NCERT Solutions for Class 11 Maths

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3.

• Sequences and Series Class 11 Ex 9.1
• Sequences and Series Class 11 Ex 9.2
• Sequences and Series Class 11 Ex 9.4

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3

Ex 9.3 Class 11 Maths Question 1.
Find the 20^th and n^th terms of the G.P. \(\frac { 5 }{ 2 } ,\frac { 5 }{ 4 } ,\frac { 5 }{ 8 } \), ….
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 \(-\frac { 2 }{ 7 } \), x, \(-\frac { 7 }{ 2 } \) 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.
\(\sqrt { 7 } ,\quad \sqrt { 21 } ,\quad 3\sqrt { 7 } \), …. 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 \(\frac { 39 }{ 10 } \) and 10 their product is 1. Find the common ratio and the terms.
Solution:
Let the first three terms of G.P. be \(\frac { a }{ r } \),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, \(\frac { 1 }{ 2 } \)
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 \(\frac { 1 }{ { r }^{ n } } \)
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 \(\frac { { a }^{ n+1 }+{ b }^{ n+1 } }{ { a }^{ n }+{ b }^{ n } } \) 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 \(\left( 3+2\sqrt { 2 } \right) :\left( 3-2\sqrt { 2 } \right) \).
Solution:
Let a and b be the two numbers such that a + b = 6 \(\sqrt { ab } \)

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 \(A\pm \sqrt { \left( A+G \right) \left( A-G \right) } \).
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.

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2.

• Binomial Theorem Class 11 Ex 8.1

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Binomial Expansion Calculator – Expand binomials using the binomial expansion method step-by-step.

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 \({ \left( 9x-\frac { 1 }{ 3\sqrt { x } } \right) }^{ 18 }\), x ≠ 0
Solution.

Find the middle terms in the expansions of
Ex 8.2 Class 11 Maths Question 7.
\({ \left( 3-\frac { { x }^{ 3 } }{ 6 } \right) }^{ 7 }\)
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.
\({ \left( \frac { x }{ 3 } +9y \right) }^{ 10 }\)
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.

Expanding Binomial Calculator is a free online tool.

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.

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1.

• Linear Inequalities Class 11 Ex 6.2
• Linear Inequalities Class 11 Ex 6.3

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}.

The linear inequality solver she used can give test data that are not certain to produce the required path due to rounding errors.

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, \(\frac { 2x-3x }{ 6 } >1\quad or\quad -\frac { x }{ 6 } >1\)
Multiplying both sides by -6, we get
x < -6
∴ The solution is (- ∞, – 6).

Ex 6.1 Class 11 Maths Question 11.
\(\frac { 3\left( x-2 \right) }{ 5 } \le \frac { 5\left( 2-x \right) }{ 3 } \)
Solution.
The inequality is \(\frac { 3\left( x-2 \right) }{ 5 } \le \frac { 5\left( 2-x \right) }{ 3 } \)
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.
\(\frac { 1 }{ 2 } \left( \frac { 3x }{ 5 } +4 \right) \ge \frac { 1 }{ 3 } \left( x-6 \right) \)
Solution.
The inequality is \(\frac { 1 }{ 2 } \left( \frac { 3x }{ 5 } +4 \right) \ge \frac { 1 }{ 3 } \left( x-6 \right) \)
or, \(\frac { 1 }{ 2 } \left( \frac { 3x+20 }{ 5 } \right) \ge \frac { 1 }{ 3 } \left( x-6 \right) \)
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.
\(\frac { x }{ 4 } <\frac { \left( 5x-2 \right) }{ 3 } -\frac { \left( 7x-3 \right) }{ 5 } \)
Solution.
The inequality is \(\frac { x }{ 4 } <\frac { \left( 5x-2 \right) }{ 3 } -\frac { \left( 7x-3 \right) }{ 5 } \)
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.
\(\frac { \left( 2x-1 \right) }{ 3 } \ge \frac { \left( 3x-2 \right) }{ 4 } -\frac { 2-x }{ 5 } \)
Solution.
The inequality is \(\frac { \left( 2x-1 \right) }{ 3 } \ge \frac { \left( 3x-2 \right) }{ 4 } -\frac { 2-x }{ 5 } \)
Multiplying each term by L.C.M. of 3,4, 5, i.e., by 60
\(\frac { \left( 2x-1 \right) }{ 3 } \times 60\ge \frac { \left( 3x-2 \right) }{ 4 } \times 60-\frac { 2-x }{ 5 } \times 60\)
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(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.
\(\frac { x }{ 2 } \ge \frac { \left( 5x-2 \right) }{ 3 } -\frac { \left( 7x-3 \right) }{ 5 } \)
Solution.
The inequality is \(\frac { x }{ 2 } \ge \frac { \left( 5x-2 \right) }{ 3 } -\frac { \left( 7x-3 \right) }{ 5 } \)

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.

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.

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1.

