📐 CHAPTER 5: LINEAR INEQUALITIES
Subject: Mathematics | Class: XI | Kendriya Vidyalaya Sangathan
📖 INTRODUCTION
A linear inequality is an inequality that involves a linear function. It contains one of the symbols: <, >, ≤, ≥
🔹 Examples:
- 3x + 5 > 7
- 2x - 3 ≤ 9
- 4y + 1 ≥ 13
- x - 2y < 6
🔹 Rules for Solving Inequalities
- Rule 1: Adding or subtracting the same number from both sides does not change the inequality.
a < b ⇒ a + c < b + c - Rule 2: Multiplying or dividing both sides by a positive number does not change the inequality.
a < b ⇒ ac < bc (c > 0) - Rule 3: Multiplying or dividing both sides by a negative number reverses the inequality.
a < b ⇒ ac > bc (c < 0)
🔢 TYPES OF INEQUALITIES
🔹 Linear Inequalities in One Variable
Inequalities of the form ax + b < 0, ax + b > 0, ax + b ≤ 0, ax + b ≥ 0
🔹 Linear Inequalities in Two Variables
Inequalities of the form ax + by < c, ax + by > c, ax + by ≤ c, ax + by ≥ c
🔹 Solution Set
The set of all values that satisfy the inequality is called the solution set.
🔹 Graphical Representation
Inequalities in two variables can be represented graphically on a Cartesian plane.
📐 ABSOLUTE VALUE INEQUALITIES
🔹 Key Properties:
- |x| < a ⇒ -a < x < a (a > 0)
- |x| ≤ a ⇒ -a ≤ x ≤ a (a > 0)
- |x| > a ⇒ x < -a or x > a (a > 0)
- |x| ≥ a ⇒ x ≤ -a or x ≥ a (a > 0)
🔹 Examples:
- |x - 2| < 3 ⇒ -3 < x - 2 < 3 ⇒ -1 < x < 5
- |2x + 1| ≥ 5 ⇒ 2x + 1 ≤ -5 or 2x + 1 ≥ 5 ⇒ x ≤ -3 or x ≥ 2
📝 MCQ QUESTIONS
Q1. Solve 7x < 24 when x ∈ N.
(a) {1, 2, 3, 4} (b) {0, 2, 3, 4} (c) {1, 2, 3} (d) {0, 2, 3, 4}
Answer: (c) {1, 2, 3}
Solution: 7x < 24 ⇒ x < 24/7 ≈ 3.43
x ∈ N = {1, 2, 3}
Q2. Solve 3 - 2x < 9 when x ∈ R. Express the solution in interval form.
(a) [-3, ∞) (b) (-3, ∞) (c) [-3, ∞] (d) [-2, ∞)
Answer: (b) (-3, ∞)
Solution: 3 - 2x < 9 ⇒ -2x < 6 ⇒ x > -3 ⇒ x ∈ (-3, ∞)
Q3. If (x - 3)/(x - 2) > 0, then x belongs to:
(a) (-∞, 2) ∪ (3, ∞) (b) (-∞, -3) ∪ (-5, ∞) (c) (-∞, 2] ∪ [5, ∞) (d) (2, 3)
Answer: (a) (-∞, 2) ∪ (3, ∞)
Solution: The expression is positive when both numerator and denominator are positive or both negative.
