NCERT Solutions  Class 11 Maths Chapter 6  Linear Inequalities
CBSE Solutions  Maths Class 11
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NCERT  Class 11 Maths
Book:  National Council of Educational Research and Training (NCERT) 

Board:  Central Board of Secondary Education (CBSE) 
Class:  11th 
Subject:  Maths 
Chapter:  6 
Chapters Name:  Linear Inequalities 
Medium:  English 
Linear Inequalities  Class 11 Maths  NCERT Books Solutions
NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1
Ex 6.1 Class 11 Maths Question 1.
Solve 24x < 100, when
(i) x is a natural number
(ii) x is an integer.
Solution.
Given inequality is 24x < 100
Dividing both sides by 24, we get
\(x<\frac { 100 }{ 24 } =\frac { 25 }{ 6 } \)
This inequality is true when
(i) x is a natural number, {1, 2, 3, 4} satisfies this inequality.
(ii) x is an integer, {…., 4, 3,2, 1, 0, 1, 2, 3, 4} satisfies this inequality.
Ex 6.1 Class 11 Maths Question 2.
Solve – 12x > 30, when
(i) x is a natural number
(ii) x is an integer
Solution.
Given inequality is 12x > 30 Dividing both sides by 12, we get
\(x<\frac { 30 }{ 12 } =\frac { 5 }{ 2 } \)
(i) This inequality is not true for any natural number.
(ii) Integers that satisfy this inequality are (…, 5, 4, 3}.
Ex 6.1 Class 11 Maths Question 3.
Solve 5x – 3 < 7, when
(i) x is an integer
(ii) x is a real number
Solution.
Given inequality is 5x – 3 < 7
Transposing 3 to R.H.S., we get
5x < 7 + 3 or 5x < 10
Dividing both sides by 5, we get
x < 2
(i) When x is an integer, {…. 2, 1, 0, 1} satisfies this inequality.
(ii) When x is real number, the solution is (∞, 2).
Ex 6.1 Class 11 Maths Question 4.
Solve 3x + 8 > 2, when
(i) x is an integer
(ii) x is a real number.
Solution.
Given inequality is 3x + 8 > 2
Transposing 8 to R.H.S., we get
3x > 2 – 8 = 6
Dividing both sides by 3, we get
x > 2
(i) When x is an integer, the solution is (1, 0, 1, 2, 3,…}
(ii) When x is real, the solution is (2, ∞).
Solve the inequalities in Exercises 5 to 16 for real x.
Ex 6.1 Class 11 Maths Question 5.
4x + 3 < 5x + 7
Solution.
The inequality is 4x + 3 < 5x + 7
Transposing 5x to L.H.S. and 3 to R.H.S., we get
4x – 5x < 7 – 3 or x < 4 Dividing both sides by 1, we get x > 4
∴ The solution is ( 4, ∞).
Ex 6.1 Class 11 Maths Question 6.
3x – 7 > 5x – 1
Solution.
The inequality is 3x – 7 > 5x 1
Transposing 5x to L.H.S. and 7 to R.H.S., we get
3x – 5x > 1 + 7 or 2x > 6
Dividing both sides by 2, we get
x < 3
∴ The solution is (∞, 3).
Ex 6.1 Class 11 Maths Question 7.
3(x – 1) < 2(x – 3)
Solution.
The inequality is 3(x – 1) < 2(x – 3) or 3x – 3 < 2x – 6
Transposing 2x to L.H.S. and 3 to R.H.S., we get
3x – 2x < – 6 + 3
⇒ x<3
∴ The solution is ( ∞, 3],
Ex 6.1 Class 11 Maths Question 8.
3(2 x) > 2(1 x)
Solution.
The inequality is 3(2 – x) > 2(1 – x) or 6 – 3x > 2 – Zx
Transposing 2x to L.H.S. and 6 to R.H.S., we get
3x + 2x > 2 – 6 or x > 4
Multiplying both sides by 1, we get
x ≤ 4
∴ The solution is ( ∞, 4],
Ex 6.1 Class 11 Maths Question 9.
\(x+\frac { x }{ 2 } +\frac { x }{ 3 } <11\)
Solution.