• Conic Sections Class 11 Ex 11.2
• Conic Sections Class 11 Ex 11.3
• Conic Sections Class 11 Ex 11.4

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1

In each of the following Exercises 1 to 5, find the equation of the circle with

Ex 11.1 Class 11 Maths Question 1.
centre (0, 2) and radius 2
Solution:
Here h = 0,k = 2 and r = 2
The equation of circle is,
(x-h)² + (y- k)² = r²
∴ (x – 0)² + (y – 2)² = (2)²
⇒ x² + y² + 4 – 4y = 4
⇒ x² + y² – 4y = 0

Ex 11.1 Class 11 Maths Question 2.
centre (-2,3) and radius 4
Solution:
Here h=-2,k = 3 and r = 4
The equation of circle is,
(x – h)² + (y – k)² = r²
∴(x + 2)² + (y – 3)² = (4)²
⇒ x² + 4 + 4x + y² + 9 – 6y = 16
⇒ x² + y² + 4x – 6y – 3 = 0

Ex 11.1 Class 11 Maths Question 3.
centre \(\left( \frac { 1 }{ 2 } ,\quad \frac { 1 }{ 4 } \right) \) and radius \(\frac { 1 }{ 12 } \)
Solution:
here h = \(\frac { 1 }{ 2 } \), k = \(\frac { 1 }{ 4 } \) and r = \(\frac { 1 }{ 12 } \)
The equation of circle is,

Ex 11.1 Class 11 Maths Question 4.
centre (1, 1) and radius \(\sqrt { 2 } \)
Solution:
Here h = l, k=l and r = \(\sqrt { 2 } \)
The equation of circle is,
(x – h)² + (y – k)² = r²
∴ (x – 1)² + (y – 1)² = \(\left( \sqrt { 2 } \right) \)²
⇒ x² + 1 – 2x + y² +1 – 2y = 2
⇒ x² + y² – 2x – 2y = 0

Ex 11.1 Class 11 Maths Question 5.
centre (-a, -b) and radius \(\sqrt { { a }^{ 2 }-{ b }^{ 2 } } \).
Solution:
Here h=-a, k = -b and r = \(\sqrt { { a }^{ 2 }-{ b }^{ 2 } } \)
The equation of circle is, (x – h)² + (y – k)² = r²
∴ (x + a)² + (y + b)² = \(\left( \sqrt { { a }^{ 2 }-{ b }^{ 2 } } \right) \)
⇒ x² + a² + 2ax + y² + b² + 2by = a² -b²
⇒ x² + y² + 2ax + 2 by + 2b² = 0

In each of the following exercises 6 to 9, find the centre and radius of the circles.

Ex 11.1 Class 11 Maths Question 6.
(x + 5)² + (y – 3)² = 36
Solution:
The given equation of circle is,
(x + 5)² + (y – 3)² = 36
⇒ (x + 5)² + (y – 3)² = (6)²
Comparing it with (x – h)2² + (y – k)² = r², we get
h = -5, k = 3 and r = 6.
Thus the co-ordinates of the centre are (-5, 3) and radius is 6.

Ex 11.1 Class 11 Maths Question 7.
x² + y² – 4x – 8y – 45 = 0
Solution:
The given equation of circle is
x² + y² – 4x – 8y – 45 = 0
∴ (x² – 4x) + (y² – 8y) = 45
⇒ [x² – 4x + (2)²] + [y² – 8y + (4)²] = 45 + (2)² + (4)²
⇒ (x – 2)² + (y – 4)² = 45 + 4 + 16
⇒ (x – 2)² + (y – 4)² = 65
⇒ (x – 2)² + (y – 4)²= \(\left( \sqrt { 65 } \right) ^{ 2 }\)
Comparing it with (x – h)² + (y – k)² = r², we
have h = 2,k = 4 and r = \(\sqrt { 65 } \).
Thus co-ordinates of the centre are (2, 4) and radius is \(\sqrt { 65 } \).

Ex 11.1 Class 11 Maths Question 8.
x² + y² – 8x + 10y – 12 = 0
Solution:
The given equation of circle is,
x² + y² – 8x + 10y -12 = 0
∴ (x² – 8x) + (y² + 10y) = 12
⇒ [x² – 8x + (4)²] + [y² + 10y + (5)²] = 12 + (4)² + (5)²
⇒ (x – 4)² + (y + 5)² = 12 + 16 + 25
⇒ (x – 4)² + (y + 5)² = 53
⇒ (x – 4)² + (y + 5)² = \(\left( \sqrt { 53 } \right) ^{ 2 }\)
Comparing it with (x – h)² + (y – k)² = r², we have h = 4, k = -5 and r = \(\sqrt { 53 } \)
Thus co-ordinates of the centre are (4, -5) and radius is \(\sqrt { 53 } \).

