NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.2: Class 11 Mathematics Chapter 5, titled Linear Inequalities, is a fundamental topic that helps students understand how to deal with relationships between quantities that are not necessarily equal. Unlike linear equations, which represent exact relationships, linear inequalities express conditions where one quantity is less than, greater than, or equal to another, providing a more flexible approach to problem-solving.
Exercise 5.2 of this chapter details about solving and graphing linear inequalities, both in one and two variables. It provides students with essential skills for handling inequalities in real-world contexts, such as budgeting, planning, and optimization problems. NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.2 offer a detailed breakdown of each problem, allowing students to follow the correct methods and improve their problem-solving techniques. Through clear step-by-step explanations, these solutions enhance students' understanding, providing the support they need to excel in both their exams and practical applications of linear inequalities. Students can check the NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.2 below.Download NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.2 PDF
Solve the following inequalities graphically in a two-dimensional plane.
1. x + y < 5
Solution:
Given, x + y < 5 Consider,X | 0 | 5 |
y | 5 | 0 |
2. 2x + y ≥ 6
Solution:
Given 2x + y ≥ 6 Now, draw a solid line 2x + y = 6 in the graph (∵ 2x + y = 6 is included in the given question.) Now, consider 2x + y ≥6 Select a point (0, 0). ⇒ 2 × (0) + 0 ≥ 6 ⇒ 0 ≥ 6 (This is false.) ∴ The solution region of the given inequality is above the line 2x + y = 6. (Away from the origin.) The graph is as follows:3. 3x + 4y ≤ 12
Solution:
Given, 3x + 4y ≤ 12 Now, draw a solid line 3x + 4y = 12 in the graph (∵ 3x + 4y = 12 is included in the given question.) Now, consider 3x + 4y ≤ 12 Select a point (0, 0). ⇒ 3 × (0) + 4 × (0) ≤ 12 ⇒ 0 ≤ 12 (This is true.) ∴ The solution region of the given inequality is below the line 3x + 4y = 12. (That is, the origin is included in the region.) The graph is as follows:4. y + 8 ≥ 2x
Solution:
Given, y + 8 ≥ 2x Now, draw a solid line y + 8 = 2x in the graph (∵ y + 8 = 2x is included in the given question.) Now, consider y + 8 ≥ 2x Select a point (0, 0). ⇒ (0) + 8 ≥ 2 × (0) ⇒ 0≤ 8 (This is true.) ∴ The solution region of the given inequality is above the line y + 8 = 2x. (That is, the origin is included in the region.) The graph is as follows:5. x – y ≤ 2
Solution:
Given, x – y ≤ 2 Now, draw a solid line x – y = 2 in the graph (∵ x – y = 2 is included in the given question.) Now, consider x – y ≤ 2 Select a point (0, 0). ⇒ (0) – (0) ≤ 2 ⇒ 0 ≤ 2 (This is true.) ∴ The solution region of the given inequality is above the line x – y = 2. (That is, the origin is included in the region.) The graph is as follows:6. 2x – 3y > 6
Solution:
Given, 2x – 3y > 6 Now, draw a dotted line 2x – 3y = 6 in the graph (∵ 2x – 3y = 6 is excluded in the given question.) Now, consider 2x – 3y > 6 Select a point (0, 0). ⇒ 2 × (0) – 3 × (0) > 6 ⇒ 0 > 6 (This is false.) ∴ The solution region of the given inequality is below the line 2x – 3y > 6. (Away from the origin.) The graph is as follows:7. – 3x + 2y ≥ – 6
Solution:
Given, – 3x + 2y ≥ – 6 Now, draw a solid line – 3x + 2y = – 6 in the graph (∵– 3x + 2y = – 6 is included in the given question.) Now, consider – 3x + 2y ≥ – 6 Select a point (0, 0). ⇒ – 3 × (0) + 2 × (0) ≥ – 6 ⇒ 0 ≥ – 6 (This is true.) ∴ The solution region of the given inequality is above the line – 3x + 2y ≥ – 6. (That is, the origin is included in the region) The graph is as follows:8. y – 5x < 30
Solution:
Given, y – 5x < 30 Now, draw a dotted line 3y – 5x = 30 in the graph (∵3y – 5x = 30 is excluded in the given question.) Now, consider 3y – 5x < 30 Select a point (0, 0). ⇒ 3 × (0) – 5 × (0) < 30 ⇒ 0 < 30 (This is true.) ∴ The solution region of the given inequality is below the line 3y – 5x < 30. (That is, the origin is included in the region.) The graph is as follows: