📝 Absolute value function properties (15 MCQs)

📖 From Calculus • 1. Basics before calculus • 15 questions available

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What is Absolute value function properties?

Definition:
The absolute value function $f(x) = |x|$ outputs the non-negative magnitude of $x$ , defined as $x$ if $x \geq 0$ and $-x$ if $x < 0$ , with properties including evenness, non-negativity, and the triangle inequality.

Example:
For $x = -3$ , $| -3 | = 3$ , and for $x = 4$ , $|4| = 4$ ; its graph is V-shaped with vertex at (0,0).

Reason:
These properties are essential in distance calculations, error bounds, and optimization problems, as the absolute value measures magnitude regardless of direction.

Easy

Medium

Hard

📝 All Absolute value function properties MCQs

Q1. According to the definition of absolute value, what is \ $|x|\$ when \ $x < 0\$ ?

A.#NAME? ✅

B.x

C.|x|

D.0

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: By definition, the absolute value of a negative number is its opposite, which removes the minus sign. When \ $x<0\$ we have \ $|x| = -x\$ . This matches option A, making it the correct choice.

Q2. If \ $a = -5\$ , what is the sign of \ $|a| + a\$ ?

A.Positive

B.Negative

C.Zero ✅

D.Undefined

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The expression \ $|a| + a\$ combines the non‑negative magnitude of \ $a\$ with the original value. For \ $a = -5\$ , \ $|a| = 5\$ and adding \ $-5\$ yields \ $0\$ . Zero is neither positive nor negative, so the correct answer is option C.

Q3. Which statement correctly compares the graph of \ $y = |x|\$ and \ $y = \\sqrt{x^2}\$ ?

A.They are identical for all real \

x\

. ✅

B.They differ for \

x<0\

C.They differ for \

x>0\

D.They intersect only at the origin.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Both functions \ $y = |x|\$ and \ $y = \\sqrt{x^{2}}\$ produce the same non‑negative output for any real \ $x\$ . The square root returns the principal (non‑negative) root, which is exactly the absolute value. Therefore the two graphs coincide for all real numbers, matching option A.

Q4. Given real numbers \ $p\$ and \ $q\$ , which identity always holds?

A.\

|p+q| = |p| + |q|\

B.\

|pq| = |p||q|\

✅

C.\

|p/q| = |p| - |q|\

D.\

|-p| = -|p|\

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The property \ $|ab| = |a||b|\$ holds for any real numbers because multiplying the numbers first and then taking the magnitude yields the same result as taking the magnitudes separately and then multiplying. Options B reflects this identity, while the others are false, so B is correct.

Q5. If \ $|x| \\le 3\$ , which interval describes all possible \ $x\$ ?

A.[-3,3] ✅

B.33

C.(-∞,3]

D.[0,3]

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The inequality \ $|x| \\le 3\$ means the distance of \ $x\$ from zero is at most three units. This describes all numbers between \ $-3\$ and \ $3\$ , inclusive, which is the interval \ $[-3,3]\$ . Option A captures this interval precisely, making it the correct choice.

Q6. Suppose \ $|2t-7| = 5\$ . Which of the following could be the value of \ $t\$ ?

A.1

B.3

C.6 ✅

D.8

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Solving \ $|2t-7| = 5\$ gives two linear equations: \ $2t-7 = 5\$ and \ $2t-7 = -5\$ . These yield \ $t = 6\$ and \ $t = 1\$ respectively. Both values satisfy the original equation, and among the options, choice C (\ $6\$ ) is a valid solution.

Q7. The function \ $f(x)=|x-2|+|x+3|\$ attains its minimum value at which \ $x\$ ?

A.-\\frac{1}{2} ✅

B.0

C.2

D.-3

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The function \ $f(x)=|x-2|+|x+3|\$ represents the sum of distances from \ $x\$ to the points \ $2\$ and \ $-3\$ on the number line. This sum is minimized at the median of the two points, which is \ $(-3+2)/2 = -\\frac{1}{2}\$ . Hence option A is correct.

Q8. Which statement correctly explains why \ $\\sqrt{x^2}=|x|\$ for all real \ $x\$ ?

A.Because squaring removes sign, and the principal square root returns non‑negative value. ✅

B.Because \

\\sqrt{x^2}=x\

always.

C.Because absolute value is defined as the square root of the square.

