Uncategorized

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

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

(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.

Solution.
The inequality is 
Simplifying, 
Multiplying both sides by , we get
x < 6
∴ The solution is (- ∞, 6),

Ex 6.1 Class 11 Maths Question 10.

Solution.
The inequality is 
Transposing  to L.H.S., we get

Simplifying, 
Multiplying both sides by -6, we get
x < -6
∴ The solution is (- ∞, – 6).

Ex 6.1 Class 11 Maths Question 11.

Solution.
The inequality is 
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.

Solution.
The inequality is 
or, 
Multiplying both sides by 30,
3(3x + 20) ≥ 10(x – 6) or, 9x + 60 ≥ 10x-60
Transposing 10 x to L.H.S. and 60 to R.H.S., we get
∴ 9x-10x ≥ -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.

Solution.
The inequality is 
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.

Solution.
The inequality is 
Multiplying each term by L.C.M. of 3,4, 5, i.e., by 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(1-x) < 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.

Solution.
The inequality is 

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

He has to obtain atleast 60 marks,
∴ 
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

This average must be atleast 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, 
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, 
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.

We hope the NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1, drop a comment below and we will get back to you at the earliest.

Solve the following inequalities graphically in two-dimensional 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

We hope the NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.2, drop a comment below and we will get back to you at the earliest.

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 l₁ : 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 l₂ : 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 l₃ : 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 l₁ : 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 l₂ : 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 l₁: 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 l₂ : 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
l₁ : 2x – y = 1, passes through  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
l₂ : 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 l₁: 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 l₂ : 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
l₁ : 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
l₂ : 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 l₁ : 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 l₂ : 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 y-axis.
∴ 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 l₁ : 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 l₂ : y = 2, line is parallel to x-axis 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 l₃ : x = 1, line parallel to y-axis 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
l₁ : 3x + 4y = 60, l₂ : 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 y-axis and the i/-axis itself.
Inequality (4) represents the region above x-axis and the x-axis 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
l₁ : 2x + y = 4, l₂ : x + y = 3 and l₃ : 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 l₁: x – 2y = 3 passes through (3, 0) and 
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 l₂ : 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 Y-axis and all the points lying on it.

(iv) The line l₃ : y = 1 is the line parallel to X-axis 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 l₁ : 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 l₂ : 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 l₃ : 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 l₁ : 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 l₂ : 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
l₃ : x = 15 represented by EF and the points lying on EF.

(iv) x ≥ 0 is the region lying on the right side of Y-axis and all the points on Y-axis.

(v) y ≥ 0 is the region lying above the X-axis and all the points on X-axis.
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) l₁ : 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) l₂ : 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) l₃ : 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 Y-axis and the points lying on x = 0.

(v) y ≥ 0 is the region above X-axis, and the points lying on y = 0.
∴ The shaded area in the figure represents the given inequalities.

We hope the NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3, drop a comment below and we will get back to you at the earliest.

Ex 10.1 Class 11 Maths Question 1.
Draw a quadrilateral in the Cartesian plane, whose vertices are (- 4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its area.
Solution:
The figure of quadrilateral whose vertices are A(- 4, 5), B(0, 7), C(5, -5) and D(-4, -2) is shown in the below figure.

Ex 10.1 Class 11 Maths Question 2.
The base of an equilateral triangle with side 2a lies along they-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.
Solution:
Since base of an equilateral triangle lies along y-axis.

Ex 10.1 Class 11 Maths Question 3.
Find the distance between P(x₁ y₁) and Q(x₂, y₂) when :
(i) PQ is parallel to the y-axis,
(ii) PQ is parallel to the x-axis.
Solution:
We are given that co-ordinates of P is (x₁, y₁) and Q is (x₂, y₁).
Distance between the points P(x₁, y₁) and Q(x₂, y₁) is

Ex 10.1 Class 11 Maths Question 4.
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Solution:
Let the point be P(x, y). Since it lies on the x-axis ∴ y = 0 i.e., required point be (x, 0).
Since the required point is equidistant from points A(7, 6) and B(3, 4) ⇒ PA = PB

Ex 10.1 Class 11 Maths Question 5.
Find the slope of a line, which passes through the origin and the mid-point of the line segment joining the points P(0, -4) and B(8,0).
Solution:
We are given that P(0, -4) and B(8, 0).
Let A be the midpoint of PB, then

Ex 10.1 Class 11 Maths Question 6.
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle.
Solution:
Let A(4, 4), B(3, 5) and C(-1, -1) be the vertices of ∆ABC.
Let m₁ and m₂ be the slopes of AB and AC respectively.

Ex 10.1 Class 11 Maths Question 7.
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Solution:
The given line makes an angle of 90° + 30° = 120° with the positive direction of x-axis.
Hence, m = tan 120° = – .

Ex 10.1 Class 11 Maths Question 8.
Find the value of x for which the points (x, -1), (2, 1) and (4,5) are collinear.
Solution:
Let A(x, -1), B(2, 1) and C(4, 5) be the given collinear points. Then by collinearity of A, B, C, we have slope of AB = slope of BC

Ex 10.1 Class 11 Maths Question 9.
Without using distance formula, show that points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.
Solution:
Let A(-2, -1), B(4, 0), C(3, 3) and D(-3, 2) be the vertices of the given quadrilateral ABCD. Then,

Ex 10.1 Class 11 Maths Question 10.
Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).
Solution:
We are given that the points are A(3, -1) and B(4, -2)

Ex 10.1 Class 11 Maths Question 11.
The slope of a line is double of the slope of another line. If tangent of the angle between them is , find the slopes of the lines.
Solution:
Let m₁ and m₂ be the slopes of two lines.

Ex 10.1 Class 11 Maths Question 12.
A line passes through (x₁, y₂) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).
Solution:
A line passes through (x₁, y₁) and (h, k). Also, the slope of the line is m.

Ex 10.1 Class 11 Maths Question 13.
If three points (h, 0), (a, b) and (0, k) lie on a line, show that 
Solution:
Let A(h, 0), B(o, b) and C(0, k) be the given collinear points.
∴ Slope of AB = Slope of BC

Ex 10.1 Class 11 Maths Question 14.
Consider the following population and year graph, find the slope of the line AB and using it, find what will be the population in the year 2010?
Solution:
Slope of AB + 
Let the population in year 2010 is y, and co-ordinate of C is (2010, y) then, slope of AB = slope of BC

We hope the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.1, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.1, drop a comment below and we will get back to you at the earliest.

Ex 10.2 Class 11 Maths Question 1.
Write the equations for the x-and y-axes.
Solution:
We know that the ordinate of each point on the x-axis is 0.
If P(x, y) is any point on the x-axis, then y = 0.
∴ Equation of x-axis is y = 0.
Also, we know that the abscissa of each point on the y-axis is 0. If P(x, y) is any point on the y-axis, then x = 0.
∴ Equation of y-axis is x = 0.

Ex 10.2 Class 11 Maths Question 2.
Passing through the point (-4,3) with slope .
Solution:
We know that the equation of a line with slope m and passing through the point (x₀, y₀) is given by (y – y₀) = m (x – x₀).

Ex 10.2 Class 11 Maths Question 3.
Passing through (0, 0) with slope m.
Solution:
We know that the equation of a line with slope m and passing through the point (x₀, y₀) is given by (y – y₀) = m(x – x₀)
Here, slope = m, x₀ = 0, y₀ = 0 Required equation is (y – 0) = m(x – 0)
⇒ y = mx.

Ex 10.2 Class 11 Maths Question 4.
Passing through (2,2^3) and inclined with the x-axis at an angle of 75°.
Solution:
We know that the equation of a line with slope m and passing through the point (X₀, y₀) is given by (y – y₀) = m(x – x₀)

Ex 10.2 Class 11 Maths Question 5.
Intersecting the x-axis at a distance of 3 units to the left of origin with slope -2.
Solution:
We know that the equation of a line with slope m and passing through the point
(x₀, y₀) is given by (y – y₀) = m(x – x₀).
Here, m = – 2, x₀ = – 3, y₀ = 0
y-0 = -2(x + 3) ⇒ 2x + y + 6 = 0

Ex 10.2 Class 11 Maths Question 6.
Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.
Solution:
We know that the equation of line with slope m and passing through the point (x₀, y₀) is given by (y – y₀) = m(x – x₀)

Ex 10.2 Class 11 Maths Question 7.
Passing through the points (-1,1) and (2, -4).
Solution:
Let the given points be A(-1, 1) and B(2, -4).
We know that the equation of a line passing through the given points (xx, y,) and (x2, y2) is given by

Ex 10.2 Class 11 Maths Question 8.
Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.
Solution:
Here, we are given that p = 5 and ⍵ = 30°.

Ex 10.2 Class 11 Maths Question 9.
The vertices of ∆PQR are P(2, 1), Q(-2, 3) and ff(4, 5). Find equation of the median through the vertex R.
Solution:
The vertices of ∆PQR are P( 2, 1), Q(-2, 3) and R(4, 5).
Let S be the midpoint of PQ.

Ex 10.2 Class 11 Maths Question 10.
Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2,5) and (-3,6).
Solution:
Let M(2, 5) and N(-3, 6) be the end points of the given line segment.

Ex 10.2 Class 11 Maths Question 11.
A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.
Solution:
Let A(1, 0) and B( 2, 3) be the given points and D divides the line segment in the ratio 1 : n.

Ex 10.2 Class 11 Maths Question 12.
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3).
Solution:
Let the required line make intercepts a on the x-axis and y-axis.
Then its equation is 
⇒ x + y = a … (i)
Since (i) passes through the point (2, 3), we have
2 + 3 = a ⇒ a = 5
So, required equation of the line is
 ⇒ x + y = 5.

Ex 10.2 Class 11 Maths Question 13.
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Solution:
Let the intercepts made by the line on the x-axis and y-axis be o and 9 – a respectively.
Then its equation is

Since it passes through point (2, 2), we have 
⇒ 2(9 – a) + 2a = a(9 – a)
⇒ 18 – 2a + 2a = 9a – 9a²
⇒ 18 = 9a – a² v a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a(a – 6) – 3 (a – 6) = 0 ⇒ a = 3, 6
Now, if a = 3 ⇒ b = 9 – 3 = 6 and if a = 6 ⇒ b = 9 – 6 = 3
So, required equation is

i.e., 2x + y – 6 = 0 or x + 2y – 6 = 0.

Ex 10.2 Class 11 Maths Question 14.
Find equation of the line through the point (0, 2) making an angle  with the positive x-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Solution:
Here, m = tan  = 
The equation of the line passing through point (0, 2) is y -2 = (x – 0)
⇒  + y – 2 = 0
The slope of line parallel to
 + y – 2 = 0 is .
Since, it passes through (0, -2).
So, the equation of line is
y + 2= (x – 0)
⇒  + y + 2 = 0.

Ex 10.2 Class 11 Maths Question 15.
The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.
Solution:

Ex 10.2 Class 11 Maths Question 16.
The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms C.
Solution:
Assuming L along x-axis and C along y-axis, we have two points (124.942, 20) and (125.134, 110). By two point form, the point (L, C) satisfies the equation

Ex 10.2 Class 11 Maths Question 17.
The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs. 14/litre and 1220 litres of milk each week at Rs. 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17/litre?
Solution:
Assuming L (litres) along x-axis and R(rupees) along y-axis, we have two points (980,14) and (1220,16).
By two point form, the point (L, R) satisfies the equation.

Ex 10.2 Class 11 Maths Question 18.
P(a, b) is the mid-point of a lone segment between axes. Show that equation of the line is .
Solution:
Let the line AB makes intercepts c and d on the x-axis and y-axis respectively.

Ex 10.2 Class 11 Maths Question 19.
Point R(h, k) divides a line segment between the axes in the ratio 1:2. Find equation of the line.
Solution:
Let AB be the given line segment making intercepts a and b on the x-axis & y-axis respectively.
Then, the equation of line AB is 

So, these points are A(a, 0) and B(0, b).
Now, R(h,k) divides the line segment Ab in the ratio 1 : 2.

Ex 10.2 Class 11 Maths Question 20.
By using the concept of equation of a line, prove that the three points (3, 0), (- 2, – 2) and (8, 2) are collinear.
Solution:
Let the given points be A(3, 0), B(-2, -2) and C(8, 2). Then the equation of the line passing through A and B is

Clearly the point C(8, 2) satisfy the equation 2x – 5y – 6 = 0.
(∵ 2(8) – 5(2) – 6 = 16 – 10 – 6 = 0)
Hence, the given points lie on the same straight line whose equation is 2x – 5y – 6 = 0.

We hope the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2, drop a comment below and we will get back to you at the earliest.