The inequality is \(x+\frac { x }{ 2 } +\frac { x }{ 3 } <11\)
Simplifying, \(\frac { 6x+3x+2x }{ 6 } <11\quad or\quad \frac { 11x }{ 6 } <11\)
Multiplying both sides by \(\frac { 6 }{ 11 } \), we get
x < 6
∴ The solution is ( ∞, 6),
Ex 6.1 Class 11 Maths Question 10.
\(\frac { x }{ 3 } >\frac { x }{ 2 } +1\)
Solution.
The inequality is \(\frac { x }{ 3 } >\frac { x }{ 2 } +1\)
Transposing \(\frac { x }{ 2 } \) to L.H.S., we get
\(\frac { x }{ 3 } \frac { x }{ 2 } >1\)
Simplifying, \(\frac { 2x3x }{ 6 } >1\quad or\quad \frac { x }{ 6 } >1\)
Multiplying both sides by 6, we get
x < 6
∴ The solution is ( ∞, – 6).
Ex 6.1 Class 11 Maths Question 11.
\(\frac { 3\left( x2 \right) }{ 5 } \le \frac { 5\left( 2x \right) }{ 3 } \)
Solution.
The inequality is \(\frac { 3\left( x2 \right) }{ 5 } \le \frac { 5\left( 2x \right) }{ 3 } \)
Multiply both sides by the L.C.M. of 5, 3 i.e., by 15.
3 x 3(x – 2) ≤ 5 x 5(2 – x)
or, 9(x – 2) ≤ 25(2 – x)
Simplifying, 9x – 18 ≤ 50 – 25x
Transposing 25x to L.H.S. and 18 to R.H.S.
∴ 9x + 25x ≤ 50 + 18 or 34x ≤ 68
Dividing both sides by 34, we get
x < 2
∴ Solution is ( ∞, 2].
Ex 6.1 Class 11 Maths Question 12.
\(\frac { 1 }{ 2 } \left( \frac { 3x }{ 5 } +4 \right) \ge \frac { 1 }{ 3 } \left( x6 \right) \)
Solution.
The inequality is \(\frac { 1 }{ 2 } \left( \frac { 3x }{ 5 } +4 \right) \ge \frac { 1 }{ 3 } \left( x6 \right) \)
or, \(\frac { 1 }{ 2 } \left( \frac { 3x+20 }{ 5 } \right) \ge \frac { 1 }{ 3 } \left( x6 \right) \)
Multiplying both sides by 30,
3(3x + 20) ≥ 10(x – 6) or, 9x + 60 ≥ 10x60
Transposing 10 x to L.H.S. and 60 to R.H.S., we get
∴ 9x10x ≥ 60 – 60 or x ≥120
Multiplying both sides by 1, we get
x < 120
∴ The solution is ( ∞, 120].
Ex 6.1 Class 11 Maths Question 13.
2(2x + 3) – 10 < 6(x – 2)
Solution.
The inequality is 2(2x + 3) – 10 < 6(x – 2)
Simplifying, 4x + 6 10 < 6x 12
or, 4x – 4 < 6x – 12
Transposing 6x to L.H.S. and – 4 to R.H.S., we get
∴ 4x – 6x < 12 + 4 or 2x < – 8 Dividing both sides by 2, we get x>4
∴ The solution is (4, ∞).
Ex 6.1 Class 11 Maths Question 14.
37 – (3x + 5) ≥ 9x – 8(x – 3)
Solution.
The inequality is 37 – (3x + 5) ≥ 9x – 8(x – 3)
Simplifying, 37 – 3x – 5 ≥ 9x – 8x + 24
or 32 – 3x ≥ x + 24
Transposing x to L.H.S. and 32 to R.H.S., We get
3x – x ≥ 24 – 32 or 4x ≥ 8
Dividing both sides by – 4, we get
x < 2
∴ The solution is ( ∞, 2].
Ex 6.1 Class 11 Maths Question 15.
\(\frac { x }{ 4 } <\frac { \left( 5x2 \right) }{ 3 } \frac { \left( 7x3 \right) }{ 5 } \)
Solution.
