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Class 11th Chapter -8 Binomial Theorem | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter – 8 Binomial Theorem NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 can be really helpful in the preparation of Binomial Theorem Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Binomial Theorem NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter - 8 Binomial Theorem| NCERT MATHS SOLUTION |

Chapter -8 Binomial Theorem EX 8.1 NCERT Solutions

Expand each of the expressions in Exercises 1 to 5.
Ex 8.1 Class 11 Maths Question 1.
${ \left( 1-2x \right) }^{ 5 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 1

Ex 8.1 Class 11 Maths Question 2.
${ \left( \frac { 2 }{ x } -\frac { x }{ 2 } \right) }^{ 5 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 2

Ex 8.1 Class 11 Maths Question 3.
${ \left( 2x-3 \right) }^{ 6 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 3

Ex 8.1 Class 11 Maths Question 4.
${ \left( \frac { x }{ 3 } +\frac { 1 }{ x } \right) }^{ 5 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 4

Ex 8.1 Class 11 Maths Question 5.
${ \left( x+\frac { 1 }{ x } \right) }^{ 6 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 5

Using binomial theorem, evaluate each of the following
Ex 8.1 Class 11 Maths Question 6.
${ \left( 96 \right) }^{ 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 6

Ex 8.1 Class 11 Maths Question 7.
${ \left( 102 \right) }^{ 5 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 7

Ex 8.1 Class 11 Maths Question 8.
${ \left( 101 \right) }^{ 4 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 8

Ex 8.1 Class 11 Maths Question 9.
${ \left( 99 \right) }^{ 5 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 9

Ex 8.1 Class 11 Maths Question 10.
Using Binomial Theorem, indicate which number is larger ${ \left( 1.1\right) }^{ 10000 }$ or 1000.
Solution.
Splitting 1.1 and using binomial theorem to write the first few terms we have
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 10

Ex 8.1 Class 11 Maths Question 11.
Find ${ \left( a+b \right) }^{ 4 }-{ \left( a-b \right) }^{ 4 }$ . Hence, evaluate ${ \left( \sqrt { 3 } +\sqrt { 2 } \right) }^{ 4 }-{ \left( \sqrt { 3 } -\sqrt { 2 } \right) }^{ 4 }$ .
Solution.
By binomial theorem, we have
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 11

Ex 8.1 Class 11 Maths Question 12.
Find ${ \left( x+1 \right) }^{ 6 }+{ \left( x-1 \right) }^{ 6 }$ . Hence or otherwise evaluate ${ \left( \sqrt { 2 } +1 \right) }^{ 6 }+{ \left( \sqrt { 2 } -1 \right) }^{ 6 }$ .
Solution.
By using binomial theorem, we have
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 12

Ex 8.1 Class 11 Maths Question 13.
Show that ${ 9 }^{ n+1 }-8n-9$ is divisible by 64, whenever n is a positive integer.
Solution.
We have to prove that ${ 9 }^{ n+1 }-8n-9=64k$
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1 13

Ex 8.1 Class 11 Maths Question 14.
Prove that $\sum _{ r=0 }^{ n }{ { 3 }^{ r } }$ 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.

Chapter -8 Binomial Theorem EX 8.2 NCERT Solutions

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
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 1

Ex 8.2 Class 11 Maths Question 2.
a⁵b⁷in (a-2b)¹²
Solution.
Suppose a⁵b⁷ occurs in the (r + 1)^th term of the expansion (a – 2b)¹².
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 2

Write the general term in the expansion of
Ex 8.2 Class 11 Maths Question 3.
(x²– y)⁶
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 3

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
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 4

Ex 8.2 Class 11 Maths Question 5.
Find the 4th term in the expansion of (x – 2y)¹².
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 5

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.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 6

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
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 7

Ex 8.2 Class 11 Maths Question 8.
${ \left( \frac { x }{ 3 } +9y \right) }^{ 10 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 8

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.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 9

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.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 10

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

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.
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2 12

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.

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Class 11th Chapter -7 Permutations and Combinations | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter -7 Permutations and Combinations NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 Linear Permutations and Combinations can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Permutations and Combinations NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter -7 Permutations and Combinations| NCERT MATHS SOLUTION |

Chapter 7 Permutations and Combinations EX 7.1 NCERT Solutions

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 $\boxed { } \boxed { } \boxed { }$ 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 $\boxed { } \boxed { } \boxed { }$ 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 $\boxed { } \boxed { } \boxed { } \boxed { }$ 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 $\boxed { 6 } \boxed { 7 }$ $\boxed { } \boxed { } \boxed { }$ 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 $\boxed { } \boxed { }$ 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.

Chapter 7 Permutations and Combinations EX 7.2 NCERT Solutions

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.
$\frac { 8! }{ 6!\times 2! }$
Solution.
$\frac { 8! }{ 6!\times 2! } =\frac { 8\times 7\times 6! }{ 6!\times 2! } =4\times 7=28$

Ex 7.2 Class 11 Maths Question 4.
$\frac { 1 }{ 6! } +\frac { 1 }{ 7! } =\frac { x }{ 8! }$ , findx.
Solution.
We have, $\frac { 1 }{ 6! } +\frac { 1 }{ 7! } =\frac { x }{ 8! }$
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.2 1

Ex 7.2 Class 11 Maths Question 5.
Evaluate $\frac { n! }{ \left( n-r \right) ! }$ , when
(i) n = 6, r = 2
(ii) n = 9, r = 5
Solution.
(i) $\frac { 6! }{ \left( 6-2 \right) ! } =\frac { 6! }{ 4! } =\frac { 6\times 5\times 4! }{ 4! } =30$

(ii) $\frac { 9! }{ \left( 9-5 \right) ! } =\frac { 9! }{ 4! } =\frac { 9\times 8\times 7\times 6\times 5\times 4! }{ 4! } =15120$

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.

Chapter 7 Permutations and Combinations EX 7.3 NCERT Solutions

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
$=\frac { 9! }{ 6! } =9\times 8\times 7=504$

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.
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 1

Ex 7.3 Class 11 Maths Question 7.
Find r if
(i) ⁵P_r = 2⁶P_r-1
(ii) ⁵P_r = ⁶P_r-1
Solution.
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 2

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
$=\frac { 6! }{ 2! } =6\times 5\times 4\times 3=360$

(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
$=\frac { 11! }{ 4!4!2! } =\frac { 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4! }{ 4\times 3\times 2\times 2\times 4! }$
= 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 $\frac { 8! }{ 4!2! }$ 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 = $\frac { 10! }{ 2! }$
$=\frac { 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2! }{ 2! } =1814400$

(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 $\frac { 8! }{ 2! } =20160$ 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 $\frac { 10! }{ 2! } =1814400$ 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.

Chapter 7 Permutations and Combinations EX 7.4 NCERT Solutions

Ex 7.4 Class 11 Maths Question 1.
lf ⁿC₈ = ⁿC₂, find ⁿC₂.
Solution.
We have, ⁿC₈ = ⁿC₂
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 1

Ex 7.4 Class 11 Maths Question 2.
Determine n if
(i) ²ⁿC₃: ⁿC₃ =12 : 1
(ii) ²ⁿC₃: ⁿC₃= 11 : 1
Solution.
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 2

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
$=\frac { 21! }{ 2!19! } =\frac { 21\times 20 }{ 2 } =210$

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 = $\frac { 5! }{ 2!3! } \times \frac { 4! }{ 3!1! }$
$=\frac { 5\times 4 }{ 2 } \times 4=40$

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
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 3

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
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 4

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
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 5

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
$=\frac { 5! }{ 2!3! } \times \frac { 6! }{ 3!3! } =\frac { 5\times 4 }{ 2 } \times \frac { 6\times 5\times 4 }{ 3\times 2 } =200$

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 = $\frac { 7! }{ 3!4! } =\frac { 7\times 6\times 5 }{ 6 } =35$

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.

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Class 11th Chapter – 5 Complex Numbers and Quadratic Equations | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter -5 Complex Numbers and Quadratic Equations NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 Complex Numbers and Quadratic Equations solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Complex Numbers and Quadratic Equations NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter 5 Complex Numbers and Quadratic Equations| NCERT MATHS SOLUTION |

Chapter 5 Complex Numbers and Quadratic Equations EX 5.1 NCERT Solutions

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.
Ex 5.1 Class 11 Maths Question 1.
$\left( 5i \right) \left( -\frac { 3 }{ 5 } i \right)$
Solution.
$\left( 5i \right) \left( -\frac { 3 }{ 5 } i \right)$
= -3i² = -3(-1) [∵ i² = -1]
= 3 = 3 + 0i

Ex 5.1 Class 11 Maths Question 2.
i⁹+ i¹⁹
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 1

Ex 5.1 Class 11 Maths Question 3.
i^-39
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 2

Ex 5.1 Class 11 Maths Question 4.
3(7 + i7) + i(7 + i7)
Solution.
3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i²
= 21 + (21 + 7)i + (-1)7 = 21 – 7 + 28i
= 14 + 28i.

Ex 5.1 Class 11 Maths Question 5.
(1 – i) – (- 1 +i6)
Solution.
(1 – i) – (-1 + i6) = 1 – i + 1 – 6i
= (1 +1) – i(1 + 6)
= 2 – 7i

Ex 5.1 Class 11 Maths Question 6.
$\left( \frac { 1 }{ 5 } +i\frac { 2 }{ 5 } \right) -\left( 4+i\frac { 5 }{ 2 } \right)$
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 3

Ex 5.1 Class 11 Maths Question 7.
$\left[ \left( \frac { 1 }{ 3 } +i\frac { 7 }{ 3 } \right) +\left( 4+i\frac { 1 }{ 3 } \right) \right] -\left( -\frac { 4 }{ 3 } +i \right)$
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 4

Ex 5.1 Class 11 Maths Question 8.
(1 -i)⁴
Solution.
(1 -i)⁴ = [(1 – i)²]² = [1 – 2i + i²]²
= [1 – 2i + (-1)]²
= (-2i)² = 4i² = 4(-1) = – 4
= – 4 + 0i

Ex 5.1 Class 11 Maths Question 9.
${ \left( \frac { 1 }{ 3 } +3i \right) }^{ 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 5

Ex 5.1 Class 11 Maths Question 10.
${ \left( -2-\frac { 1 }{ 3 } i \right) }^{ 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 6

Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.
Ex 5.1 Class 11 Maths Question 11.
4 – 3i
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 7

Ex 5.1 Class 11 Maths Question 12.
$\sqrt { 5 } +3i$
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 8

Ex 5.1 Class 11 Maths Question 13.
-i
Solution.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 9

Ex 5.1 Class 11 Maths Question 14.
Express the following expression in the form of a + ib:
$\frac { \left( 3+i\sqrt { 5 } \right) \left( 3-i\sqrt { 5 } \right) }{ \left( \sqrt { 3 } +\sqrt { 2 } i \right) -\left( \sqrt { 3 } -i\sqrt { 2 } \right) }$
Solution.
We have, $\frac { \left( 3+i\sqrt { 5 } \right) \left( 3-i\sqrt { 5 } \right) }{ \left( \sqrt { 3 } +\sqrt { 2 } i \right) -\left( \sqrt { 3 } -i\sqrt { 2 } \right) }$
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 10

We hope the NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1, drop a comment below and we will get back to you at the earliest.

Chapter 5 Complex Numbers and Quadratic Equations EX 5.2 NCERT Solutions

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.
Ex 5.2 Class 11 Maths Question 1.
$z=-1-i\sqrt { 3 }$
Solution.
We have, $z=-1-i\sqrt { 3 }$
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 1

Ex 5.2 Class 11 Maths Question 2.
$z=-\sqrt { 3 } +i$
Solution.
We have, $z=-\sqrt { 3 } +i$
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 2

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:
Ex 5.2 Class 11 Maths Question 3.
1 – i
Solution.
We have, z = 1 – i
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 3

Ex 5.2 Class 11 Maths Question 4.
-1 + i
Solution.
We have, z = -1 + i
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 4

Ex 5.2 Class 11 Maths Question 5.
-1 – i
Solution.
We have, z = -1 – i
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 5

Ex 5.2 Class 11 Maths Question 6.
-3
Solution.
We have, z = -3, i.e., z = -3 + 0i
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 6

Ex 5.2 Class 11 Maths Question 7.
$\sqrt { 3 } +i$
Solution.
We have, $z=\sqrt { 3 } +i$
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 7

Ex 5.2 Class 11 Maths Question 8.
i
Solution.
We have, z = i, i.e., z = 0 + 1.i
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 8

We hope the NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2, drop a comment below and we will get back to you at the earliest.

Chapter 5 Complex Numbers and Quadratic Equations EX 5.3 NCERT Solutions

Solve each of the following equations:
Ex 5.3 Class 11 Maths Question 1.
x² + 3 = 0
Solution.
We have, x² + 3 = 0 ⇒ x² = -3
⇒ $x=\pm \sqrt { -3 }$ ⇒ x = $\pm \sqrt { 3 } i$

Ex 5.3 Class 11 Maths Question 2.
2x² + x + 1 = 0
Solution.
We have, 2x² + x + 1 = 0
Comparing the given equation with the general form ax² + bx + c = 0, we get
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 1

Ex 5.3 Class 11 Maths Question 3.
x² + 3x + 9 = 0
Solution.
We have, x² + 3x + 9 = 0
Comparing the given equation with the general form ax² + bx + c = 0,we get a = 1, b = 3, c = 9
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 2

Ex 5.3 Class 11 Maths Question 4.
-x² + x – 2 = 0
Solution.
We have, -x² + x – 2 = 0
Comparing the given equation with the general form ax² + bx + c = 0,we get
a = 1, b = 1, c = -2
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 3

Ex 5.3 Class 11 Maths Question 5.
x² + 3x + 5 = 0
Solution.
We have, x² + 3x + 5 = 0
Comparing the given equation with the general form ax² + bx + c = 0, we get a = 1, b = 3, c = 5.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 4

Ex 5.3 Class 11 Maths Question 6.
x² – x + 2 = 0
Solution.
We have, x² – x + 2 = 0
Comparing the given equation with the general form ax² + bx + c = 0, we get a = 1, b = -1, c = 2.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 5

Ex 5.3 Class 11 Maths Question 7.
$\sqrt { 2 } { x }^{ 2 }+x+\sqrt { 2 } =0$
Solution.
We have, $\sqrt { 2 } { x }^{ 2 }+x+\sqrt { 2 } =0$
Comparing the given equation with the general form ax² + bx + c = 0, we get
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 6

Ex 5.3 Class 11 Maths Question 8.
$\sqrt { 3 } { x }^{ 2 }+\sqrt { 2 } x+3\sqrt { 3 } =0$
Solution.
We have, $\sqrt { 3 } { x }^{ 2 }+\sqrt { 2 } x+3\sqrt { 3 } =0$
Comparing the given equation with the general form ax² + bx + c = 0, we get
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 7

Ex 5.3 Class 11 Maths Question 9.
${ x }^{ 2 }+x+\frac { 1 }{ \sqrt { 2 } } =0$
Solution.
We have, ${ x }^{ 2 }+x+\frac { 1 }{ \sqrt { 2 } } =0$
Comparing the given equation with the general form ax² + bx + c = 0, we get
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 8

Ex 5.3 Class 11 Maths Question 10.
${ x }^{ 2 }+\frac { x }{ \sqrt { 2 } } +1=0$
Solution.
We have, ${ x }^{ 2 }+\frac { x }{ \sqrt { 2 } } +1=0$
Comparing the given equation with the general form ax² + bx + c = 0, we get
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 9

We hope the NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3, drop a comment below and we will get back to you at the earliest.

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Class 11th Chapter -4 Principle of Mathematical Induction | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter – 4 Principle of Mathemarical Induction NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 Principle of Mathemarical Induction solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Principle of Mathemarical Induction NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter 4 Principle of Mathemarical Induction| NCERT MATHS SOLUTION |

Chapter 4 Principle of Mathemarical Induction EX 4.1 NCERT Solutions

Prove the following by using the principle of mathematical induction for aline n ∈ N :
Ex 4.1 Class 11 Maths Question 1.
$1+{ 3 }^{ 2 }+{ 3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ 3 }^{ n }=frac { left( { 3 }^{ n }-1 right) }{ 2 }$
Solution.
Let the given statement be P(n) i.e.,
P(n) : $1+{ 3 }^{ 2 }+{ 3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ 3 }^{ n }=frac { left( { 3 }^{ n }-1 right) }{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 1

Ex 4.1 Class 11 Maths Question 2.
${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ n }^{ 3 }={ left( frac { nleft( n+1 right) }{ 2 } right) }^{ 2 }$
Solution.
Let the given statement be P(n) i.e.,
P(n) : ${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ n }^{ 3 }={ left( frac { nleft( n+1 right) }{ 2 } right) }^{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 3