Ex 10.3 Class 11 Maths Question 1.
Reduce the following equations into slope- intercept form and find their slopes and the y-intercepts.
(i) x + 7y = 0,
(ii) 6x + 3y-5 = 0,
(iii) y=0
Solution:
(i) We have given an equation x + 7y = 0, which can be written in the form
⇒ 7y = – x ⇒ y = x + 0 … (1)
Also, the slope intercept form is y-mx + c …(2)
On comparing (1) and (2), we get
m = , c = 0
Hence the slope is  and the y-intercept = 0.

(ii) We have given an equation 6x + 3y – 5 = 0, which can be written in the form 3y = – 6x + 5
⇒ y = – 2x +  …(1)
Also, the slope intercept form is y = mx + c … (2)
On comparing (1) and (2), we get
m = – 2 and c = 
i.e. slope = – 2 and the y-intercept = 

(iii) We have given an equation y = 0
y = 0·x + 0 … (1)
Also, the slope intercept form is y = mx + c … (2) On comparing (1) and (2), we get
m = 0, c = 0.
Hence, slope is 0 and the y-intercept is 0.

Ex 10.3 Class 11 Maths Question 2.
Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0,
(ii) 4x – 3y = 6,
(iii) 3y + 2 = 0
Solution:
(i) Given equation is 3x + 2y – 12 = 0 We have to reduce the given equation into intercept form, i.e.,  …(1)
Now given, 3x + 2y = 12
⇒  ⇒  …(2)
On comparing (1) and (2), we get a = 4, b = 6 Hence, the intercepts of the line are 4 and 6.

(ii) Given equation is 4x – 3y = 6
We have to reduce the given equation into intercept form, i.e.,  …(1)
 …(2)
On comparing (1) and (2), we get
a =  and b = – 2
Hence, the intercepts of the line are  and -2.

(iii) Given equation is 3y + 2 = 0
We have to reduce the given equation into intercept form, i.e., 
3y = -2
⇒ y = 
The above equation shows that, it is not the required equation of the intercept form as it is parallel to x-axis.
We observe that y-intercept of the line is , but there is no intercept on x-axis.

Ex 10.3 Class 11 Maths Question 3.
Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.
(i) x –  + 8 = 0,
(ii) y-2 = 0,
(iii) x-y = 4.
Solution:
(i) Given equation is x –  + 8 = 0
 x –  = -8
 -x +  = 8 … (i)
Also, 

Now dividing both the sides of (1) by 2, we get

⇒ – cos 60°x + sin 60° y = 4.
⇒ {cos (180° – 60°)) x + {sin (180° – 60°)|y = 4
⇒ cos 120° x + sin 120° y = 4
∴ x cos 120° + y sin 120° = 4 is the required equation in normal form
∵ The normal form is x coso) + y sin⍵ = p
So, ⍵ = 120° and p = 4
⍵ Distance of the line from origin is 4 and the angle between perpendicular and positive x-axis is 120°.

(ii) Given equation is y – 2 = 0
⇒ y = 2
⇒ 0 · x + l · y = 2
⇒ x cos 90° + y sin 90° = 2 is the required equation in normal form
∵ The normal form is x cos⍵ + y sin⍵ = p
So, ⍵ = 90° and p = 2
⍵ Distance of the line from origin is 2 and the angle between perpendicular and positive x-axis is 90°.

(iii) Given equation is x – y = 4 … (1)
Also 

Now dividing both the sides of (1) bt , we get

is the required equation in normal form.
∵ The normal form is x cos⍵ + y sin⍵ = p
So, P =  and ⍵ = 315°
∴ Distance of the line from the origin is  and the angle between perpendicular and the positive x-axis is 315°.

Ex 10.3 Class 11 Maths Question 4.
Find the distance of the point (-1, 1) from the line 12 (x+ 6) = 5(y — 2).
Solution:
The equation of line is 12(x + 6) = 5(y – 2) …(i)
⇒ 12x + 72 = 5y-10
⇒ 12x – 5y + 82 = 0
∴ Distance of the point (-1, 1) from the line (i)

Ex 10.3 Class 11 Maths Question 5.
Find the points on the x-axis, whose distances from the line  are 4 units.
Solution:
We have a equation of line , which can be written as
4x + 3y – 12 = 0 … (i)
Let (a, 0) be the point on x-axis whose distance from line (i) is 4 units.

Ex 10.3 Class 11 Maths Question 6.
Find the distance between parallel lines
(i) 15x+8y-34 = 0and 15x + 8y+31 =0
(ii) |(x + y) + p = 0 and |(x + y) – r = 0.
Solution:
If lines are Ax + By + Q = 0
and Ax + By + C₂ = 0, then distance between

Ex 10.3 Class 11 Maths Question 7.
Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).
Solution:
We have given an equation of line 3x – 4y + 2 = 0
Slop of the line(i) = 
Thus, slope of any line parallel to the given line (i) is  and passes through (-2, 3), then its equation is

Ex 10.3 Class 11 Maths Question 8.
Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
Solution:
Given equation is x – 7y + 5 = 0 … (i)
Slope of this line = 
∴ Slope of any line perpendicular to the line (i) is -7 and passes through (3, 0) then
y – 0 = -7(x – 3)
[∵ Product of slope of perpendicular lines is -1]
⇒ y = -7x + 21
⇒ 7x + y – 21 = 0, is the required equation of line.

Ex 10.3 Class 11 Maths Question 9.
Find angles between the lines  + y = 1 and x +  = 1.
Solution:
The given equations are
 + y = 1 … (i)
x +  = 1 … (ii)
Since we have to find an angle between the two lines i.e., firstly we have to find the slopes of (i) and (ii).

Ex 10.3 Class 11 Maths Question 10.
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.
Solution:
Given points are (h, 3) and (4,1).
∴ Slope of the line joining (h, 3) & (4,1)

Ex 10.3 Class 11 Maths Question 11.
Prove that the line through the point (x₁ y₁) and parallel to the line Ax + By + C = 0 is A(x-x₁) + B(y-y₁) = 0.
Solution:
Given equation of a line is Ax + By + C = 0
∴ Slope of the above line = 
i.e. slope of any line parallel to given line and passing through (x₁, y₁) is 
Then equation is (y – y₂) =  (x – x₁)
=> B(y – y₁) = -A(x – x₁)
=> A(x – x₁) + B(y – y₁) = 0.
Hence proved.

Ex 10.3 Class 11 Maths Question 12.
Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.
Solution:
We have given a point (2, 3), through which two lines are passing and intersects at an angle of 60°.
Let m be the slope of the other line

Ex 10.3 Class 11 Maths Question 13.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
Solution:
suppose the given points are A and B.
Let M be the mid point of AB.

Ex 10.3 Class 11 Maths Question 14.
Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.
Solution:
We have, 3x – 4y – 16 = 0
Slope of the kine(i) = 
Then equation of any line ⊥ from (-1, 3) to the given line(i) is

Ex 10.3 Class 11 Maths Question 15.
The perpendicular from the origin to the line y = mx + c meets it at the point (-1,2). Find the values of m and c.
Solution:
Given, the perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2)
∴ 2 = m (-1) + c … (i)
⇒ c – m = 2

Ex 10.3 Class 11 Maths Question 16.
If p and q are the lengths of perpendiculars from the origin to the lines x cosθ – y sinθ = k cos 2θ and x secθ + y cosecθ = k, respectively, prove that p² + 4q² = k².
Solution:
Given p and q are the lengths of perpendiculars from the origin to the lines x cos θ – ysinθ=k cos 2θ and xsecθ+y cosec θ = k.

Ex 10.3 Class 11 Maths Question 17.
In the triangle ABC with vertices A(2, 3), 8(4, -1) and C( 1, 2), find the equation and length of altitude from the vertex A.
Solution:
We have given a AABC with the vertices, A (2, 3), B (4, -1) and C (1, 2)

Ex 10.3 Class 11 Maths Question 18.
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that .
Solution:
Given, p be the length of perpendicular from the origin to the line whose intercepts

We hope the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3, drop a comment below and we will get back to you at the earliest.

Ex 9.1 Class 11 Maths Question 1.
a_n = n(n + 2)
Solution:
We haven a_n = n(n + 2)
subtituting n = 1, 2, 3, 4, 5, we get
a₁ = 9(1 + 2) = 1 x 3 = 3
a₂ = 2(2 + ) = 2 x 4 = 8
a₃ = 3(3 + 2) = 3 x 5 = 15
a₄ = 4(4 + 2) = 4 x 6 = 24
a₅ = 5(5 + 2) = 5 x 7 = 35
∴ The first five terms are 3, 8, 15, 24, 35.

Ex 9.1 Class 11 Maths Question 2.
a_n = 
Solution:

Ex 9.1 Class 11 Maths Question 3.
a_n = 2ⁿ
Solution:

Ex 9.1 Class 11 Maths Question 4.
a_n = 
Solution:

Ex 9.1 Class 11 Maths Question 5.
a_n = (- 1)^n-1 5ⁿ⁺¹
Solution:
We have, a_n = (- 1)^n-1 5ⁿ⁺¹
Substituting n = 1, 2, 3, 4, 5, we get
a₁ =(-1)^1-1 5¹⁺¹ = (-1)^° 5² = 25
a₂ =(-1)^2-1 5²⁺¹ = (-1)¹ 5³ = 125
a₃ =(-1)^3-1 5³⁺¹ = (-1)² 5⁴ = 625
a₄ =(-1)^4-1 5⁴⁺¹ = (-1)³ 5⁵ = -3125
a₅ =(-1)^5-1 5⁵⁺¹ = (-1)⁴ 5⁶ = 15625
∴ The first five terms are 25, – 125, 625, -3125, 15625.

Ex 9.1 Class 11 Maths Question 6.
a_n = n
Solution:
Substituting n = 1, 2, 3, 4, 5, we get


Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

Ex 9.1 Class 11 Maths Question 7.
a_n = 4n – 3; a₁₇, a₂₄
Solution:
We have a_n = 4n – 3

Ex 9.1 Class 11 Maths Question 8.
a_n = ; a₇
Solution:
We have, a_n = ; a₇

Ex 9.1 Class 11 Maths Question 9.
a_n = (-1)^{n – 1} n³; a₉
Solution:
We have, a_n = (-1)^{n – 1} n³

Ex 9.1 Class 11 Maths Question 10.
a_n = ; a ₂₀
Solution:
We have, a_n = 

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

Ex 9.1 Class 11 Maths Question 11.
a₁ = 3, a_n = 3a_n-1+2 for all n>1
Solution:
We have given a₁ = 3, a_n = 3a_n-1+2
⇒ a₁ = 3, a₂ = 3a₁ + 2 = 3.3 + 2 = 9 + 2 = 11,
a₃ = 3a₂ + 2 = 3.11 + 2 = 33 + 2 = 35,
a₄ = 3a₃ + 2 = 3.35 + 2 = 105 + 2 = 107,
a₅ = 3a₄ + 2 = 3.107 + 2 = 321 + 2 = 323,
Hence, the first five terms of the sequence are 3, 11, 35, 107, 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + ………..

Ex 9.1 Class 11 Maths Question 12.
a₁ = -1, a_n = , n ≥ 2
Solution:
We have given

Hence the first five terms of the given sequence are – 1, -1/2, -1/6, -1/24, -1/120.
The corresponding series is

Ex 9.1 Class 11 Maths Question 13.
a₁ = a₂ = 2, a_n = a_n-1 – 1, n >2
Solution:
We have given a₁ = a₂ = 2, a_n = a_n-1 – 1, n >2
a₁ = 2, a₂ = 2, a₃= a₂ – 1 = 2 – 1 = 1,
a₄ = a₃ – 1 = 1 – 1 = 0 and a₅ = a₄ – 1 = 0 – 1 = -1
Hence the first five terms of the sequence are 2, 2, 1, 0, -1
The corresponding series is
2 + 2 + 1 + 0 + (-1) + ……

Ex 9.1 Class 11 Maths Question 14.
Find Fibonacci sequence is defined by 1 = a₁ = a₂ and a_n = a_n-1 + a_n-2, n > 2
Find , for n = 1, 2, 3, 4, 5
Solution:
We have,

We hope the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1, drop a comment below and we will get back to you at the earliest.

Ex 9.2 Class 11 Maths Question 1.
Find the sum of odd integers from 1 to 2001.
Solution:
We have to find 1 + 3 + 5 + ……….. + 2001
This is an A.P. with first term a = 1, common difference d = 3-1 = 2 and last term l = 2001
∴ l = a + (n-1 )d ⇒ 2001 = 1 + (n -1)2 ⇒ 2001 = 1 + 2n – 2 ⇒ 2n = 2001 + 1

Ex 9.2 Class 11 Maths Question 2.
Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Solution:
We have to find 105 +110 +115 + ……..+ 995
This is an A.P. with first term a = 105, common difference d = 110 -105 = 5 and last term 1 = 995

Ex 9.2 Class 11 Maths Question 3.
In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is -112.
Solution:
Let a = 2 be the first term and d be the common difference.

Ex 9.2 Class 11 Maths Question 4.
How many terms of the A.P. -6, , -5, ……. are needed to give the sum – 25 ?
Solution:
Let a be the first term and d be the common difference of the given A.P., we have

Ex 9.2 Class 11 Maths Question 5.
In an A.P., if pth term is  and qth term is  prove that the sum of first pq terms is , where p±q.
Solution:
Let a be the first term & d be the common difference of the A.P., then

Ex 9.2 Class 11 Maths Question 6.
If the sum of a certain number of terms of the A.P. 25, 22, 19, …. is 116. Find the last term.
Solution:
Let a be the first term and d be the common difference.
We have a = 25, d = 22 – 25 = -3, S_n = 116

Ex 9.2 Class 11 Maths Question 7.
Find the sum to n terms of the A.P., whose k^th term is 5k + 1.
Solution:
We have a_k = 5k +1
By substituting the value of k = 1, 2, 3 and 4,
we get

Ex 9.2 Class 11 Maths Question 8.
If the sum of n terms of an A.P. is (pn + qn²), where p and q are constants, find the common difference.
Solution:
We have S_n = pn + qn², where S„ be the sum of n terms.

Ex 9.2 Class 11 Maths Question 9.
The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18^th terms.
Solution:
Let a₁, a₂ & d₁ d₂ be the first terms & common differences of the two arithmetic progressions respectively. According to the given condition, we have

Ex 9.2 Class 11 Maths Question 10.
If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
Solution:
Let the first term be a and common difference be d.
According to question

Ex 9.2 Class 11 Maths Question 11.
Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that
Solution:
Let the first term be A & common difference be D. We have

Ex 9.2 Class 11 Maths Question 12.
The ratio of the sums of m and n terms of an A.P. is m²: n². Show that the ratio of m^th and n^th term is (2m -1): (2n -1).
Solution:
Let the first term be a & common difference be d. Then

Ex 9.2 Class 11 Maths Question 13.
If the sum of n terms of an A.P. is 3n² + 5n and its mth term is 164, find the value of m.
Solution:
We have S_n = 3n² + 5n, where Sn be the sum of n terms.

Ex 9.2 Class 11 Maths Question 14.
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Solution:
Let A₁, A₂, A₃, A₄, A₅ be numbers between 8 and 26 such that 8, A₁, A₂, A₃, A₄, A₅, 26 are in A.P., Here a = 8, l = 26, n = 7

Ex 9.2 Class 11 Maths Question 15.
If  is the A.M. between a and b, then find the value of n.
Solution:
We have 

Ex 9.2 Class 11 Maths Question 16.
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7^thand (m – 1 )^th numbers is 5:9. Find the value of m.
Solution:
Let the sequence be 1, A₁, A₂, ……… A_m, 31 Then 31 is (m + 2)^th term, a = 1, let d be the common difference

Ex 9.2 Class 11 Maths Question 17.
A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs. 5 every month, what amount he will pay in the 30th instalment?
Solution:
Here, we have an A.P. with a = 100 and d = 5
∴ a_n= a + 29d = 100 + 29(5) = 100 + 145 = 245
Hence he will pay Rs. 245 in 30^th instalment.

Ex 9.2 Class 11 Maths Question 18.
The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.
Solution:
Let there are n sides of a polygon.

We hope the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2, drop a comment below and we will get back to you at the earliest.

Ex 9.3 Class 11 Maths Question 1.
Find the 20^th and n^th terms of the G.P. , ….
Solution:

Ex 9.3 Class 11 Maths Question 2.
Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.
Solution:
We have, a_s = 192, r = 2

Ex 9.3 Class 11 Maths Question 3.
The 5^th, 8^th and 11th terms of a G.P. are p, q and s, respectively. Show that q² = ps.
Solution:
We are given

Ex 9.3 Class 11 Maths Question 4.
The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7^th term.
Solution:
We have a= -3, a₄ = (a₂)²
Ex 9.3 Class 11 Maths Question 5.
Which term of the following sequences:


Solution:

Ex 9.3 Class 11 Maths Question 6.
For what values of x, the numbers , x,  are in G.P.?
Solution:
Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

Ex 9.3 Class 11 Maths Question 7.
0.14, 0.015, 0.0015, …. 20 items.
Solution:
In the given G.P.

Ex 9.3 Class 11 Maths Question 8.
, …. n terms
Solution:
In the given G.P.

Ex 9.3 Class 11 Maths Question 9.
1, -a, a²,- a³ … n terms (if a ≠ -1)
Solution:
In the given G.P.. a = 1, r = -a

Ex 9.3 Class 11 Maths Question 10.
x³, x⁵, ⁷, ….. n terms (if ≠±1).
Solution:
In the given G.P., a = x³, r = x²

Ex 9.3 Class 11 Maths Question 11.
Evaluate
Solution:

Ex 9.3 Class 11 Maths Question 12.
The sum of first three terms of a G.P. is  and 10 their product is 1. Find the common ratio and the terms.
Solution:
Let the first three terms of G.P. be ,a,ar, where a is the first term and r is the common ratio.

Ex 9.3 Class 11 Maths Question 13.
How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?
Solution:
Let n be the number of terms we needed. Here a = 3, r = 3, S_n = 120

Ex 9.3 Class 11 Maths Question 14.
The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Solution:
Let a₁, a₂, a₃, a₄, a₅, a₆ be the first six terms of the G.P.

Ex 9.3 Class 11 Maths Question 15.
Given a G.P. with a = 729 and 7^th term 64, determine S₇.
Solution:
Let a be the first term and the common ratio be r.

Ex 9.3 Class 11 Maths Question 16.
Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
Solution:
Let a₁ a₂ be first two terms and a₃ a₅ be third and fifth terms respectively.
According to question

Ex 9.3 Class 11 Maths Question 17.
If the 4^th, 10^th and 16^th terms of a G.P. are x, y and z respectively. Prove that x, y, z are in G.P.
Solution:
Let a be the first term and r be the common ratio, then according to question

Ex 9.3 Class 11 Maths Question 18.
Find the sum to n terms of the sequence, 8, 88, 888, 8888 ………
Solution:
This is not a G.P., however we can relate it to a G.P. by writing the terms as S_n= 8 +88 + 888 + 8888 + to n terms

Ex 9.3 Class 11 Maths Question 19.
Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 
Solution:
On multiplying the corresponding terms of sequences, we get 256, 128, 64, 32 and 16, which forms a G.P. of 5 terms

Ex 9.3 Class 11 Maths Question 20.
Show that the products of the corresponding terms of the sequences a, ar, ar², ………… ar^n-1 and A, AR, AR², …….. , AR^n-1 form a G.P., and find the common ratio.
Solution:
On multiplying the corresponding terms, we get aA, aArR, aAr²R²,…… aAr^n-1R^n-1. We can see that this new sequence is G.P. with first term aA & the common ratio rR.

Ex 9.3 Class 11 Maths Question 21.
Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4^th by 18.
Solution:
Let the four numbers forming a G.P. be a, ar, ar², ar³
According to question,

Ex 9.3 Class 11 Maths Question 22.
If the p^th, q^th and r^th terms of a G.P. are a, b and c, respectively. Prove that a^q-r b^r-p c^p-q = 1.
Solution:
Let A be the first term and R be the common ratio, then according to question

Ex 9.3 Class 11 Maths Question 23.
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P² – (ab)ⁿ.
Solution:
Let r be the common ratio of the given G.P., then b = n^th term = ar^n-1

Ex 9.3 Class 11 Maths Question 24.
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)^th to (2n)^th term is 
Solution:
Let the G.P. be a, ar, ar², ……
Sum of first n terms = a + ar + ……. + ar^n-1

Ex 9.3 Class 11 Maths Question 25.
If a, b,c and d are in G.P., show that
(a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²
Solution:
We have a, b, c, d are in G.P.
Let r be a common ratio, then

Ex 9.3 Class 11 Maths Question 26.
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution:
Let G₁, G₂ be two numbers between 3 and 81 such that 3, G₁ G₂,81 is a G.P.

Ex 9.3 Class 11 Maths Question 27.
Find the value of n so that  may be the geometric mean between a and b.
Solution:

Ex 9.3 Class 11 Maths Question 28.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio .
Solution:
Let a and b be the two numbers such that a + b = 6 

Ex 9.3 Class 11 Maths Question 29.
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are .
Solution:
Let a and b be the numbers such that A, G are A.M. and G.M. respectively between them.

Ex 9.3 Class 11 Maths Question 30.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?
Solution:
There were 30 bacteria present in the culture originally and it doubles every hour. So, the number of bacteria at the end of successive hours form the G.P. i.e., 30, 60, 120, 240, …….

Ex 9.3 Class 11 Maths Question 31.
What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Solution:
We have, Principal value = Rs. 500 Interest rate = 10% annually

Ex 9.3 Class 11 Maths Question 32.
If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Solution:
Let α & β be the roots of a quadratic equation such that A.M. & G.M. of α, β are 8 and 5 respectively.

We hope the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3, drop a comment below and we will get back to you at the earliest.

Find the sunt to n terms of each of the series in Exercises 1 to 7.

Ex 9.4 Class 11 Maths Question 1.
1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ………
Solution:
In the given series, there is a sum of multiple of corresponding terms of two A.P’s. The two A.P’s are

Ex 9.4 Class 11 Maths Question 2.
1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + ……
Solution:
In the given series, there is a sum of multiple of corresponding terms of two A.P’s. The three A.P’s are

Ex 9.4 Class 11 Maths Question 3.
3 x 1² + 5 x 2² + 7 x 3² + …..
Solution:
In the given series there is sum of multiple of corresponding terms of two A.P’s. The two A.P’s are
(i) 3, 5, 7, …………… and
(ii) 1², 2², 3², ………………….
Now the n^th term of sum is an = (nth term of the sequence formed by first A.P.) x (n^th term of the sequence formed by second A.P.) = (2 n + 1) x n² = 2n³ + n² Hence, the sum to n terms is,

Ex 9.4 Class 11 Maths Question 4.
 …….
Solution:
In the given series there is sum of multiple of corresponding terms of two A.P’s. The two A.P’s are

Ex 9.4 Class 11 Maths Question 5.
5² + 6² + 7² + ………….. + 20²
Solution:
The given series can be written in the following way

Ex 9.4 Class 11 Maths Question 6.
3 x 8 + 6 x 11 + 9 x 25 + ………….
Solution:
In the given series, there is sum of multiple of corresponding terms of two A.P/s. The two A.P/s are
(i) 3, 6, 9, ………….. and
(ii) 8, 11, 14, ……………….
Now the nth term of sum is an = (n^th term of the sequence formed by first A.P.) x (n^th term of the sequence formed by second A.P.)

Ex 9.4 Class 11 Maths Question 7.
1² + (1² + 2²) + (1² + 2² + 3²) + ………….
Solution:
In the given series
a_n = 1² + 2² + …………….. + n²

Find the sum to n terms of the series in Exercises 8 to 10 whose n^th terms is given by

Ex 9.4 Class 11 Maths Question 8.
n(n + 1)(n + 4)
Solution:
We have

Ex 9.4 Class 11 Maths Question 9.
n² + 2ⁿ
Solution:
We have a_n = n² + 2ⁿ
Hence, the sum to n terms is,

Ex 9.4 Class 11 Maths Question 10.
(2n – 1)²
Solution:
We have

We hope the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4, drop a comment below and we will get back to you at the earliest.

Expand each of the expressions in Exercises 1 to 5.
Ex 8.1 Class 11 Maths Question 1.

Solution.

Ex 8.1 Class 11 Maths Question 2.

Solution.

Ex 8.1 Class 11 Maths Question 3.

Solution.

Ex 8.1 Class 11 Maths Question 4.

Solution.

Ex 8.1 Class 11 Maths Question 5.

Solution.

Using binomial theorem, evaluate each of the following
Ex 8.1 Class 11 Maths Question 6.

Solution.

Ex 8.1 Class 11 Maths Question 7.

Solution.

Ex 8.1 Class 11 Maths Question 8.

Solution.

Ex 8.1 Class 11 Maths Question 9.

Solution.

Ex 8.1 Class 11 Maths Question 10.
Using Binomial Theorem, indicate which number is larger or 1000.
Solution.
Splitting 1.1 and using binomial theorem to write the first few terms we have

Ex 8.1 Class 11 Maths Question 11.
Find . Hence, evaluate .
Solution.
By binomial theorem, we have

Ex 8.1 Class 11 Maths Question 12.
Find . Hence or otherwise evaluate .
Solution.
By using binomial theorem, we have

Ex 8.1 Class 11 Maths Question 13.
Show that  is divisible by 64, whenever n is a positive integer.
Solution.
We have to prove that 

Ex 8.1 Class 11 Maths Question 14.
Prove that  ⁸C_{r = 4ⁿ
Solution.
We have,}

We hope the NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1, drop a comment below and we will get back to you at the earliest.