The inequality is \(\frac { x }{ 4 } <\frac { \left( 5x2 \right) }{ 3 } \frac { \left( 7x3 \right) }{ 5 } \)
Multiplying each term by the L.C.M. of 4, 3, 5, i.e., by 60, we get
15x < 100x – 40 – 84x + 36
or, 15x < 100x – 84x 40 + 36
or, 15x < 16x – 4
Tranposing 16x to L.H.S., we get
15x – 16x < 4 or x < 4 Multiplying both sides by 1, we get x >4
∴ The solution is (4, ∞).
Ex 6.1 Class 11 Maths Question 16.
\(\frac { \left( 2x1 \right) }{ 3 } \ge \frac { \left( 3x2 \right) }{ 4 } \frac { 2x }{ 5 } \)
Solution.
The inequality is \(\frac { \left( 2x1 \right) }{ 3 } \ge \frac { \left( 3x2 \right) }{ 4 } \frac { 2x }{ 5 } \)
Multiplying each term by L.C.M. of 3,4, 5, i.e., by 60
\(\frac { \left( 2x1 \right) }{ 3 } \times 60\ge \frac { \left( 3x2 \right) }{ 4 } \times 60\frac { 2x }{ 5 } \times 60\)
or, 20(2x – 1) ≥ (3x – 2) x 15 – (2 – x) x 12
or, 40x – 20 ≥ 45x – 30 – 24 + 12x
or, 40x – 20 ≥ 57x – 54
Transposing 57x to L.H.S. and 20 to R.H.S., we get
40x – 57x ≥ 54 + 20 or 17x ≥ 34
Dividing both sides by 17, we get
x <2
∴ The solution is ( ∞, 2].
Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.
Ex 6.1 Class 11 Maths Question 17.
3x – 2 < 2x + 1
Solution.
The inequality is 3x – 2 < 2x + 1
Transposing 2x to L.H.S. and 2 to R.H.S, we get
3x 2x < 1 + 2 or, x <3
Ex 6.1 Class 11 Maths Question 18.
5x – 3 ≥ 3x – 5
Solution.
The inequality is 5x – 3 ≥ 3x – 5
Transposing 3x to L.H.S. and 3 to R.H.S., we get
∴ 5x – 3x ≥ 5 + 3 or, 2x ≥ 2
Dividing both sides by 2, we get
Ex 6.1 Class 11 Maths Question 19.
3(1 – x) < 2(x + 4)
Solution.
3(1x) < 2(x + 4)
Simplifying 3 – 3x < 2x + 8
Transposing 2x to L.H.S. and 3 to R.H.S., we get
3x – 2x < 8 – 3 or 5x < 5
Dividing both sides by 5, we get
Ex 6.1 Class 11 Maths Question 20.
\(\frac { x }{ 2 } \ge \frac { \left( 5x2 \right) }{ 3 } \frac { \left( 7x3 \right) }{ 5 } \)
Solution.
The inequality is \(\frac { x }{ 2 } \ge \frac { \left( 5x2 \right) }{ 3 } \frac { \left( 7x3 \right) }{ 5 } \)
Ex 6.1 Class 11 Maths Question 21.
Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution.
Let Ravi gets x marks in third unit test.
∴ Average marks obtained by Ravi
\(=\frac { 70+75+x }{ 3 } \)
He has to obtain atleast 60 marks,
∴ \(\frac { 70+75+x }{ 3 } \ge 60\quad or,\quad \frac { 145+x }{ 3 } \ge 60
\)
Multiplying both sides by 3,
145 + x ≥ 60 x 3 = 180
Transposing 145 to R.H.S., we get
x ≥ 180 – 145 = 35
∴ Ravi should get atleast 35 marks in the third unit test.
Ex 6.1 Class 11 Maths Question 22.
To receive Grade ‘A’ in the course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.
Solution.
Let Sunita obtained x marks in the fifth examination.
∴ Average marks of 5 examinations
\(=\frac { 87+92+94+95+x }{ 5 } =\frac { 368+x }{ 5 } \)
This average must be atleast 90
∴ \(\frac { 368+x }{ 5 } \ge 90\)
Multiplying both sides by 5
368 + x ≥ 5 x 90 = 450
Transposing 368 to R.H.S., we get
x ≥ 450 – 368 = 82
∴ Sunita should obtain atleast 82 marks in the fifth examination.
Ex 6.1 Class 11 Maths Question 23.
Find all pairs of consecutive odd positive integers both of which are smaller than 10, such that their sum is more than 11.