D.Because \

\\sqrt{x^2}= -x\

when \

x<0\

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The equality \ $\\sqrt{x^{2}} = |x|\$ follows because squaring any real number eliminates its sign, and the principal square root returns the non‑negative value. Therefore the expression always yields the absolute value of \ $x\$ . Option A correctly describes this reasoning.

Q9. If \ $|a| < |b|\$ and \ $b>0\$ , which statement must be true?

A.|a| < b

B.#NAME?

C.a^2 < b^2 ✅

D.a < b

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Since \ $b>0\$ , we have \ $|b| = b\$ . The inequality \ $|a| < |b|\$ then implies \ $|a| < b\$ . Squaring both sides preserves the inequality because both sides are non‑negative, giving \ $a^{2} < b^{2}\$ . Thus statement C must be true.

Q10. Given that \ $|x| = 4\$ and \ $|y| = 5\$ , what is the maximum possible value of \ $|xy|\$ ?

A.1

B.9

C.20 ✅

D.40

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The product property of absolute value tells us \ $|xy| = |x||y|\$ . With \ $|x| = 4\$ and \ $|y| = 5\$ , the maximum (and indeed the exact) value of \ $|xy|\$ is \ $4 \\times 5 = 20\$ . Option C reflects this product, making it correct.

Q11. The graph of \ $y = |x-1| + 2\$ is obtained from the graph of \ $y = |x|\$ by which sequence of transformations?

A.Shift right 1, up 2 ✅

B.Shift left 1, up 2

C.Shift right 1, down 2

D.Shift left 1, down 2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The graph of \ $y = |x-1| + 2\$ can be obtained by first shifting the basic V‑shaped graph of \ $y = |x|\$ one unit to the right (replacing \ $x\$ with \ $x-1\$ ) and then moving the entire graph upward by two units. This sequence matches option A.

Q12. How many real solutions does the equation \ $|x| = x^2 - 4\$ have?

A.0

B.1

C.2 ✅

D.3

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: To count real solutions of \ $|x| = x^{2} - 4\$ , we consider the cases \ $x \\ge 0\$ and \ $x < 0\$ separately, leading to the quadratic equations \ $x^{2} - x - 4 = 0\$ and \ $x^{2} + x - 4 = 0\$ . Each equation yields one admissible root, giving a total of two real solutions. Hence option C is correct.

Q13. Which of the following steps correctly justifies the triangle inequality \ $|a+b| \\le |a|+|b|\$ for real numbers?

A.Square both sides and use \

(a+b)^2 \\le a^2 + b^2\

B.Apply the definition \

|x| = \\sqrt{x^2}\

and use the Cauchy‑Schwarz inequality. ✅

C.Observe that \

|a+b| = |a|+|b|\

when a and b have same sign.

D.Use the fact that \

|a| \\ge a\

and \

|b| \\ge b\

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Writing \ $|a+b| = \\sqrt{(a+b)^{2}}\$ and applying the Cauchy‑Schwarz inequality \ $(a^{2}+b^{2})(1^{2}+1^{2}) \\ge (a+b)^{2}\$ leads directly to \ $|a+b| \\le |a|+|b|\$ . This reasoning correctly justifies the triangle inequality, corresponding to option B.

Q14. Find all real numbers \ $x\$ satisfying \ $|3x-4| > 2|x+1|\$ .

A.x < \\frac{2}{5}

B.\\frac{2}{5} < x < 6

C.x > 6

D.x < \\frac{2}{5} \\text{ or } x > 6 ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Squaring the inequality \ $|3x-4| > 2|x+1|\$ gives \ $(3x-4)^{2} > 4(x+1)^{2}\$ , which simplifies to \ $5x^{2} - 32x + 12 > 0\$ . The quadratic factors to \ $(x-6)(x-\\frac{2}{5}) > 0\$ , so the solution set is \ $x < \\frac{2}{5}\$ or \ $x > 6\$ . Option D expresses this union correctly.

Q15. Given real numbers \ $a,b\$ with \ $|a|=3\$ and \ $|b|=4\$ , which of the following statements is always true?

A.|a+b| = 7

B.|a-b| \\ge 5

C.|a|+|b| = |a+b|

D.|ab| = 12 ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: With \ $|a| = 3\$ and \ $|b| = 4\$ , the product property gives \ $|ab| = |a||b| = 12\$ for any choices of signs of \ $a\$ and \ $b\$ . None of the other statements hold for all sign combinations. Therefore option D, stating \ $|ab| = 12\$ , is always true.

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