Ex 8.2 Class 11 Maths Question 1.
Find the coefficient of x⁵ in (x + 3)⁸
Solution.
Suppose x⁵ occurs in the (r + 1)^th term of the expansion (x + 3)⁸

Ex 8.2 Class 11 Maths Question 2.
a⁵ b⁷in (a-2b)¹²
Solution.
Suppose a⁵ b⁷ occurs in the (r + 1)^th term of the expansion (a – 2b)¹².

Write the general term in the expansion of
Ex 8.2 Class 11 Maths Question 3.
(x² – y)⁶
Solution.

Ex 8.2 Class 11 Maths Question 4.
(x² – yx)¹², x ≠ 0
Solution.
We have given, (x² – yx)¹² = (x² + (- yx))¹², x ≠ 0

Ex 8.2 Class 11 Maths Question 5.
Find the 4th term in the expansion of (x – 2y)¹².
Solution.

Ex 8.2 Class 11 Maths Question 6.
Find the 13th term in the expansion of , x ≠ 0
Solution.

Find the middle terms in the expansions of
Ex 8.2 Class 11 Maths Question 7.

Solution.
As the exponent 7 is odd, so there will be two middle terms in the expansion

Ex 8.2 Class 11 Maths Question 8.

Solution.

Ex 8.2 Class 11 Maths Question 9.
In the expansion of (1 + a)^m+n, prove that coefficients of a^m and aⁿ are equal.
Solution.

Ex 8.2 Class 11 Maths Question 10.
The coefficients of the (r – 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1 )ⁿ are in the ratio 1: 3: 5. Find n and r.
Solution.

Ex 8.2 Class 11 Maths Question 11.
Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^2n-1.
Solution.

Ex 8.2 Class 11 Maths Question 12.
Find a positive value of m for which the coefficient of x² in the expansion (1 + x)^m is 6.
Solution.

We hope the NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2, drop a comment below and we will get back to you at the earliest.

Ex 7.1 Class 11 Maths Question 1.
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
Solution.
There will be as many ways as there are ways of filling 3 vacant  places in succession by the five given digits.
(i) When repetition is allowed then each place can be filled in five different ways. Therefore, by the multiplication principle the required number of 3- digit numbers is 5 x 5 x 5 i.e., 125.
(ii) When repetition is not allowed then first place can be filled in 5 different ways, second place can be filled in 4 different ways & third place can be filled in 3 different ways. Therefore by the multiplication principle the required number of three digit numbers is 5 x 4 x 3 i.e, 60.

Ex 7.1 Class 11 Maths Question 2.
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Solution.
There will be as many ways as there are ways of filling 3 vacant places  in succession by the 6 given digits. In this case we start filling in unit’s place, because the options for this place are 2, 4 & 6 only and this can be done in 3 ways. Ten’s and hundred’s place can be filled in 6 different ways. Therefore, by the multiplication principle, the required number of 3-digit even numbers is 6 x 6 x 3 i.e., 108.

Ex 7.1 Class 11 Maths Question 3.
How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Solution.
There will be as many ways as there are ways of filling 4 vacant places  in succession by the first 10 letters of the English alphabet, when repetition is not allowed then first place can be filled in 10 different ways, second place can be filled in 9 different ways, third place can be filled in 8 different ways and fourth place can be filled in 7 different ways. Therefore, by the multiplication principle the required number of 4 letter codes are 10 x 9 x 8 x 7 i.e., 5040.

Ex 7.1 Class 11 Maths Question 4.
How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Solution.
The 5 digit telephone numbers of the form   can be constructed using the digits 0 to 9. When repetition is not allowed then at first & second place 6 & 7 are fixed respectively. Therefore, third, fourth and fifth place can be filled in 8, 7 and 6 ways respectively. So, by the multiplication principle the required numbers of 5-digit telephone numbers is 8 x 7 x 6 i.e., 336.

Ex 7.1 Class 11 Maths Question 5.
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Solution.
When a coin is tossed there are two possible outcomes i.e. head or tail. When the coin is tossed three times then the total possible outcomes are 2 x 2 x 2 i.e., 8.

Ex 7.1 Class 11 Maths Question 6.
Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Solution.
There will be as many signals as there are ways of filling in 2 vacant places  in succession by the 5 flags of different colours. The upper vacant place can be filled in 5 different ways by anyone of the 5 flags ; following which the lower vacant place can be filled in 4 different ways by anyone of the remaining 4 different flags. Hence by the multiplication principle the required number of signals is 5 x 4 = 20.

We hope the NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.1, drop a comment below and we will get back to you at the earliest.

Ex 7.2 Class 11 Maths Question 1.
Evaluate
(i) 8!
(ii) 4!-3!
Solution.
(i) 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320
(ii) 4! – 3! = (1 x 2 x 3 x 4) – (1 x 2 x 3) = 24 – 6 = 18

Ex 7.2 Class 11 Maths Question 2.
Is 3! + 4! = 7! ?
Solution.
3! + 4! = (1 x 2 x 3) + (1 x 2 x 3 x 4) = 6 + 24 = 30 … (i)
7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 ………(ii)
From (i) & (ii), we get 3! + 4! ≠ 7!.

Ex 7.2 Class 11 Maths Question 3.

Solution.

Ex 7.2 Class 11 Maths Question 4.
, findx.
Solution.
We have, 

Ex 7.2 Class 11 Maths Question 5.
Evaluate , when
(i) n = 6, r = 2
(ii) n = 9, r = 5
Solution.
(i) 

(ii) 

We hope the NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.2, drop a comment below and we will get back to you at the earliest.

Ex 7.3 Class 11 Maths Question 1.
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Solution.
Total digits are 9. We have to form 3 digit numbers without repetition.
∴ The required 3 digit numbers = ⁹P₃

Ex 7.3 Class 11 Maths Question 2.
How many 4-digit numbers are there with no digit repeated?
Solution.
The 4-digit numbers are formed from digits 0 to 9. In four digit numbers 0 is not taken at thousand’s place, so thousand’s place can be filled in 9 different ways. After filling thousand’s place, 9 digits are left. The remaining three places can be filled in 9P3 ways.
So the required 4-digit numbers
= 9 x ⁹P₃
= 9 x 504 = 4536.