Ex 4.1 Class 11 Maths Question 3.
$1+frac { 1 }{ left( 1+2 right) } +frac { 1 }{ left( 1+2+3 right) } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 1+2+3+cdot cdot cdot cdot cdot cdot cdot cdot +n right) } =frac { 2 }{ left( n+1 right) }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1+frac { 1 }{ left( 1+2 right) } +frac { 1 }{ left( 1+2+3 right) } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 1+2+3+.cdot cdot cdot cdot cdot cdot cdot cdot +n right) } =frac { 2 }{ left( n+1 right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 5

Ex 4.1 Class 11 Maths Question 4.
$1.2.3+2.3.4+cdot cdot cdot cdot cdot cdot cdot cdot +nleft( n+1 right) left( n+2 right) =frac { nleft( n+1 right) left( n+2 right) left( n+3 right) }{ 4 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1.2.3+2.3.4+cdot cdot cdot cdot cdot cdot cdot cdot +nleft( n+1 right) left( n+2 right) =frac { nleft( n+1 right) left( n+2 right) left( n+3 right) }{ 4 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 7

Ex 4.1 Class 11 Maths Question 5.
$1.3+{ 2.3 }^{ 2 }+{ 3.3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ n.3 }^{ n }=frac { left( 2n-1 right) { 3 }^{ n+1 }+3 }{ 4 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1.3+{ 2.3 }^{ 2 }+{ 3.3 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ n.3 }^{ n }=frac { left( 2n-1 right) { 3 }^{ n+1 }+3 }{ 4 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 9

Ex 4.1 Class 11 Maths Question 6.
$1.2+2.3+3.4+cdot cdot cdot cdot cdot cdot cdot cdot +n.left( n+1 right) =left[ frac { nleft( n+1 right) left( n+2 right) }{ 3 } right]$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1.2+2.3+3.4+cdot cdot cdot cdot cdot cdot cdot cdot +n.left( n+1 right) =left[ frac { nleft( n+1 right) left( n+2 right) }{ 3 } right]$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 11

Ex 4.1 Class 11 Maths Question 7.
$1.3+3.5+5.7+cdot cdot cdot cdot cdot cdot cdot cdot +left( 2n-1 right) left( 2n+1 right) =frac { nleft( { 4n }^{ 2 }+6n-1 right) }{ 3 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1.3+3.5+5.7+cdot cdot cdot cdot cdot cdot cdot cdot +left( 2n-1 right) left( 2n+1 right) =frac { nleft( { 4n }^{ 2 }+6n-1 right) }{ 3 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 13

Ex 4.1 Class 11 Maths Question 8.
$1.2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +n.{ 2 }^{ n }=left( n-1 right) { 2 }^{ n+1 }+2$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1.2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+cdot cdot cdot cdot cdot cdot cdot cdot +n.{ 2 }^{ n }=left( n-1 right) { 2 }^{ n+1 }+2$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 16

Ex 4.1 Class 11 Maths Question 9
$frac { 1 }{ 2 } +frac { 1 }{ 4 } +frac { 1 }{ 8 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ { 2 }^{ n } } =1-frac { 1 }{ { 2 }^{ n } }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $frac { 1 }{ 2 } +frac { 1 }{ 4 } +frac { 1 }{ 8 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ { 2 }^{ n } } =1-frac { 1 }{ { 2 }^{ n } }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 17

Ex 4.1 Class 11 Maths Question 10.
$frac { 1 }{ 2.5 } +frac { 1 }{ 5.8 } +frac { 1 }{ 8.11 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 3n-1 right) left( 3n+2 right) } =frac { n }{ left( 6n+4 right) }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $frac { 1 }{ 2.5 } +frac { 1 }{ 5.8 } +frac { 1 }{ 8.11 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 3n-1 right) left( 3n+2 right) } =frac { n }{ left( 6n+4 right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 19

Ex 4.1 Class 11 Maths Question 11.
$frac { 1 }{ 1.2.3 } +frac { 1 }{ 2.3.4 } +frac { 1 }{ 3.4.5 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ nleft( n+1 right) left( n+2 right) } =frac { nleft( n+3 right) }{ 4left( n+1 right) left( n+2 right) }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $frac { 1 }{ 1.2.3 } +frac { 1 }{ 2.3.4 } +frac { 1 }{ 3.4.5 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ nleft( n+1 right) left( n+2 right) } =frac { nleft( n+3 right) }{ 4left( n+1 right) left( n+2 right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 21

Ex 4.1 Class 11 Maths Question 12.
$a+ar+{ ar }^{ 2 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ ar }^{ n-1 }=frac { aleft( { r }^{ n }-1 right) }{ r-1 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $a+ar+{ ar }^{ 2 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ ar }^{ n-1 }=frac { aleft( { r }^{ n }-1 right) }{ r-1 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 24

Ex 4.1 Class 11 Maths Question 13.
$left( 1+frac { 3 }{ 1 } right) left( 1+frac { 5 }{ 4 } right) left( 1+frac { 7 }{ 9 } right) cdot cdot cdot cdot cdot cdot cdot cdot left( 1+frac { left( 2n+1 right) }{ { n }^{ 2 } } right) ={ left( n+1 right) }^{ 2 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $left( 1+frac { 3 }{ 1 } right) left( 1+frac { 5 }{ 4 } right) left( 1+frac { 7 }{ 9 } right) cdot cdot cdot cdot cdot cdot cdot cdot left( 1+frac { left( 2n+1 right) }{ { n }^{ 2 } } right) ={ left( n+1 right) }^{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 26

Ex 4.1 Class 11 Maths Question 14.
$left( 1+frac { 1 }{ 1 } right) left( 1+frac { 1 }{ 2 } right) left( 1+frac { 1 }{ 3 } right) cdot cdot cdot cdot cdot cdot cdot cdot left( 1+frac { 1 }{ n } right) =left( n+1 right)$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $left( 1+frac { 1 }{ 1 } right) left( 1+frac { 1 }{ 2 } right) left( 1+frac { 1 }{ 3 } right) cdot cdot cdot cdot cdot cdot cdot cdot left( 1+frac { 1 }{ n } right) =left( n+1 right)$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 28

Ex 4.1 Class 11 Maths Question 15.
${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ left( 2n-1 right) }^{ 2 }=frac { nleft( 2n-1 right) left( 2n+1 right) }{ 3 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : ${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+cdot cdot cdot cdot cdot cdot cdot cdot +{ left( 2n-1 right) }^{ 2 }=frac { nleft( 2n-1 right) left( 2n+1 right) }{ 3 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 30

Ex 4.1 Class 11 Maths Question 16.
$frac { 1 }{ 1.4 } +frac { 1 }{ 4.7 } +frac { 1 }{ 7.10 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 3n-2 right) left( 3n+1 right) } =frac { n }{ left( 3n+1 right) }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $frac { 1 }{ 1.4 } +frac { 1 }{ 4.7 } +frac { 1 }{ 7.10 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 3n-2 right) left( 3n+1 right) } =frac { n }{ left( 3n+1 right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 32

Ex 4.1 Class 11 Maths Question 17.
$frac { 1 }{ 3.5 } +frac { 1 }{ 5.7 } +frac { 1 }{ 7.9 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 2n+1 right) left( 2n+3 right) } =frac { n }{ 3left( 2n+3 right) }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $frac { 1 }{ 3.5 } +frac { 1 }{ 5.7 } +frac { 1 }{ 7.9 } +cdot cdot cdot cdot cdot cdot cdot cdot +frac { 1 }{ left( 2n+1 right) left( 2n+3 right) } =frac { n }{ 3left( 2n+3 right) }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 34

Ex 4.1 Class 11 Maths Question 18.
$1+2+3+cdot cdot cdot cdot cdot cdot cdot cdot +n<frac { 1 }{ 8 } { left( 2n+1 right) }^{ 2 }$
Solution.
Let the given statement be P(n), i.e.,
P(n) : $1+2+3+cdot cdot cdot cdot cdot cdot cdot cdot +n<frac { 1 }{ 8 } { left( 2n+1 right) }^{ 2 }$
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 36

Ex 4.1 Class 11 Maths Question 19.
n(n+1 )(n + 5) is a multiple of 3.
Solution.
Let the given statement be P(n), i.e.,
P(n): n(n + l)(n + 5) is a multiple of 3.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 38

Ex 4.1 Class 11 Maths Question 20.
${ 10 }^{ 2n-1 }+1$ is divisible by 11.
Solution.
Let the given statement be P(n), i.e.,
P(n): ${ 10 }^{ 2n-1 }+1$ is divisible by 11
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 40

Ex 4.1 Class 11 Maths Question 21.
${ x }^{ 2n }-{ y }^{ 2n }$ is divisible by x + y.
Solution.
Let the given statement be P(n), i.e.,
P(n): ${ x }^{ 2n }-{ y }^{ 2n }$ is divisible by x + y.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 41

Ex 4.1 Class 11 Maths Question 22.
${ 3 }^{ 2n+2 }-8n-9$ is divisible by 8.
Solution.
Let the given statement be P(n), i.e.,
P(n): ${ 3 }^{ 2n+2 }-8n-9$ is divisible by 8.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 42

Ex 4.1 Class 11 Maths Question 23.
${ 41 }^{ n }-{ 14 }^{ n }$ is a multiple of 27.
Solution.
Let the given statement be P(n), i.e.,
P(n): ${ 41 }^{ n }-{ 14 }^{ n }$ is a multiple of 27.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 43

Ex 4.1 Class 11 Maths Question 24.
$left( 2n+7 right) <{ left( n+3 right) }^{ 2 }$
Solution.
Let the given statement be P(n), i.e.,
P(n): $left( 2n+7 right) <{ left( n+3 right) }^{ 2 }$
First we prove that the statement is true for n = 1.
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction 44

We hope the NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction, drop a comment below and we will get back to you at the earliest.

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Class 11th Chapter -3 Trigonometric Functions | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter -3 Trigonometric Fuctions NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 Trigonometric Fuctions solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Trigonometric Fuctions NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter 3 Trigonometric Fuctions | NCERT MATHS SOLUTION |

Chapter 3 Trigonometric Fuctions EX 3.1 NCERT Solutions

Ex 3.1 Class 11 Maths Question 1.
Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) -47°30′
(iii) 240°
(iv) 520°
Solution.
We have, 180° = π Radians
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 1
Ex 3.1 Class 11 Maths Question 2.
Find the degree measures corresponding to the following radian measures $\left( Use\quad \pi =\frac { 22 }{ 7 } \right)$
(i) $\frac { 11 }{ 16 }$
(ii) -4
(iii) $\frac { 5\pi }{ 3 }$
(iv) $\frac { 7\pi }{ 6 }$
Solution.
We have π Radians = 180°
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 2

Ex 3.1 Class 11 Maths Question 3.
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second ?
Solution.
Number of revolutions made by wheel in one minute = 360
As we know that, 1 Revolution = 27 π Radians
∴ 360 Revolutions = 720 π Radians
∴ In 1 minute wheel can make = 720 π Radians
⇒ In 60 seconds wheel can make = 720 π Radians
⇒ In 1 second wheel can make
$\frac { 720\pi }{ 3 } Radians=12\pi \quad Radians$

Ex 3.1 Class 11 Maths Question 4.
Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm $\left( Use\quad \pi =\frac { 22 }{ 7 } \right)$
Solution.
Let O be the centre and AB be the arc length of the circle.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 3

Ex 3.1 Class 11 Maths Question 5.
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of Xhe chord.
Solution.
Let AB be the minor arc of the chord.
AB = 20 cm, OA = OB = 20 cm
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 4

Ex 3.1 Class 11 Maths Question 6.
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their, radii.
Solution.
Let r₁ r₂ and θ₁, θ₂ be the radii and angles subtended at the centre of two circles respectively.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 5

Ex 3.1 Class 11 Maths Question 7.
Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 6

We hope the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, drop a comment below and we will get back to you at the earliest.