x - 3 > 0 and x - 2 > 0 ⇒ x > 3
x - 3 < 0 and x - 2 < 0 ⇒ x < 2
∴ x ∈ (-∞, 2) ∪ (3, ∞)
Q4. If x > 0 and y < 0, then (x, y) lies in:
(a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant
Answer: (d) IV quadrant
Q5. If -2 < 2x - 1 < 2, then the value of x lies in the interval:
(a) (1/2, 3/2) (b) (-1/2, 3/2) (c) (3/2, 1/2) (d) (3/2, 1/2)
Answer: (b) (-1/2, 3/2)
Solution: -2 < 2x - 1 < 2 ⇒ -1 < 2x < 3 ⇒ -1/2 < x < 3/2
Q6. The solution of |2/(x - 4)| > 1, where x ≠ 4 is:
(a) (2, 6) (b) (2, 4) ∪ (4, 6) (c) (2, 4) ∪ (4, ∞) (d) (-∞, 4) ∪ (4, 6)
Answer: (b) (2, 4) ∪ (4, 6)
Solution: |2/(x-4)| > 1 ⇒ 2/|x-4| > 1 ⇒ |x-4| < 2
⇒ -2 < x - 4 < 2 ⇒ 2 < x < 6, x ≠ 4
∴ x ∈ (2, 4) ∪ (4, 6)
Q7. The solution of 0 < 3(x - 2)/5 < 15 is:
(a) 2 < x < 27 (b) 27 < x < -2 (c) -27 < x < 2 (d) -27 < x < -2
Answer: (a) 2 < x < 27
Solution: 0 < 3(x-2)/5 < 15 ⇒ 0 < x - 2 < 25 ⇒ 2 < x < 27
Q8. Solve: f(x) = {(x - 1)x(2 - x)}/(x - 3) ≥ 0
(a) (-∞, 1] ∪ (2, ∞) (b) (-∞, 1] ∪ [2, 3) (c) (-∞, 1] ∪ (3, ∞) (d) None of these
Answer: (b) (-∞, 1] ∪ [2, 3)
Q9. The solution of the inequality 3(2 - x) ≥ 2(1 - x) for real x is:
(a) x < 4 (b) x > 4 (c) x ≤ 4 (d) x ≥ 4
Answer: (c) x ≤ 4
Solution: 6 - 3x ≥ 2 - 2x ⇒ 6 - 2 ≥ 3x - 2x ⇒ 4 ≥ x ⇒ x ≤ 4
Q10. The solution to |3x - 1| + 1 < 3 is:
(a) 2 < x < 3/4 (b) -1/3 < x < 1 (c) -1/3 < x < 1/4 (d) -3 < x < 3
Answer: (b) -1/3 < x < 1
Solution: |3x - 1| + 1 < 3 ⇒ |3x - 1| < 2
⇒ -2 < 3x - 1 < 2 ⇒ -1 < 3x < 3 ⇒ -1/3 < x < 1
Q11. Solve: 3x + 5 < x - 7, when x is a real number.
(a) x < -12 (b) x > -6 (c) x < -6 (d) None of these
Answer: (c) x < -6
Solution: 3x + 5 < x - 7 ⇒ 2x < -12 ⇒ x < -6
Q12. If |x - 3| < 7 and x is a real number, then x belongs to:
(a) (4, 10) (b) (-10, 4) (c) (4, -10) (d) None of these
Answer: (d) None of these
Solution: |x - 3| < 7 ⇒ -7 < x - 3 < 7 ⇒ -4 < x < 10 ⇒ x ∈ (-4, 10)
Q13. The solution set of -x² > 16, x is a real number, is:
(a) (-4, 4) (b) (0, 4) (c) (-4, 0) (d) Empty set
Answer: (d) Empty set
Solution: -x² > 16 ⇒ x² < -16 (impossible as x² ≥ 0)
Q14. Set of points in the second quadrant is represented by:
(a) x > 0 and y < 0 (b) x < 0 and y < 0 (c) x > 0 and y > 0 (d) x < 0 and y > 0
Answer: (d) x < 0 and y > 0
Q15. The graph of x < y:
(a) Contains origin (b) Contains (-2, 3) (c) Contains (2, 2) (d) None of these
Answer: (b) Contains (-2, 3)
Solution: For (-2, 3): -2 < 3 ✓
Q16. If -1 < x < 2 then:
(a) x can be 2 (b) x can be -1 (c) x can be 0 (d) None of these
Answer: (c) x can be 0
Q17. x² - 1 ≥ 8 then x belongs to:
(a) R - (-3, 3) (b) (-3, 3) (c) (0, 3) (d) (-3, 0)
Answer: (a) R - (-3, 3)
Solution: x² - 1 ≥ 8 ⇒ x² ≥ 9 ⇒ |x| ≥ 3 ⇒ x ≤ -3 or x ≥ 3
Q18. The open upper half plane of x-axis can be expressed as:
(a) x > 0 and y < 0 (b) x > 0 (c) y > 0 (d) x < 0
Answer: (c) y > 0
Q19. The word "linear" stands for:
(a) One term (b) One power (c) One degree (d) None of these
Answer: (c) One degree
Q20. |x - 3| < 0. Then solution set of x is:
(a) (0, 3) (b) (-3, 0) (c) (-3, 3) (d) None of these
Answer: (d) None of these
Solution: |x - 3| < 0 has no solution as |x - 3| ≥ 0
📚 CASE BASED QUESTIONS
Q21. A beaker contains 640 litres of 8% solution of boric acid. This is to be diluted by adding 2% solution of boric acid to it.