Solution.
Let x be the smaller of the two odd positive integers. Then the other integer is x + 2. We should havex + 2< 10 and x + (x + 2) > 11 or, 2x + 2 > 11
or, 2x > 11 – 2 or, 2x > 9 or, \(x>\frac { 9 }{ 2 } \)
Hence, if one number is 5 (odd number), then the other is 7. If the smaller number is 7, then the other is 9. Hence, possible pairs are (5, 7) and (7, 9).
Ex 6.1 Class 11 Maths Question 24.
Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.
Solution.
Let x be the smaller of the two positive even integers then the other one is x + 2, then we should have x > 5
and x + x + 2 < 23 or, 2x + 2 < 23
or, 2x < 21 or, \(x<\frac { 21 }{ 2 } \)
Thus, the value of x may be 6,8,10 (even integers) Hence, the pairs may be (6, 8), (8,10), (10,12).
Ex 6.1 Class 11 Maths Question 25.
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Solution.
Let the shortest side measures x cm
The longest side will be 3x cm.
Third side will be (3x – 2) cm.
According to the problem, x + 3x + 3x – 2 ≥ 61
or, 7x – 2 ≥ 61 or, 7x ≥ 63 or, x ≥ 9
Hence, the minimum length of the shortest side is 9 cm.
Ex 6.1 Class 11 Maths Question 26.
A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths for the shortest board if the third piece is to be atleast 5 cm longer than the second?
[Hint: If x is the length of the shortest board, then x, (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5].
Solution.
Let x be the length of the shortest board, then x + 3 is the second length and 2x is the third length. Thus, x + (x + 3) + 2x ≤ 91
or 4x + 3 ≤ 91 or 4x ≤ 88 or x ≤ 22
According to the problem, 2x ≥ (x + 3) + 5 or x ≥ 8
∴ Atleast 8 cm but not more than 22 cm are the possible lengths for the shortest board.
NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.2
Solve the following inequalities graphically in twodimensional plane
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 Ex 6.3
Solve the following system of inequalities graphically:
Ex 6.3 Class 11 Maths Question 1.
x ≥ 3, y ≥ 2
Solution.
x ≥ 3, y ≥ 2
(i) AB represents the line x = 3
Putting x = 0 in x ≥ 3
0 ≥ 3, which is not true.
Therefore, origin does not lie in the region of x ≥ 3
Its graph is shaded on the right side of AB.
(ii) CD represents the line y = 2
Putting y = 0 in y ≥ 2 0 ≥ 2, which is not true.
Therefore, origin does not lie in the region of y ≥ 2.
Its graph is shaded above the CD.
Solution of system x ≥ 3 and y ≥ 2 is shown as the shaded region.
Ex 6.3 Class 11 Maths Question 2.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution.
Inequalities are 3x + 2y ≤ 12, x ≥ 1, y ≥ 2
(i) The line l_{1} : 3x + 2y = 12 passes through (4, 0), (0, 6)
AB represents the line, 3x + 2y = 12.
Consider the inequality 3x + 2y ≤ 12
Putting x = 0, y = 0 in 3x + 2y ≤ 12
0 + 0 ≤ 12, which is true.
Therefore, origin lies in the region 3x + 2y ≤ 12
∴ The region lying below the line AB including all the points lying on it.
(ii) The line l_{2} : x = 1 passes through (1, 0). This line is represented by EF. Consider the inequality x ≥ 1
Putting x = 0, 0 ≥ 1, which is not true.
Therefore, origin does not lie in the region of x ≥ 1
The region lies on the right of EF and the points on EF from the inequality x ≥ 1.
(iii) The line l_{3} : y = 2 passes through (0, 2). This line is represented by CD.
Consider the inequality y ≥ 2
Putting y = 0 in y ≥ 2, we get 0 ≥ 2 which is false.
∴ Origin does not lie in the region of y ≥ 2. y ≥ 2 is represented by the region above CD and all the points on this line. Hence, the region satisfying the inequalities.
3x + 2y ≥ 12, x ≥ 1, y ≥ 2 is the APQR.
Ex 6.3 Class 11 Maths Question 3.
2x + y ≥ 6, 3x + 4y ≤ 12
Solution.