Chapter 3 Trigonometric Fuctions EX 3.2 NCERT Solutions

Find the values of other five trigonometric functions in Exercises 1 to 5.
Ex 3.2 Class 11 Maths Question 1.
$\cos { x } =\frac { -1 }{ 2 }$ , x lies in third quadrant.
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 1

Ex 3.2 Class 11 Maths Question 2.
$\sin { x } =\frac { 3 }{ 5 }$ , x lies in second quadrant.
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 2

Ex 3.2 Class 11 Maths Question 3.
$\cot { x= } \frac { 3 }{ 4 }$ , xlies in third quadrant.
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 3

Ex 3.2 Class 11 Maths Question 4.
$\sec { x } =\frac { 13 }{ 5 }$ , x lies in fourth quadrant.
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 4

Ex 3.2 Class 11 Maths Question 5.
$\tan { x } =-\frac { 5 }{ 12 }$ , x lies in second quadrant.
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 4

Find the values of the trigonometric functions in Exercises 6 to 10.
Ex 3.2 Class 11 Maths Question 6.
sin 765°
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 5

Ex 3.2 Class 11 Maths Question 7.
cosec (-1410°)
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 6

Ex 3.2 Class 11 Maths Question 8.
$tan\quad \frac { 19\pi }{ 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 7

Ex 3.2 Class 11 Maths Question 9.
$sin\left( -\frac { 11\pi }{ 3 } \right)$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 8

Ex 3.2 Class 11 Maths Question 10.
$cot\left( -\frac { 15\pi }{ 4 } \right)$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 9

We hope the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2, drop a comment below and we will get back to you at the earliest.

Chapter 3 Trigonometric Fuctions EX 3.3 NCERT Solutions

Ex 3.3 Class 11 Maths Question 1.
Prove that: ${ sin }^{ 2 }\frac { \pi }{ 6 } +{ cos }^{ 2 }\frac { \pi }{ 3 } -{ tan }^{ 2 }\frac { \pi }{ 4 } =-\frac { 1 }{ 2 }$
Solution.
L.H.S. = ${ sin }^{ 2 }\frac { \pi }{ 6 } +{ cos }^{ 2 }\frac { \pi }{ 3 } -{ tan }^{ 2 }\frac { \pi }{ 4 } =-\frac { 1 }{ 2 }$
$=\left[ { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( 1 \right) }^{ 2 } \right] =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } -1=\frac { -1 }{ 2 } =\quad R.H.S.$

Ex 3.3 Class 11 Maths Question 2.
$2{ sin }^{ 2 }\frac { \pi }{ 6 } +{ cosec }^{ 2 }\frac { 7\pi }{ 6 } { cos }^{ 2 }\frac { \pi }{ 3 } =\frac { 3 }{ 2 }$
Solution.
L.H.S. = $2{ sin }^{ 2 }\frac { \pi }{ 6 } +{ cosec }^{ 2 }\frac { 7\pi }{ 6 } { cos }^{ 2 }\frac { \pi }{ 3 }$
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 1

Ex 3.3 Class 11 Maths Question 3.
${ cot }^{ 2 }\frac { \pi }{ 6 } +cosec\frac { 5\pi }{ 6 } +3{ tan }^{ 2 }\frac { \pi }{ 6 } =6$
Solution.
L.H.S. = ${ cot }^{ 2 }\frac { \pi }{ 6 } +cosec\frac { 5\pi }{ 6 } +3{ tan }^{ 2 }\frac { \pi }{ 6 }$
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 2

Ex 3.3 Class 11 Maths Question 4.
$2{ sin }^{ 2 }\frac { 3\pi }{ 4 } +2{ cos }^{ 2 }\frac { \pi }{ 4 } +2{ sec }^{ 2 }\frac { \pi }{ 3 } =10$
Solution.
L.H.S. = $2{ sin }^{ 2 }\frac { 3\pi }{ 4 } +2{ cos }^{ 2 }\frac { \pi }{ 4 } +2{ sec }^{ 2 }\frac { \pi }{ 3 }$
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 3

Ex 3.3 Class 11 Maths Question 5.
Find the value of:
(i) sin 75°
(ii) tan 15°
Solution.
(i) sin (75°) = sin (30° + 45°)
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 4

(ii) tan 15° = tan (45° – 30°)
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 5

Prove the following:
Ex 3.3 Class 11 Maths Question 6.
$cos\left( \frac { \pi }{ 4 } -x \right) cos\left( \frac { \pi }{ 4 } -y \right) -sin\left( \frac { \pi }{ 4 } -x \right) sin\left( \frac { \pi }{ 4 } -y \right)$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 6

Ex 3.3 Class 11 Maths Question 7.
$\frac { tan\left( \frac { \pi }{ 4 } +x \right) }{ tan\left( \frac { \pi }{ 4 } -x \right) } ={ \left( \frac { 1+tan\quad x }{ 1-tan\quad x } \right) }^{ 2 }$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 7

Ex 3.3 Class 11 Maths Question 8.
$\frac { cos\left( \pi +x \right) cos\left( -x \right) }{ sin\left( \pi -x \right) cos\left( \frac { \pi }{ 2 } +x \right) } ={ cot }^{ 2 }x$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 8

Ex 3.3 Class 11 Maths Question 9.
$cos\left( \frac { 3\pi }{ 2 } +x \right) cos\left( 2\pi +x \right) \left[ cot\left( \frac { 3\pi }{ 2 } -x \right) +cot\left( 2\pi +x \right) \right] =1$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 9