Ex 7.3 Class 11 Maths Question 3.
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Solution.
For 3-digit even numbers unit place can be filled by 2, 4, 6 i.e in 3 ways. Then the remaining two places can be filled in ⁵P₂ ways.
∴ The required 3-digit even numbers
= 3 x ⁵P₂
= 60

Ex 7.3 Class 11 Maths Question 4.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Solution.
The 4-digit numbers can be formed from digits 1 to 5 in ⁵P₄ways.
∴ The required 4 digit numbers = ⁵P₄ = 120 For 4-digit even numbers unit place can be filled by 2,4, i.e., in 2 ways. Then the remaining three places can be filled in ⁴P₃ ways.
∴ The required 4-digit even numbers
= 2 x ⁴P₃ = 2 x 24 = 48

Ex 7.3 Class 11 Maths Question 5.
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
Solution.
From a committee of 8 persons, we can choose a chairman and a vice chairman

Ex 7.3 Class 11 Maths Question 6.
Find n if  ^n-1P₃: ⁿP₄ = 1 : 9.
Solution.

Ex 7.3 Class 11 Maths Question 7.
Find r if
(i) ⁵P_r = 2⁶P_r-1
(ii) ⁵P_r = ⁶P_r-1
Solution.

Ex 7.3 Class 11 Maths Question 8.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Solution.
No. of letters in the word EQUATION = 8
∴ No. of words that can be formed
= ⁸P₈ = 8!
=40320

Ex 7.3 Class 11 Maths Question 9.
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
Solution.
No. of letters in the word MONDAY = 6
(i) When 4 letters are used at a time.
Then, the required number of words
= ⁶P₄

(ii) When all letters are used at a time. Then the required number of words
= ⁶P₆ = 6!
= 720

(iii) All letters are used but first letter is a vowel.
So the first letter can be either A or O.
So there are 2 ways to fill the first letter & remaining places can be filled in ⁵P₅ ways.
∴ The required number of words
= 2 x ⁵P₅
= 2 x 5! =240.

Ex 7.3 Class 11 Maths Question 10.
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Solution.
There are 11 letters, of which I appears 4 times, S appears 4 times, P appears 2 times & M appears 1 time.
∴ The required number of arrangements

= 10 x 10 x 9 x 7 x 5 = 34650                       … (i)
When four I’s come together, we treat them as a single object. This single object with 7 remaining objects will account for 8 objects. These 8 objects in which there are 4S’s & 2P’s
can be rearranged in  ways i.e. in 840 ways      … (ii)
Number of arrangements when four I’s do not come together = 34650 – 840 = 33810.

Ex 7.3 Class 11 Maths Question 11.
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S,
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?
Solution.
There are 12 letters of which T appears 2 times
(i) When words start with P and end with S, then there are 10 letters to be arranged of which T appears 2 times.
∴ The required words = 

(ii) When vowels are taken together i.e E U A I O we treat them as a single object. This single object with remaining 7 objects will account for 8 objects, in which there w are 2Ts, which can be rearranged in  ways. Corresponding to each of these arrangements the 5 vowels E, U, A, I, O can be rearranged in 5! = 120 ways. Therefore, by multiplication principle, the required number of arrangements = 20160 x 120 = 2419200.

(iii) When there are always 4 letters between P & S
∴ P & S can be at
1^st & 6^th place
2^nd & 7^th place
3^rd& 8^th place
4^th & 9^th place
5^th & 10^th place
6^th & 11^th place
7^th & 12^th place.
So, P & S will be placed in 7 ways & can be arranged in 7 x 2! = 14
The remaining 10 letters with 2T’s, can be arranged in  ways.
∴ The required number of arrangements = 14 x 1814400= 25401600.

We hope the NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3, drop a comment below and we will get back to you at the earliest.

Ex 7.4 Class 11 Maths Question 1.
lf ⁿC₈ = ⁿC₂, find ⁿC₂.
Solution.
We have, ⁿC₈ = ⁿC₂

Ex 7.4 Class 11 Maths Question 2.
Determine n if
(i) ²ⁿC₃: ⁿC₃ =12 : 1
(ii) ²ⁿC₃: ⁿC₃= 11 : 1
Solution.

Ex 7.4 Class 11 Maths Question 3.
How many chords can be drawn through 21 points on a circle?
Solution.
A chord is formed by joining two points on a circle.
∴ Required number of chords = ²ⁿC₂

Ex 7.4 Class 11 Maths Question 4.
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Solution.
3 boys can be selected from 5 boys in ⁵C₃ ways & 3 girls can be selected from 4 girls in ⁴C₃ ways.
∴ Required number of ways of team selection = ⁵C₃ x ⁴C₃ = 

Ex 7.4 Class 11 Maths Question 5.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution.
No. of ways of selecting 3 red balls =⁶C₃
No. of ways of selecting 3 white balls = ⁵C₃
No. of ways of selecting 3 blue balls = ⁵C₃
∴ Required no. of ways of selecting 9 balls

Ex 7.4 Class 11 Maths Question 6.
Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Solution.
Total no. of cards = 52
No. of ace cards = 4
No. of non-ace cards = 48
∴ One ace card out of 4 can be selected in ⁴C₁ ways.
Remaining 4 cards out of 48 cards can be selected in ⁴⁸C₄ways.
∴ Required no. of ways of selecting 5 cards

Ex 7.4 Class 11 Maths Question 7.
In Kbw many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Solution.
Total players = 17, No. of bowlers = 5,
No. of non-bowlers = 12
No. of ways of selecting 4 bowlers = ⁵C₄
No. of ways of selecting 7 non-bowlers = ¹²C₇
∴ Required no. of ways of selecting a cricket team

Ex 7.4 Class 11 Maths Question 8.
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Solution.
No. of ways of selecting 2 black balls = ⁵C₂
No. of ways of selecting 3 red balls = ⁶C₃
∴ Required no. of ways of selecting 2 black & 3 red balls = ⁵C₂ x ⁶C₃

Ex 7.4 Class 11 Maths Question 9.
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Solution.
Total no. of courses = 9
No. of compulsory courses = 2
So, the student will choose 3 courses out of 7 courses [non compulsory courses].
∴ Required no. of ways a student can choose a programme = ⁷C₃ = 

We hope the NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4, drop a comment below and we will get back to you at the earliest.