The inequalities are 2x + y ≥ 6, 3x + 4y ≤ 12
(i) The line l_{1} : 2x + y = 6 passes through (3, 0), (0, 6)
AB represents the line 2x + y = 6
Putting x = 0, y = 0 in 2x + y ≥ 6 0 ≥ 6, which is false.
∴ Origin does not lie in the region of 2x + y ≥ 6 Therefore, the region lying above the line AB and all the points on AB represents the inequality 2x + y ≥ 6
(ii) The line l_{2} : 3x + 4y = 12 passes through (4, 0) and (0, 3).
This line is represented by CD.
Consider the inequality 3x + 4y ≤ 12
Putting x = 0, y = 0 in 3x + 4y ≤ 12, we get 0 ≤ 12, which is true.
∴ 3x + 4y ≤ 12 represents the region below the line CD (towards origin) and all the points lying on it.
The common region is the solution of 2x + 3y ≥ 6 are 3x + 4y ≤ 12 represented by the
shaded region in the graph.
Ex 6.3 Class 11 Maths Question 4.
x + y ≥ 4, 2x – y > 0
Solution.
The inequalities are , x + y ≥ 4, 2x – y > 0
(i) The line l_{1}: x + y = 4 passes through (4, 0) and (0, 4). This line is represented by AB. Consider the inequality x + y ≥ 4
Putting x = 0, y = 0 in x + y ≥ 4, we get 0 ≥ 4, which is false.
Origin does not lie in this region.
Therefore, ,r + y > 4 is represented by the region above the line x + y = 4 and all points lying on it.
(ii) The line l_{2} : 2x – y = 0 passes through (0, 0) and (1, 2).
This line is represented by CD.
Consider the inequality 2x – y > 0
Putting x = 1, y = 0, we get 2 > 0, which is true
This shows (1, 0) lies in the region.
i.e. region lying below the line 2x – y = 0
represents 2x — y > 0
∴ The common region to both inequalities is shaded region as shown in the figure.
Ex 6.3 Class 11 Maths Question 5.
2x – y> 1, x – 2y < 1
Solution.
The inequalities are 2x – y > 1 and x – 2y < 1
(i) Let us draw the graph of line
l_{1} : 2x – y = 1, passes through \(\left( \frac { 1 }{ 2 } ,0 \right) \) and
(0, 1) which is represented by AB. Consider the inequality 2x – y > 1.
Putting x = y = 0, we get 0 > 1, which is false.
Therefore, origin does not lie in region of 2x – y > 1 i.e., 2x – y > 1 represents the area below the line AB excluding all the points lying on 2x – y = 1.
(ii) Let us draw the graph of the line
l_{2} : x – 2y = 1, passes through (1, 0) and (0,1/2) which is represented by CD.
Consider the inequality x – 2y < 1
Putting x = y = 0, we have 0 < 1, which is false.
Therefore, origin does not lie in region of x – 2y < 1 i.e., x – 2y < 1 represents the area above the line CD excluding all the points lying on x – 2y = 1
⇒ The common region of both the inequality is the shaded region as shown in figure.
Ex 6.3 Class 11 Maths Question 6.
x + y ≤ 6, x + y ≥ 4
Solution.
The inequalities are
x + y ≤ 6 and x + y ≥ 4
(i) The line l_{1}: x + y = 6 passes through (6, 0) and (0, 6). It is represented by AB.
Consider the inequality x + y ≤ 6
Putting x = 0, y = 0 in x + y ≤ 6 0 ≤ 6, which is true.
∴ Origin lies in the region of x + y ≤ 6
∴ x + y ≤ 6 is represented by the region below the line x + y = 6 and all the points lying on it.
(ii) The line l_{2} : x + y = 4 passes through (4, 0) and (0, 4). It is represented by CD.
Consider the inequality x + y ≥ 4
Putting x = 0, y = 0 in x + y ≥ 4 or, 0 + 0 ≥ 4, which is false.
∴ Origin does not lie in the region of x + y ≥ 4
∴ x + y ≥ 4 is represented by the region above the line x + y = 4 and all the points lying on it.
∴ The solution region is the shaded region between AB and CD as shown in the figure.
Ex 6.3 Class 11 Maths Question 7.
2x + y ≥ 8, x + 2y ≥ 10
Solution.
