NCERT Solutions For Class 11 Maths Chapter 6 Linear Inequalities Ex 6.2
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| Textbook | NCERT |
| Board | CBSE |
| Category | NCERT Solutions |
| Class | Class 11 |
| Subject | Maths |
| Chapter | Chapter 6 |
| Exercise | Class 11 Maths Chapter 6 Linear Inequalities Exercise 6.2 |
| Number of Questions Solved | 10 |
NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.2
NCERT TEXTBOOK EXERCISES
Ex 6.2 Class 11 Maths Question 1.
x + y < 5
Solution.
Consider the equation x + y = 5. It passes through the points (0, 5) and (5, 0). The line x + y = 5 is represented by AB. Consider the inequality x + y < 5
Put x = 0, y = 0
0 + 0 = 0 < 5, which is true. So, the origin O lies in the plane x + y < 5
∴ Shaded region represents the inequality x + y < 5
Ex 6.2 Class 11 Maths Question 2.
2x + y ≥ 6
Solution.
Consider the equation 2x + y = 6
The line passes through (0, 6), (3, 0).
The line 2x + y = 6 is represented by AB.
Now, consider 2x + y ≥ 6
Put x = 0, y = 0
0 + 0 ≥ 6, which does not satisfy this inequality.
∴ Origin does not lie in the region of 2x + y ≥ 6.
The shaded region represents the inequality 2x + y ≥ 6
Ex 6.2 Class 11 Maths Question 3.
3x + 4y ≤ 12
Solution.
We draw the graph of the equation 3x + 4y = 12. The line passes through the points (4, 0), (0, 3). This line is represented by AB. Now consider the inequality 3x + 4y ≤ 12
Putting x = 0, y = 0 0 + 0 = 0 ≤ 12, which is true
∴ Origin lies in the region of 3x + 4y ≤ 12 The shaded region represents the inequality 3x + 4y ≤ 12
Ex 6.2 Class 11 Maths Question 4.
y + 8 ≥ 2x
Solution.
Given inequality is y + 8 ≥ 2x
Let us draw the graph of the line, y+ 8 = 2x
The line passes through the points (4, 0), (0, -8).
This line is represented by AB.
Now, consider the inequality y + 8 ≥ 2x.
Putting x = 0, y = 0
0 + 8 ≥ 0, which is true
∴ Origin lies in the region of y + 8 ≥ 2x
The shaded region represents the inequality y + 8 ≥ 2x.
Ex 6.2 Class 11 Maths Question 5.
x – y ≤ 2
Solution.
Given inequality is x – y ≤ 2
Let us draw the graph of the line x – y = 2
The line passes through the points (2, 0), (0, -2)
This line is represented by AB.
∴ Origin lies in the region of x – y ≤ 2
The shaded region represents the inequality x – y ≤ 2.
Ex 6.2 Class 11 Maths Question 6.
2x – 3y > 6
Solution.
We draw the graph of line 2x – 3y = 6.
The line passes through (3, 0), (0, -2)
AB represents the equation 2x – 3y = 6
Now consider the inequality 2x – 3y > 6
Putting x = 0, y = 0
0 – 0 > 6, which is not true
∴ Origin does not lie in the region of 2x – 3y > 6.
The shaded region represents the inequality 2x – 3y > 6
Ex 6.2 Class 11 Maths Question 7.
-3x + 2y ≥ -6.
Solution.
Let us draw the line -3x + 2y = -6
The line passes through (2, 0), (0, -3)
The line AB represents the equation -3x + 2y = -6
Now consider the inequality -3x+ 2y ≥ -6
Putting x = 0, y = 0
0 + 0 ≥ -6, which is true.
∴ Origin lies in the region of -3x + 2y ≥ -6
The shaded region represents the inequality -3x + 2y ≥ – 6
Ex 6.2 Class 11 Maths Question 8.
3y- 5x < 30
Solution.
Given inequality is 3y – 5x < 30
Let us draw the graph of the line 3y – 5x = 30
The line passes through (-6, 0), (0, 10)
The line AB represents the equation 3y – 5x = 30
Now, consider the inequality 3y – 5x < 30
Putting x = 0, y = 0
0 – 0 < 30, which is true.
∴ Origin lies in the region of 3y – 5x < 30
The shaded region represents the inequality 3y – 5x < 30
Ex 6.2 Class 11 Maths Question 9.
y<- 2
Solution.
Given inequality is y < -2 ………(1)
Let us draw the graph of the line y = -2
AB is the required line.
Putting y = 0 in (1), we have
0 < -2, which is not true.
The solution region is the shaded region below the line.
Hence, every point below the line (excluding the line) is the solution area.
Ex 6.2 Class 11 Maths Question 10.
x > -3
Solution.
Let us draw the graph of x = -3
∴ AB represents the line x = -3
By putting x = 0 in the inequality x > -3
We get, 0 > -3, which is true.
∴ Origin lies in the region of x > -3.
Graph of the inequality x > -3 is shown in the figure by the shaded area
NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Exercise 6.2 PDF
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