Ex 3.3 Class 11 Maths Question 10.
sin(n +1 )x sin(n + 2)x + cos(n +1 )x cos(n + 2)x = cosx
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 10

Ex 3.3 Class 11 Maths Question 11.
$cos\left( \frac { 3\pi }{ 4 } +x \right) -cos\left( \frac { 3\pi }{ 4 } -x \right) =-\sqrt { 2 } sinx$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 11

Ex 3.3 Class 11 Maths Question 12.
sin²6x – sin²4x= sin²x sin10x
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 12

Ex 3.3 Class 11 Maths Question 13.
cos²2x – cos²6x = sin 4x sin 8x
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 13

Ex 3.3 Class 11 Maths Question 14.
sin2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 14

Ex 3.3 Class 11 Maths Question 15.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 15

Ex 3.3 Class 11 Maths Question 16.
$\frac { cos9x-cos5x }{ sin17x-sin3x } =-\frac { sin2x }{ cos10x }$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 16

Ex 3.3 Class 11 Maths Question 17.
$\frac { sin5x+sin3x }{ cos5x+cos3x } =tan4x$
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 17

Ex 3.3 Class 11 Maths Question 18.
$\frac { sinx-siny }{ cosx+cosy } =tan\left( \frac { x-y }{ 2 } \right)$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 18

Ex 3.3 Class 11 Maths Question 19.
$\frac { sinx+sin3x }{ cosx+cos3x } =tan2x$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 19

Ex 3.3 Class 11 Maths Question 20.
$\frac { sinx-sin3x }{ { sin }^{ 2 }x-{ cos }^{ 2 }x } =2sinx$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 20

Ex 3.3 Class 11 Maths Question 21.
$\frac { cos4x+cos3x+cos2x }{ sin4x+sin3x+sin2x } =cot3x$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 21

Ex 3.3 Class 11 Maths Question 22.
cot x cot 2x – cot 2x cot 3x – cot3x cotx = 1
Solution.
We know that 3x = 2x + x.
Therefore,
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 22

Ex 3.3 Class 11 Maths Question 23.
$tan4x=\frac { 4tanx\left( 1-{ tan }^{ 2 }x \right) }{ 1-6{ tan }^{ 2 }x+{ tan }^{ 4 }x }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 23

Ex 3.3 Class 11 Maths Question 24.
cos 4x = 1 – 8 sin²x cos²x
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 24

Ex 3.3 Class 11 Maths Question 25.
cos 6x = 32 cos6 x – 48 cos⁴x + 18 cos² x -1
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 25

We hope the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3, drop a comment below and we will get back to you at the earliest.

Chapter 3 Trigonometric Fuctions EX 3.4 NCERT Solutions

Find the principal and general solutions of the following equations:
Ex 3.4 Class 11 Maths Question 1.
$tanx=\sqrt { 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 1

Ex 3.4 Class 11 Maths Question 2.
sec x = 2
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 2

Ex 3.4 Class 11 Maths Question 3.
$cotx=-\sqrt { 3 }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 3

Ex 3.4 Class 11 Maths Question 4.
cosec x = -2
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 5

Find the general solution for each of the following equations:
Ex 3.4 Class 11 Maths Question 5.
cos 4x = cos 2x
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 6

Ex 3.4 Class 11 Maths Question 6.
cos 3x + cos x – cos 2x=0
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 7

Ex 3.4 Class 11 Maths Question 7.
sin 2 x + cos x = 0
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 8

Ex 3.4 Class 11 Maths Question 8.
sec²2x = 1 – tan 2x
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 9

Ex 3.4 Class 11 Maths Question 9.
sin x + sin 3x + sin 5x = 0
Solution.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 10

We hope the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4, drop a comment below and we will get back to you at the earliest.

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Class 11th Chapter -2 Relations and Functions | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter 2 Relations and functions NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 Relations and functions solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Relations and functions NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter 1 Relations and Functions | NCERT MATHS SOLUTION |

Chapter 1 - Relations and functions EX 2.1 NCERT Solutions

Ex 2.1 Class 11 Maths Question 1.
If $\left( \frac { x }{ 3 } +1,y-\frac { 2 }{ 3 } \right) =\left( \frac { 5 }{ 3 } ,\frac { 1 }{ 3 } \right)$ , find the values of x and y.
Solution.
Since the ordered pairs are equal. So, the corresponding elements are equal
∴ $\frac { x }{ 3 } +1=\frac { 5 }{ 3 }$ and $y-\frac { 2 }{ 3 } =\frac { 1 }{ 3 }$
⇒ $\frac { x }{ 3 } =\frac { 5 }{ 3 } -1$ and $y=\frac { 1 }{ 3 } +\frac { 2 }{ 3 }$ ⇒ x = 2 and y = 1.

Ex 2.1 Class 11 Maths Question 2.
If the set A has 3 elements and the set B {3, 4, 5}, then find the number of elements in (A x B).
Solution.
According to question, n(A) = 3 and n(B) = 3.
∴ n(A x B) = n(A) x n(B) = 3 x 3 = 9
∴ There are total 9 elements in (A x B).

Ex 2.1 Class 11 Maths Question 3.
If G = {7, 8} and H = {5, 4, 2}, find G x H and H x G.
Solution.
We have G = {7, 8} and H = {5, 4, 2} Then, by the definition of the cartesian product, we have
G x H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H x G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.

Ex 2.1 Class 11 Maths Question 4.
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = {n, m}, then P x Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then Ax B is a non-empty set of ordered pairs (x, y) such
that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A x (B ∩φ) = φ
Solution.
(i) False, if P = {m, n} and Q = {n, m}
Then P x Q = {(m, n), (m, m), (n, n), (n, m)}.
(ii) True, by the definition of cartesian product.
(iii) True, We have A = {1, 2} and B = {3, 4}
Now, B ∩ φ = φ ∴ A x (B ∩ φ) = A x φ = φ.