Ex 11.1 Class 11 Maths Question 9.
2x² + 2y² – x = 0
Solution:
The given equation of circle is,
2x² + 2y² – x = 0

Ex 11.1 Class 11 Maths Question 10.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution:
The equation of the circle is,
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (4, 1)
∴ (4 – h)² + (1 – k)² = r²
⇒ 16 + h² – 8h + 1 + k² – 2k = r²
⇒ h²+ k² – 8h – 2k + 17 = r² …. (ii)
Also, the circle passes through point (6, 5)
∴ (6 – h² + (5 – k)² = r²
⇒ 36 + h² -12h + 25 + k² – 10k = r²
⇒ h² + k² – 12h – 10kk + 61 = r² …. (iii)
From (ii) and (iii), we have h² + k² – 8h – 2k +17
= h² + k²– 12h – 10k + 61
⇒ 4h + 8k = 44 => h + 2k = ll ….(iv)
Since the centre (h, k) of the circle lies on the line 4x + y = 16
∴ 4h + k = 16 …(v)
Solving (iv) and (v), we get h = 3 and k = 4.
Putting value of h and k in (ii), we get
(3)² + (4)² – 8 x 3 – 2 x 4 + 17 = r²
∴ r² = 10
Thus required equation of circle is
(x – 3)² + (y – 4)² = 10
⇒ x² + 9 – 6x + y² +16 – 8y = 10
⇒ x² + y² – 6x – 8y +15 = 0.

Ex 11.1 Class 11 Maths Question 11.
Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre is on the line x – 3y – 11 = 0.
Solution:
The equation of the circle is,
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (2, 3)
∴ (2 – h)² + (3 – k)² = r²
⇒ 4 + h² – 4h + 9 + k² – 6k = r²
⇒ h²+ k² – 4h – 6k + 13 = r² ….(ii)
Also, the circle passes through point (-1, 1)
∴ (-1 – h)² + (1 – k)² = r²
⇒ 1 + h² + 2h + 1 + k² – 2k = r²
⇒ h² + k² + 2h – 2k + 2 = r² ….(iii)
From (ii) and (iii), we have
h² + k² – 4h – 6k + 13 = h² + k² + 2h – 2k + 2
⇒ -6h – 4k = -11 ⇒ 6h + 4k = 11 …(iv)
Since the centre (h, k) of the circle lies on the line x – 3y-11 = 0.
∴ h – 3k – 11 = 0 ⇒ h -3k = 11 …(v)
Solving (iv) and (v), we get
h = \(\frac { 7 }{ 2 } \) and k = \(\frac { -5 }{ 2 } \)
Putting these values of h and k in (ii), we get
\(\left( \frac { 7 }{ 2 } \right) ^{ 2 }+\left( \frac { -5 }{ 2 } \right) ^{ 2 }-\frac { 4\times 7 }{ 2 } -6\times \frac { -5 }{ 2 } +13={ r }^{ 2 }\)
⇒ \(\frac { 49 }{ 4 } +\frac { 25 }{ 4 } -14+15+13 \) ⇒ \({ r }^{ 2 }=\frac { 65 }{ 2 } \)
Thus required equation of circle is
⇒ \(\left( x-\frac { 7 }{ 2 } \right) ^{ 2 }+\left( y+\frac { 5 }{ 2 } \right) ^{ 2 }=\frac { 65 }{ 2 } \)
⇒ \({ x }^{ 2 }+\frac { 49 }{ 4 } -7x+{ y }^{ 2 }+\frac { 25 }{ 4 } +5y=\frac { 65 }{ 2 } \)
⇒ 4x² + 49 – 28x + 4y² + 25 + 20y = 130
⇒ 4x² + 4y² – 28x + 20y – 56 = 0
⇒ 4(x² + y² – 7x + 5y -14) = 0
⇒ x² + y² – 7x + 5y -14 = 0.

Ex 11.1 Class 11 Maths Question 12.
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Solution:
Since the centre of the circle lies on x-axis, the co-ordinates of centre are (h, 0). Now the circle passes through the point (2, 3).
∴ Radius of circle

Ex 11.1 Class 11 Maths Question 13.
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the co-ordinate axes.
Solution:
Let the circle makes intercepts a with x-axis and b with y-axis.
∴ OA = a and OB = b
So the co-ordinates of A are (a, 0) and B are (0,b)
Now, the circle passes through three points 0(0, 0), A(a, 0) and B(0, b).

Ex 11.1 Class 11 Maths Question 14.
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Solution:
The equation of circle is
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2)
∴ radius of circle

Ex 11.1 Class 11 Maths Question 15.
Does the point (-2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?
Solution:
The equation of given circle is x² + y² = 25
⇒ (x – 0)² + (y – 0)² = (5)²
Comparing it with (x – h)² + (y – k)² = r², we
get
h = 0,k = 0, and r = 5
Now, distance of the point (-2.5, 3.5) from the centre (0, 0)
\(\sqrt { \left( 0+2.5 \right) ^{ 2 }+\left( 0-3.5 \right) ^{ 2 } } =\sqrt { 6.25+12.25 } \)
\(\sqrt { 18.5 } \) = 4.3 < 5.
Thus the point (-2.5, 3.5) lies inside the circle.

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.1, drop a comment below and we will get back to you at the earliest.