(i) If x litres of 2% boric acid solution is added to the beaker, find the quantity of acid in the resulting solution.
Answer: Acid in 640 litres = 8% of 640 = 51.2 litres
Acid in x litres of 2% = 2x/100 = x/50 litres
Total acid = (51.2 + x/50) litres
(ii) Find the initial quantity of water in the beaker.
Answer: Water = 640 - 51.2 = 588.8 litres
(iii) Find the initial quantity of Boric acid in the beaker.
Answer: Boric acid = 8% of 640 = 51.2 litres
Q22. Kelvin(K), degree Celsius(°C) and degree Fahrenheit(°F) are three units of temperature. Conversion formula: F = 9/5C + 32 and K = C + 273.15
(i) To maintain Celsius temperature at least 5°C, what minimum Fahrenheit temperature should be maintained?
Answer: C ≥ 5 ⇒ F = 9/5(5) + 32 = 9 + 32 = 41°F
(ii) To maintain Kelvin temperature maximum 100 K, what maximum Celsius temperature should be maintained?
Answer: K ≤ 100 ⇒ C + 273.15 ≤ 100 ⇒ C ≤ -173.15°C
(iii) Find the Celsius temperature for which Kelvin and Fahrenheit temperatures are equal.
Answer: K = F ⇒ C + 273.15 = 9/5C + 32
⇒ C - 9/5C = 32 - 273.15 ⇒ -4/5C = -241.15 ⇒ C = 301.4°C
Q23. In three examinations (each of 100 marks), Amit and Arti scored:
| Name | Test-1 | Test-2 | Test-3 |
|---|---|---|---|
| Amit | 60 | 54 | x |
| Arti | 50 | 84 | y |
(i) To attain an average of at least 65 marks, how many marks must Amit score in the third test?
Answer: (60 + 54 + x)/3 ≥ 65 ⇒ 114 + x ≥ 195 ⇒ x ≥ 81
(ii) The average of Amit is greater than the average of Arti, establish relationship between x and y.
Answer: (60+54+x)/3 > (50+84+y)/3 ⇒ 114 + x > 134 + y ⇒ x - y > 20
(iii) To get A grade, one must attain an average of at least 80 marks. Find the minimum marks Arti should score to get A grade.
Answer: (50+84+y)/3 ≥ 80 ⇒ 134 + y ≥ 240 ⇒ y ≥ 106
Since maximum marks is 100, she cannot get A grade.
Q24. IQ of a person is calculated by the formula IQ = (M/C) × 100 where M is mental age and C is chronological age.
(i) If IQ of a person of mental age 30 years is 60, find the chronological age.
Answer: 60 = (30/C) × 100 ⇒ 60C = 3000 ⇒ C = 50 years
(ii) A person's chronological age is 50 years. To have IQ at least 70, what minimum mental age should he have?