The inequalities are 2x + y ≥ 8 and x + 2y ≥ 10
(i) Let us draw the graph of the line
l_{1} : 2x + y = 8 passes through (4, 0) and (0, 8) which is represented by AB.
Consider the inequality 2x + y ≥ 8
Putting x = y = 0, we get 0 ≥ 8, which is false.
∴ Origin does not lie in the region of 2x + y ≥ 8.
i.e., 2x + y ≥ 8 represents the area above the line AB and all the points lying on 2x + y = 8.
(ii) Let us draw the graph of line
l_{2} : x + 2y = 10, passes through (10, 0) and (0, 5) which is represented by CD. Consider the inequality x + 2y ≥ 10
Putting x = 0, y = 0, we have 0 ≥ 10, which is false.
∴ The origin does not lie in region of x + 2y ≥ 10
i.e. x + 2y ≥ 10 represents the area above the line CD and all the points lying on x + 2y = 10.
⇒ The common region of both the inequality is the shaded region as shown in the figure.
Ex 6.3 Class 11 Maths Question 8.
x + y ≤ 9, y > x, x ≥ 0
Solution.
The inequalities are x + y ≤ 9, y > x and x ≥ 0
(i) Consider the inequality x + y ≤ 9
The line l_{1} : x + y = 9 passes through (9, 0) and (0, 9). AB represents this line.
Putting x = 0, y = 0 in x + y ≤ 9
0 + 0 = 0 ≤ 9, which is true.
Origin lies in this region. i.e., x + y ≤ 9 represents the area below the line AB and all the points lying on x + y = 9.
(ii) The line l_{2} : y = x, passes through the origin and (2, 2).
∴ CD represents the line y = x
Consider the inequality y – x > 0
Putting x = 0,y = 1 in y – x > 0
1 – 0 > 0, which is true.
∴ (0, 1) lies in this region.
The inequality y > x is represented by the region above the line CD, excluding all the points lying on y – x = 0.
(iii) The region x ≥ 0 lies on the right of yaxis.
∴ The common region of the inequalities is the region bounded by ΔPQO is the solution of x + y ≤ 9, y > x, x ≥ 0.
Ex 6.3 Class 11 Maths Question 9.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution.
The inequalities are 5x + 4y ≤ 20, x ≥ 1, y ≥ 2
(i) The line l_{1} : 5x + 4y = 20 passes through (4, 0) and (0, 5). This line is represented by AB.
Consider the inequality 5x + 4y ≤ 20
Putting x = 0, y = 0
0 + 0 = 0 ≤ 20, which is true.
The origin lies in this region, i.e., region below the line 5x + 4y = 20 and all the points lying on it belong to 5x + 4y ≤ 20.
(ii) The line l_{2} : y = 2, line is parallel to xaxis at a distance 2 from the origin. It is represented by EF. Putting y = 0, 0 ≥ 2 is not true.
Origin does not lie in this region.
Region above y = 2 represents the inequality y ≥ 2 including the points lying on it.
(iii) The line l_{3} : x = 1, line parallel to yaxis at a distance 1 from the origin. It is represented by CD. Putting x = 0in x – 1 ≥ 0
1 ≥ 0, which is not true.
Origin does not lie in this region.
∴ The region on the right of x = 1 and all the points lying on it belong to x ≥ 1.
∴ Shaded area bounded by ΔPQR is the solution of given inequalities.
Ex 6.3 Class 11 Maths Question 10.
3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Solution.
The inequalities are
We first draw the graphs of lines
l_{1} : 3x + 4y = 60, l_{2} : x + 3y – 30, x = 0 and y = 0.
(i) The line 3x + 4y = 60 passes through (20, 0) and (0,15) which is represented by AB. Consider the inequality 3x + 4y ≤ 60, putting x = 0, y = 0 in 3x + 4y ≤ 60, we get 0 + 0 ≤ 60, which is true.
∴ 3x + 4y ≤ 60 represents the region below AB and all the points on AB.
(ii) Further, x + 3y = 30 passes through (0,10) and (30, 0), CD represents this line.
Consider the inequality x + 3y ≤ 30
Putting x = 0, y = 0 in x + 3y < 30, w’e get 0 < 30 is true.
∴Origin lies in the region x + 3y ≤ 30. This inequality represents the region below it and the line itself.
