Questions & Answers: "Function - Modulus function"

Correct Option: Using the property |X| ≤ a ⇔ -a ≤ X ≤ a, we get -4 ≤ 3x - 2 ≤ 4. Adding 2 to all parts gives -2 ≤ 3x ≤ 6. Dividing by 3 gives -2/3 ≤ x ≤ 2. In interval notation, this is [-2/3, 2]. Incorrect Options: Option 2 uses open intervals which is incorrect for ≤; Option 3 has an incorrect lower bound; Option 4 represents the solution for |3x - 2| ≥ 4.

Correct Option: The modulus function |x - 4| is defined for all real numbers, so the Domain is R. Since the minimum value of any absolute value expression |a| is 0, the minimum value of |x - 4| is 0 (at x=4). Adding 3 to this minimum value gives 0 + 3 = 3. Therefore, the function values range from 3 to infinity, making the Range [3, ∞). Incorrect Options: Option 2 incorrectly restricts the domain; Option 3 is wrong because the modulus function cannot output negative values in this context; Option 4 fails to account for the vertical shift of +3.

Correct Option: A function is one-one if f(x1) = f(x2) implies x1 = x2. Here, f(1) = |1| = 1 and f(-1) = |-1| = 1. Since 1 ≠ -1 but they have the same image, it is not one-one. A function is onto if the Range equals the Codomain. The Range of |x| is [0, ∞), but the Codomain is given as R. Since negative real numbers in the codomain have no pre-image, it is not onto. Incorrect Options: All other options claim the function is either one-one or onto (or both), which contradicts the basic definition and range of the modulus function.

The modulus function |x - 3| is always greater than or equal to 0 for all real x. Therefore, the minimum value of |x - 3| is 0 (at x = 3). Adding 2 to this result gives f(x) ≥ 0 + 2, which means f(x) ≥ 2. Thus, the range is [2, ∞). Option B is wrong because the value 2 is included. Option C is the range of |x-3| only. Option D is wrong because modulus functions cannot output negative values in this context.

Since x² is same as |x|², we can rewrite the equation as |x|² - 5|x| + 6 = 0. Let y = |x|. Then y² - 5y + 6 = 0. Factoring gives (y - 2)(y - 3) = 0, so y = 2 or y = 3. Since y = |x|, we have |x| = 2 (which gives x = ±2) and |x| = 3 (which gives x = ±3). Total 4 real solutions: {2, -2, 3, -3}. Option B is wrong as it ignores negative roots. Option C and D are incorrect.

According to the definition of modulus, |a| = b implies a = b or a = -b. Applying this to |2x - 5| = 9: Case 1: 2x - 5 = 9 gives 2x = 14, so x = 7. Case 2: 2x - 5 = -9 gives 2x = -4, so x = -2. Both values satisfy the equation. Option B is incomplete. Option C has incorrect signs during transposition. Option D is incorrect calculation.

The inequality |x - a| < r translates to -r < x - a < r. Here, |x + 4| < 3 implies -3 < x + 4 < 3. Subtracting 4 from all sides: -3 - 4 < x < 3 - 4 results in -7 < x < -1. Since the inequality is strict ( 3. Option D has calculation errors.

By definition, the square root of a squared number must be non-negative. If x is positive, √x² = x. If x is negative, say -3, √(-3)² = √9 = 3 (which is -x). This behavior precisely matches the definition of the modulus function |x|. Option B is wrong because it fails for negative x. Option C is a common misconception; a function must return a unique value, not two values. Option D is wrong for positive x.

The inequality |a| ≥ b implies a ≥ b or a ≤ -b. For |3x - 2| ≥ 7: Case 1: 3x - 2 ≥ 7 => 3x ≥ 9 => x ≥ 3. Case 2: 3x - 2 ≤ -7 => 3x ≤ -5 => x ≤ -5/3. Combining these gives x ∈ (-∞, -5/3] ∪ [3, ∞). Option B represents |3x - 2| ≤ 7. Option C has the numbers swapped. Option D is incomplete.

In the interval 1 < x 0, so |x - 1| = x - 1. (2) x - 2 < 0, so |x - 2| = -(x - 2) = 2 - x. Therefore, f(x) = (x - 1) + (2 - x) = 1. This shows that the function is constant between its roots. Option B is for x ≥ 2. Option C is for x ≤ 1. Option D is mathematically incorrect calculation.

Given 1/|x - 3| < 1/2. Since the modulus is always positive for x ≠ 3, we can cross-multiply: 2 2. This implies x - 3 > 2 or x - 3 5 or x < 1. In interval notation, this is (-∞, 1) ∪ (5, ∞). Option B is for |x - 3| < 2. Options C and D do not cover both regions correctly.

For the function to be defined, the denominator |x| - x must not be zero. |x| - x = 0 occurs when |x| = x, which is true for all x ≥ 0. Therefore, x cannot be zero or any positive number. However, for x < 0, |x| = -x, so the denominator becomes -x - x = -2x, which is positive and non-zero. Thus, the domain is all negative real numbers, (-∞, 0). Option B and C include values where the denominator becomes zero.

The outer modulus implies |x| - 2 = 3 or |x| - 2 = -3. Case 1: |x| - 2 = 3 => |x| = 5, which gives x = 5 or x = -5. Case 2: |x| - 2 = -3 => |x| = -1. Since a modulus cannot be negative, Case 2 provides no real solutions. Thus, only x = ±5 are correct. Option B is wrong because it incorrectly treats |x| = -1 as having solutions ±1. Option C is only the incorrect part of the logic.