Answer: (M/50) × 100 ≥ 70 ⇒ 2M ≥ 70 ⇒ M ≥ 35 years
(iii) Mental age of a person is 30 years. To have IQ between 50 and 60, what should his chronological age be?
Answer: 50 < (30/C) × 100 < 60 ⇒ 50 < 3000/C < 60
⇒ 3000/60 < C < 3000/50 ⇒ 50 < C < 60
Q25. Consider the following system of linear inequalities: x - 2y ≤ 0, 2x + y ≤ 4, y ≤ 2, x ≥ 0.
(i) The graph of x - 2y ≤ 0 contains the origin?
Answer: Yes, (0,0) satisfies 0 ≤ 0 ✓
(ii) In which quadrants the solution region cannot lie?
Answer: Second and third quadrants (since x ≥ 0)
(iii) Show the region 2x + y ≤ 4.
Answer: The region below the line 2x + y = 4
📝 VERY SHORT ANSWER TYPE QUESTIONS
Q26. Minimum value of x - 3 is 7, find minimum value of x.
Answer: x - 3 ≥ 7 ⇒ x ≥ 10 ⇒ Minimum value = 10
Q27. Maximum value of 2m - 3 is 7, find the maximum value of m.
Answer: 2m - 3 ≤ 7 ⇒ 2m ≤ 10 ⇒ m ≤ 5 ⇒ Maximum value = 5
Q28. If -3 < m < 7, find the minimum value of |2m + 10|.
Answer: m ∈ (-3, 7) ⇒ 2m + 10 ∈ (4, 24)
Minimum value of |2m + 10| = 4
Q29. If |x + 2| ≤ 9, find interval of x.
Answer: -9 ≤ x + 2 ≤ 9 ⇒ -11 ≤ x ≤ 7 ⇒ x ∈ [-11, 7]
Q30. If x ≥ -3, find the minimum value of x + 5.
Answer: x + 5 ≥ 2 ⇒ Minimum value = 2
Q31. If x² ≤ 9, find the interval of x.
Answer: |x| ≤ 3 ⇒ x ∈ [-3, 3]
Q32. If |x| - 3 < 8, find the interval of x.
Answer: |x| < 11 ⇒ x ∈ (-11, 11)
Q33. If -11 < 4x - 3 ≤ 13, find interval of x.
Answer: -11 + 3 < 4x ≤ 13 + 3 ⇒ -8 < 4x ≤ 16 ⇒ -2 < x ≤ 4
x ∈ (-2, 4]
Q34. Represent solution set of x - 3 < 6 on number line.
Answer: x < 9 ⇒ Number line: Open circle at 9, shaded left
Q35. Represent solution set of x - 3 < 3 on number line.
Answer: x < 6 ⇒ Number line: Open circle at 6, shaded left
Q36. Represent solution set of x² ≤ 4 on number line.
Answer: |x| ≤ 2 ⇒ [-2, 2] ⇒ Closed circles at -2 and 2, shaded between
Q37. How many points are there in the solution set of x² < 0, x ∈ R?
Answer: No points (empty set), as x² ≥ 0
Q38. How many integral solutions are there for the inequation -5 < 3x < 6?
Answer: -5 < 3x < 6 ⇒ -5/3 < x < 2 ⇒ x = -1, 0, 1
3 integral solutions
Q39. Write any one solution of x - 2y < -6.
Answer: (1, 4): 1 - 8 = -7 < -6 ✓
Q40. Write one solution of |x - 3| < y.
Answer: (2, 2): |2-3| = 1 < 2 ✓
Q41. Represent first quadrant in terms of inequalities of x and y.
Answer: x > 0, y > 0
Q42. Represent the first and fourth quadrants together in terms of inequalities of x and y.
Answer: y < 0 (or x > 0 for first quadrant)
Q43. Represent third quadrant in terms of inequalities of x and y.
Answer: x < 0, y < 0
Q44. Write the interval in which sin²x lies for any real x.
