Class 11 Mathematics Notes Chapter 6 (Chapter 6) – Examplar Problems (English) Book

Detailed Notes with MCQs of Chapter 6: Linear Inequalities from your NCERT Exemplar. This chapter is crucial not just for your Class 11 exams but also forms the foundation for topics like Linear Programming which frequently appear in various government entrance examinations. Pay close attention to the concepts and rules.

Chapter 6: Linear Inequalities - Detailed Notes for Government Exam Preparation

1. Introduction to Inequalities:

• Definition: An inequality is a mathematical statement that compares two values or expressions using inequality symbols.
• Symbols:
  • < : Less than
  • > : Greater than
  • ≤ : Less than or equal to (Slack inequality)
  • ≥ : Greater than or equal to (Slack inequality)
• Linear Inequality: An inequality involving a linear function.
  • In one variable: ax + b < 0, ax + b > 0, ax + b ≤ 0, ax + b ≥ 0 (where a ≠ 0).
  • In two variables: ax + by < c, ax + by > c, ax + by ≤ c, ax + by ≥ c (where a ≠ 0 or b ≠ 0).

2. Algebraic Solution of Linear Inequalities in One Variable:

• Goal: To find the set of all possible values of the variable (solution set) that make the inequality true.
• Rules for Solving:
  • Rule 1: Equal numbers may be added to (or subtracted from) both sides of an inequality without changing the sign of inequality.
    • If x < y, then x + c < y + c and x - c < y - c.
  • Rule 2: Both sides of an inequality can be multiplied (or divided) by the same positive number without changing the sign of inequality.
    • If x < y and c > 0, then xc < yc and x/c < y/c.
  • Rule 3: Both sides of an inequality can be multiplied (or divided) by the same negative number, but then the sign of inequality is reversed.
    • If x < y and c < 0, then xc > yc and x/c > y/c. (This is extremely important!)
• Representation on Number Line:
  • Use an open circle o for < or > (endpoint not included).
  • Use a closed circle • for ≤ or ≥ (endpoint included).
  • Shade the part of the number line representing the solution set.
  • Example: Solve 3x - 6 ≥ 0.
    • 3x ≥ 6 (Adding 6 to both sides)
    • x ≥ 2 (Dividing by 3, a positive number)
    • Number line: A closed circle at 2, shaded to the right. Solution set: [2, ∞).
  • Example: Solve -2x + 4 > 0.
    • -2x > -4 (Subtracting 4)
    • x < 2 (Dividing by -2, reversing the inequality sign)
    • Number line: An open circle at 2, shaded to the left. Solution set: (-∞, 2).

3. System of Linear Inequalities in One Variable:

• Solve each inequality separately.
• Find the intersection of their solution sets (the values that satisfy all the inequalities simultaneously).
• Represent the common solution on the number line.
• Example: Solve 3x - 1 ≥ 5 and x + 2 < 6.
  • 3x ≥ 6 => x ≥ 2 => [2, ∞)
  • x < 4 => (-∞, 4)
  • Intersection: Values that are ≥ 2 AND < 4.
  • Solution set: [2, 4). Number line: Closed circle at 2, open circle at 4, shaded in between.

4. Graphical Solution of Linear Inequalities in Two Variables:

• Steps:
  1. Convert to Equation: Replace the inequality sign with an equal sign (=) to get the equation of the boundary line (ax + by = c).
  2. Draw the Boundary Line:
     • If the inequality is strict (< or >), draw a dashed line (points on the line are not part of the solution).
     • If the inequality is slack (≤ or ≥), draw a solid line (points on the line are part of the solution).
     • Find two points on the line (e.g., by setting x=0 and y=0) to draw it.
  3. Choose a Test Point: Select a point not on the line. The origin (0, 0) is usually the easiest, provided the line doesn't pass through it.
  4. Test the Inequality: Substitute the coordinates of the test point into the original inequality.
  5. Shade the Region:
     • If the test point satisfies the inequality, shade the entire half-plane containing the test point.
     • If the test point does not satisfy the inequality, shade the other half-plane (the one not containing the test point).
• The shaded region represents the solution set of the inequality.
• Example: Graph 2x + y ≤ 4.
  1. Equation: 2x + y = 4.
  2. Line: Solid line (due to ≤). Points: If x=0, y=4 -> (0, 4). If y=0, 2x=4 -> x=2 -> (2, 0). Draw a solid line through (0, 4) and (2, 0).
  3. Test Point: (0, 0).
  4. Test: 2(0) + 0 ≤ 4 => 0 ≤ 4 (True).
  5. Shade: Shade the half-plane containing the origin (0, 0).