Thus, we note that inequalities (1) and (2) represent the two regions below the respective lines (including the lines).
Inequality (3) represents the region on the right of yaxis and the i/axis itself.
Inequality (4) represents the region above xaxis and the xaxis itself.
∴ Shaded area in the figure is the solution area.
Ex 6.3 Class 11 Maths Question 11.
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Solution.
We have the inequalities :
We first draw the graphs of lines
l_{1} : 2x + y = 4, l_{2} : x + y = 3 and l_{3} : 2x – 3y = 6
(i) 2x + y = 4, passes through (2, 0) and (0, 4) which represents by AB
Consider the inequality 2x + y ≥ 4
Putting x = 0, y = 0 in 2x + y ≥ 4, we get 0 ≥ 4 is false.
∴ Origin does not lie in the region of 2x + y ≥ 4 This inequality represents the region above the line AB and all the points on the line AB.
(ii) Again, x + y = 3 is represented by the line CD, passes through (3, 0) and (0, 3). Consider the inequality x + y ≤ 3, putting x = 0,y = 0 in x + y ≤ 3, we get 0 ≤ 3 is true.
∴ Origin lies in the region of x + y ≤ 3
∴ x + y ≤ 3 represents the region below the line CD and all the points on the line CD.
(iii) Further, 2x – 3y = 6 is represented by EF passes through (0, 2) and (3, 0).
Consider the inequality 2x – 3y ≤ 6, putting x = 0, y = 0 in 2x – 3y ≤ 6, we get 0 ≤ 6, which is true.
∴ Origin lies in it.
∴ 2x – 3y ≤ 6 represents the region above the line EF and all the points on the line EF.
∴ Shaded triangular area in the figure is the solution of given inequalities.
Ex 6.3 Class 11 Maths Question 12.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Solution.
The inequalities are
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
(i) The line l_{1}: x – 2y = 3 passes through (3, 0) and \(\left( 0,\frac { 3 }{ 2 } \right) \)
This is represented by AB. Consider the inequality x – 2y ≤ 3, putting x = 0, y = 0 we get 0 ≤ 3, which is true.
⇒ Origin lies in the region of x – 2y ≤ 3.
Region on the above of this line and including its points represents x – 2y ≤ 3
(ii) The line l_{2} : 3x + 4y = 12 passes through (4, 0) and (0, 3). CD represents this line. Consider the inequality 3x + 4y ≥12 putting x = 0, y = 0, we get 0 ≥ 12 which is false.
∴ Origin does not lie in the region of 3x + 4y ≥ 12.
The region above the line CD and including points of the line CD represents 3x + 4y ≥ 12.
(iii) x ≥ 0 is the region on the right of Yaxis and all the points lying on it.
(iv) The line l_{3} : y = 1 is the line parallel to Xaxis at a distance 1 from it. Consider y ≥ 1 or y – 1 ≥ 0, putting y = 0 in y 1 ≥ 0
We get 1 ≱ 0, origin does not lie in the region.
y ≥ 1 is the region above y = 1 and the points lying on it.
∴ The shaded region shown in figure represents the solution of the given inequalities.
Ex 6.3 Class 11 Maths Question 13.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
Solution.
The inequalities are 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
(i) The line l_{1} : 4x + 3y = 60 passes through (15, 0), (0, 20) and it is represented by AB. Consider the inequality 4x + 3y ≤ 60
Putting x = 0, y = 0.
0 + 0 = 0 ≤ 60 which is true,
therefore, origin lies in this region.
Thus, region is below the line AB and the points lying on the line AB represents the inequality 4x + 3y ≤ 60.
(ii) The line l_{2} : y = 2x passes through (0, 0). It is represented by CD.
Consider the inequality y ≥ 2x. Putting x = 0, y = 5 in y – 2x ≥ 0 5 ≥ 0 is true.
∴ (0, 5) lies in this region.
Region lying above the line CD and including the points on the line CD represents y ≥ 2x
(iii) x ≥ 3 is the region lying on the right of line l_{3} : x = 3 and points lying on x = 3 represents the inequality x ≥ 3.
∴ The shaded area APQR in which x ≥ 0 and y ≥ 0 is true for each point, is the solution of given inequalities.
Ex 6.3 Class 11 Maths Question 14.
3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0
Solution.
The inequalities are 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0
(i) The line l_{1} : 3x + 2y = 150 passes through the points (50,0) and (0, 75). AB represents the line. Consider the inequality 3x + 2y ≤ 150.
Putting x = 0, y = 0 in 3x + 2y ≤ 150
⇒ 0 ≤ 150 which is true, shows that origin lies in this region.
The region lying below the line AB and the points lying on AB represents the inequality 3x + 2y < 150.
(ii) The line l_{2} : x + 4y = 80 passes through the points (80, 0), (0, 20). This is represented by CD.
Consider the inequality x + 4y ≤ 80 putting x = 0, y = 0, we get 0 ≤ 80, which is true.
⇒ Region lying below the line CD and the points on the line CD represents the inequality x + 4y ≤ 80
(iii) x ≤ 15 is the region lying on the left to
l_{3} : x = 15 represented by EF and the points lying on EF.
(iv) x ≥ 0 is the region lying on the right side of Yaxis and all the points on Yaxis.
(v) y ≥ 0 is the region lying above the Xaxis and all the points on Xaxis.
Thus, the shaded region in the figure is the solution of the given inequalities.
Ex 6.3 Class 11 Maths Question 15.
x+2y ≤ 10 , x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Solution.
The inequalities are x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
(i) l_{1} : x + 2y = 10 passes through (10, 0) and (0, 5). The line AB represents this equation. Consider the inequality x + 2y ≤ 10 putting x = 0, y = 0, we get 0 ≤ 10 which is true.
∴ Origin lies in the region of x + 2y ≤ 10.
∴ Region lying below the line AB and the points lying on it represents x + 2y ≤ 10
(ii) l_{2} : x + y = 1 passes through (1, 0) and (0, 1). Thus line CD represents this equation. Consider the inequality x + y ≥ 1 putting x = 0, y = 0, we get 0 ≥ 1, which is not true. Origin does not lie in the region of x + y ≥ 1.
∴ The region lying above the line CD and the points lying on it represents the inequality x + y ≥1
(iii) l_{3} : x – y = 0, passes through (0, 0). This is being represented by EF.
Consider the inequality x – y ≤0, putting x = 0, y = 1, We get 0 – 1 ≤ 0 which is true
⇒ (0,1) lies on x – y ≤ 0
The region lying above the line EF and the points lying on it represents the inequality x – y ≤ 0.
(iv) x ≥ 0 is the region lying on the right of Yaxis and the points lying on x = 0.
(v) y ≥ 0 is the region above Xaxis, and the points lying on y = 0.
∴ The shaded area in the figure represents the given inequalities.
NCERT Class 11 Maths
Class 11 Maths Chapters  Maths Class 11 Chapter 6
Chapterwise NCERT Solutions for Class 11 Maths

NCERT Solutions For Class 11 Maths Chapter 1 Sets
NCERT Solutions For Class 11 Maths Chapter 2 Functions and relations
NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric functions
NCERT Solutions For Class 11 Maths Chapter 4 Principle of mathematical induction
NCERT Solutions For Class 11 Maths Chapter 5 Quadratic equations and complex numbers
NCERT Solutions For Class 11 Maths Chapter 6 Linear Inequalities
NCERT Solutions For Class 11 Maths Chapter 7 Permutations and combinations
NCERT Solutions For Class 11 Maths Chapter 8 Binomial Theorem
NCERT Solutions For Class 11 Maths Chapter 9 Sequences and series
NCERT Solutions For Class 11 Maths Chapter 10 Straight lines
NCERT Solutions For Class 11 Maths Chapter 11 Conic sections
NCERT Solutions For Class 11 Maths Chapter 12 Introduction to threedimensional geometry
NCERT Solutions For Class 11 Maths Chapter 13 Limits and derivatives
NCERT Solutions For Class 11 Maths Chapter 14 Mathematical Reasoning
NCERT Solutions For Class 11 Maths Chapter 15 Statistics
NCERT Solutions For Class 11 Maths Chapter 16 Probability
NCERT Solutions for Class 12 All Subjects  NCERT Solutions for Class 10 All Subjects 
NCERT Solutions for Class 11 All Subjects  NCERT Solutions for Class 9 All Subjects 
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