Answer: [0, 1]
Q45. Write the interval in which sin x + 3 lies for any real x.
Answer: [2, 4]
🧠 ASSERTION & REASON QUESTIONS
Q46. Assertion: |3x - 5| > 9 implies x ∈ (-∞, -4/3) ∪ (14/3, ∞)
Reason: The region containing all the solutions of inequality is called the solution region.
Answer: (b) Assertion and reason both are correct statements and reason is not the correct explanation of the assertion.
Q47. Assertion: The solution set of x² - 4x + 3 < 0 is (1, 3).
Reason: The quadratic expression x² - 4x + 3 = (x - 1)(x - 3)
Answer: (a) Assertion and reason both are correct statements and reason is the correct explanation of the assertion.
Q48. Assertion: The inequality |x| < a has solution -a < x < a for a > 0.
Reason: Absolute value of x is defined as |x| = x if x ≥ 0 and |x| = -x if x < 0.
Answer: (a) Assertion and reason both are correct statements and reason is the correct explanation of the assertion.
Q49. Assertion: If a < b, c < 0, then a/c < b/c.
Reason: If both sides are divided by the same negative quantity, then the inequality is reversed.
Answer: (d) Reason is correct statement and assertion is wrong statement.
Explanation: If c < 0, then a/c > b/c (inequality reverses)
Q50. Assertion: If -5 ≤ 2x + 9 ≤ 2, then x ∈ [-7, -3.5]
Reason: The graphical representation of -5 ≤ 2x + 9 ≤ 2 is a line segment.
Answer: (a) Assertion and reason both are correct statements and reason is the correct explanation of the assertion.
Solution: -5 ≤ 2x + 9 ≤ 2 ⇒ -14 ≤ 2x ≤ -7 ⇒ -7 ≤ x ≤ -3.5 ✓
📝 SHORT ANSWER TYPE QUESTIONS
Q51. Solve 2m - 3 < -8 when m is a natural number.
Answer: 2m - 3 < -8 ⇒ 2m < -5 ⇒ m < -2.5
No natural number satisfies this. Solution set = { }
Q52. Solve (m - 3)(m + 2) < 0 when m is a real number.
Answer: Critical points: m = -2, 3
(m - 3)(m + 2) < 0 ⇒ m ∈ (-2, 3)
Q53. Solve |m - 5| < 6.
Answer: -6 < m - 5 < 6 ⇒ -1 < m < 11 ⇒ m ∈ (-1, 11)
Q54. Solve 4x + 3 < 5x + 7 if x is a natural number.
Answer: 4x + 3 < 5x + 7 ⇒ -4 < x ⇒ x > -4
x ∈ N ⇒ x ∈ {1, 2, 3, ...}
Q55. Solve x + 5 < 0, if x is a real number.
Answer: x < -5 ⇒ x ∈ (-∞, -5)
Q58. Solve 5x - 3 > 3x - 5, if x is a real number.
Answer: 5x - 3 > 3x - 5 ⇒ 2x > -2 ⇒ x > -1 ⇒ x ∈ (-1, ∞)
Q59. Solve (x - 3)/(x - 5) > 2, for real x.
Answer: (x - 3)/(x - 5) > 2 ⇒ (x - 3)/(x - 5) - 2 > 0
⇒ [(x - 3) - 2(x - 5)]/(x - 5) > 0 ⇒ (x - 3 - 2x + 10)/(x - 5) > 0
⇒ (-x + 7)/(x - 5) > 0 ⇒ (x - 7)/(x - 5) < 0
⇒ x ∈ (5, 7)
Q60. Solve 0 < -x < 3 for real x.
Answer: 0 < -x < 3 ⇒ -3 < x < 0 ⇒ x ∈ (-3, 0)
Q61. Solve |x - 3| > 3 for real x.
Answer: |x - 3| > 3 ⇒ x - 3 < -3 or x - 3 > 3
⇒ x < 0 or x > 6 ⇒ x ∈ (-∞, 0) ∪ (6, ∞)
Q62. We know that sin x ∈ [-1, 1], x ∈ R. Find the range of 2sin x - 3.
Answer: sin x ∈ [-1, 1] ⇒ 2sin x ∈ [-2, 2] ⇒ 2sin x - 3 ∈ [-5, -1]
Q63. Solve x + 5 < 3 on number line.
Answer: x < -2 ⇒ Open circle at -2, shaded left
Q64. Represent the solution set of 3x - 5 ≥ 7 on real number line.
Answer: 3x - 5 ≥ 7 ⇒ 3x ≥ 12 ⇒ x ≥ 4
Closed circle at 4, shaded right
Q65. Represent the solution set of |x + 5| < 6 on real number line.
Answer: -6 < x + 5 < 6 ⇒ -11 < x < 1
Open circles at -11 and 1, shaded between
Q66. The maximum value of 7x - 5 is 9. Find the maximum value of 2x + 3.
Answer: 7x - 5 ≤ 9 ⇒ 7x ≤ 14 ⇒ x ≤ 2
2x + 3 ≤ 2(2) + 3 = 7
Q67. The minimum value of x - 2 is 12, find the minimum value of 3x + 5.
Answer: x - 2 ≥ 12 ⇒ x ≥ 14
3x + 5 ≥ 3(14) + 5 = 47
Q68. Find the solutions of |2 - x| = x - 2.
Answer: |2 - x| = x - 2 ⇒ |-(x - 2)| = x - 2 ⇒ |x - 2| = x - 2
This is true when x - 2 ≥ 0 ⇒ x ≥ 2
Q69. Solve 1 ≤ |x - 2| ≤ 3 for real x.
Answer: |x - 2| ≥ 1 ⇒ x - 2 ≤ -1 or x - 2 ≥ 1 ⇒ x ≤ 1 or x ≥ 3
|x - 2| ≤ 3 ⇒ -3 ≤ x - 2 ≤ 3 ⇒ -1 ≤ x ≤ 5
Intersection: x ∈ [-1, 1] ∪ [3, 5]
Q70. Solve 10 ≤ -5(x - 2) < 20 for real x.
Answer: 10 ≤ -5x + 10 < 20 ⇒ 0 ≤ -5x < 10 ⇒ -10 < 5x ≤ 0
⇒ -2 < x ≤ 0 ⇒ x ∈ (-2, 0]
✅ ANSWERS - MCQs
| Q.No. | Answer | Q.No. | Answer |
|---|---|---|---|
| 1 | (c) | 11 | (c) |
| 2 | (b) | 12 | (d) |
| 3 | (a) | 13 | (d) |
| 4 | (d) | 14 | (d) |
| 5 | (b) | 15 | (b) |
| 6 | (a) | 16 | (a) |
| 7 | (a) | 17 | (a) |
| 8 | (b) | 18 | (c) |
| 9 | (c) | 19 | (c) |
| 10 | (b) | 20 | (d) |
📌 KEY FORMULAS SUMMARY
| Inequality Type | Solution Set |
|---|---|
| |x| < a (a > 0) | -a < x < a |
| |x| ≤ a (a > 0) | -a ≤ x ≤ a |
| |x| > a (a > 0) | x < -a or x > a |
| |x| ≥ a (a > 0) | x ≤ -a or x ≥ a |
| ax + b < 0 (a > 0) | x < -b/a |
| ax + b > 0 (a > 0) | x > -b/a |
| ax + b < 0 (a < 0) | x > -b/a (inequality reversed) |
🔹 Important Rules:
- Adding/subtracting same number: Inequality unchanged
- Multiplying/dividing by positive: Inequality unchanged
- Multiplying/dividing by negative: Inequality reversed
📚 Kendriya Vidyalaya Sangathan | Class XI Mathematics
Chapter 5: Linear Inequalities | Student Support Material