorrect option x+y = 0 would require |x| = |y|, contradicting our given condition.