5. Solution of System of Linear Inequalities in Two Variables (Graphical Method):

• Steps:
  1. Graph each inequality in the system on the same coordinate plane (following steps 1-5 from section 4 for each). Remember dashed/solid lines and correct shading for each.
  2. Identify the region that is common to all the shaded regions (the intersection of the half-planes).
• Feasible Region: The common shaded region is called the feasible region or solution region. Every point within this region (including the boundary lines if they are solid and part of the common region) satisfies all the inequalities in the system simultaneously.
• Bounded vs. Unbounded Region:
  • Bounded: The feasible region can be enclosed within a circle.
  • Unbounded: The feasible region extends indefinitely in some direction.
• Corner Points: For bounded regions, the vertices (intersection points of the boundary lines) are called corner points. These are particularly important in Linear Programming.

Key Points for Exams:

• Remember to reverse the inequality sign when multiplying or dividing by a negative number. This is a very common error.
• Distinguish between strict (<, >) and slack (≤, ≥) inequalities, especially for graphical representation (dashed vs. solid lines) and number line representation (open vs. closed circles).
• When solving systems graphically, clearly identify the common feasible region. Use different shading patterns or colors initially if it helps, then darken the final common region.
• Always use a test point that is not on the boundary line to determine the correct half-plane.
• Practice interpreting word problems and translating them into linear inequalities.

Multiple Choice Questions (MCQs):

1. The solution set of the inequality 3x + 9 ≥ 15 is:
   (A) (2, ∞)
   (B) [2, ∞)
   (C) (-∞, 2)
   (D) (-∞, 2]

2. Solve: -2x + 6 < 10.
   (A) x < -2
   (B) x > -2
   (C) x < 2
   (D) x > 2

3. The solution to the system of inequalities x - 2 > 0 and 2x < 6 is:
   (A) (2, 3)
   (B) [2, 3)
   (C) (2, 3]
   (D) (-∞, 3)

4. The graphical representation of x ≥ -1 on a number line is:
   (A) Open circle at -1, shaded right
   (B) Closed circle at -1, shaded right
   (C) Open circle at -1, shaded left
   (D) Closed circle at -1, shaded left

5. Which of the following points lies in the solution region of the inequality x + 2y ≤ 6?
   (A) (4, 2)
   (B) (5, 1)
   (C) (1, 2)
   (D) (7, 0)

6. The boundary line for the inequality 3x - y > 5 should be drawn as:
   (A) A solid line passing through (0, -5)
   (B) A dashed line passing through (0, -5)
   (C) A solid line passing through (5/3, 0)
   (D) A dashed line passing through (5/3, 0) and (0, -5)

7. When graphing y ≤ 3, the shaded region would be:
   (A) Above the solid line y = 3
   (B) Below the solid line y = 3
   (C) Above the dashed line y = 3
   (D) Below the dashed line y = 3

8. The inequality representing the solution set shown on the number line (Closed circle at 4, shaded left) is:
   (A) x < 4
   (B) x > 4
   (C) x ≤ 4
   (D) x ≥ 4

9. The solution region of the system of inequalities x ≥ 0, y ≥ 0, x + y ≤ 1 is:
   (A) A triangle with vertices (0,0), (1,0), (0,1)
   (B) An unbounded region in the first quadrant
   (C) The entire first quadrant
   (D) A square with vertices (0,0), (1,0), (0,1), (1,1)

10. Ravi needs at least 100 marks in total in subjects A and B. The marks in subject A are represented by x and in subject B by y. Which inequality represents this situation?
    (A) x + y < 100
    (B) x + y > 100
    (C) x + y ≤ 100
    (D) x + y ≥ 100

Answer Key for MCQs:

1. (B) 3x ≥ 15 - 9 => 3x ≥ 6 => x ≥ 2. Solution set: [2, ∞)
2. (B) -2x < 10 - 6 => -2x < 4. Divide by -2 and reverse the sign: x > -2.
3. (A) x > 2. x < 3. Intersection is 2 < x < 3. Solution set: (2, 3)
4. (B) ≥ means the endpoint (-1) is included (closed circle) and values are greater than or equal to -1 (shaded right).
5. (C) Test (1, 2): 1 + 2(2) = 1 + 4 = 5. Since 5 ≤ 6, the point (1, 2) lies in the solution region. (A: 4+4=8>6; B: 5+2=7>6; D: 7+0=7>6)
6. (D) The inequality is strict (>), so the line 3x - y = 5 is dashed. If x=0, y=-5. If y=0, x=5/3. The line passes through (0, -5) and (5/3, 0).
7. (B) ≤ means the line y=3 is solid. y values less than or equal to 3 are below the horizontal line y=3.
8. (C) Closed circle means included (≤ or ≥). Shaded left means less than. So, x ≤ 4.
9. (A) x ≥ 0 and y ≥ 0 restrict the region to the first quadrant. x + y ≤ 1 is the region below or on the solid line x + y = 1. The common region is the triangle formed by the origin (0,0), the x-intercept (1,0), and the y-intercept (0,1).
10. (D) "At least 100" means 100 or more. Total marks x + y must be greater than or equal to 100. So, x + y ≥ 100.

Study these notes thoroughly and practice the problems from your Exemplar book. Good luck!