Ex 2.1 Class 11 Maths Question 5.
If A = {-1, 1},find A x A x A.
Solution.
A = {-1, 1}
Then, A x A = {-1, 1} x {-1, 1} = {(-1, -1), (-1,1),(1,-1), (1,1)}
A x A x A = ((-1,-1),(-1,1),(1,-1),(1,1)} x {-1,1}
= {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1,1), (1, -1, -1), (1, -1,1), (1,1,-1), (1,1,1)}

Ex 2.1 Class 11 Maths Question 6.
If A x B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Solution.
Given, A x B = {(a, x), (a, y), (b, x), (b, y)}
If {p, q) ∈ A x B, then p ∈ A and q ∈ B
∴ A = {a, b} and B = {x, y}.

Ex 2.1 Class 11 Maths Question 7.
Let A = {1, 2}, B = (1, 2, 3, 4), C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A x (B ∩ C = (A x B) ∩ (AxC)
(ii) A x C is a subset of B x D.
Solution.
Given, A = {1, 2}, B ={1, 2, 3, 4}, C = {5, 6}, D = (5, 6, 7, 8}
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1 1

Ex 2.1 Class 11 Maths Question 8.
Let A = {1, 2} and B = {3, 4}. Write 4 x B. How many subsets will 4 x B have? List them.
Solution.
Given, A = {1, 2} and B = {3, 4}
Then, A x B = {(1, 3), (1,4), (2, 3), (2, 4)}
i. e., A x B has 4 elements. So, it has 2⁴ i.e. 16 subsets.
The subsets of A x B are as follows :
φ, {(1, 3)1, ((1, 4)), {(2, 3)|, {(2, 4)}, {(1, 3), (1,4)}, {(1,3), (2,3)},{(1,3), (2,4)), ((1,4), (2,3)},
{(1, 4), (2, 4)},{(2, 3), (2, 4)},{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 4), (2, 3), (2,4)},{(1, 3), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2,4)}.

Ex 2.1 Class 11 Maths Question 9.
Let A and B be two sets such that n (A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A x B, find A and B, where x, y and z are distinct elements.
Solution.
Given, n(A) = 3 and n(B) = 2
Now (x, 1) ∈ A x B ⇒ x ∈ A and 1 ∈ B,
(y, 2) ∈ A x B ⇒ y ∈ A and 2 ∈ B
(z, 1) ∈ A x B ⇒z ∈ A and 1 ∈ B
∴ x, y, z ∈ A and 1, 2 ∈ B
Hence, A = {x, y, z} and B = {1, 2}.

Ex 2.1 Class 11 Maths Question 10.
The Cartesian product 4×4 has 9 elements among which are found (-1, 0) and (0, 1). Find the set 4 and the remaining elements of 4 x 4.
Solution.
Since, we have n(A x A) = 9
⇒ n(A) x n(A) = 9 [ ∵ n (A x B) = n(A) x n(B)]
⇒ (n(A))² = 9 ⇒ n(A) = 3
Also, given (-1, 0) ∈ A x A ⇒ -1, 0 ∈ A ,
and (0,1) ∈ A x A ⇒ 0, 1 ∈ A
∴ -1, 0,1 ∈ A
Hence, A = {-1, 0, 1} (∵ n(A) = 3)
and the remaining elements of A x A are (-1, -1), (-1,1), (0, -1), (0,0), (1, -1), (1,0), (1,1).

We hope the NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1, drop a comment below and we will get back to you at the earliest.

Chapter 1 - Relations and functions EX 2.2 NCERT Solutions

Ex 2.2 Class 11 Maths Question 1.
Let 4 = {1,2,3, ,14}. Define a relation R from A to A by R = {(x,y): 3x – y = 0, where x, y ∈ 4}.
Write down its domain, codomain and range.
Solution.
We have A = (1, 2, 3,……..,14)
Given relation R = {(x, y) : 3x – y = 0, where x, y ∈ A}
= {(x, y): y = 3x, where x, y ∈ A)
= {(x, 3x), where x, 3x ∈ A}
= {(1, 3), (2, 6), (3, 9), (4,12)}
[∵ 1 ≤ 3x ≤ 14, ∴ $\frac { 1 }{ 3 } \le x\le \frac { 14 }{ 3 }$ ⇒ x = 1, 2, 3, 4 ]
Domain of R = {1, 2, 3, 4}
Codomain of R = {1, 2,……, 14}
Range of R = {3, 6, 9, 12}.

Ex 2.2 Class 11 Maths Question 2.
Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.
Solution.
Given relation R = {(x, y): y = x + 5, x < 4 and x, y ∈ N)
= {(x, y): y = x + 5, x ∈ (1, 2, 3) & y ∈ N}
= {(x, x + 5): x = 1, 2, 3}
Thus, R = {(1, 6), (2, 7), (3, 8)}.
Domain of R = {1, 2, 3}, Range of R = {6, 7, 8}.

Ex 2.2 Class 11 Maths Question 3.
A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ 4, y ∈B}. Write R in roster form.
Solution.
We have, A = {1, 2, 3, 5} and B = {4, 6, 9} R = {(x, y): difference between x and y is odd;
x ∈ A, y ∈B}
= {(x, y): y – x = odd; x ∈ A, y ∈ B}
Hence R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

Ex 2.2 Class 11 Maths Question 4.
The figure shows a relationship between the sets P and Q. Write this relation
(i) in set-builder form
(ii) roster form.
What is its domain and range?
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2 1
Solution.
(i) Its set builder form is
R = {(x, y): x – y = 2; x ∈ P, y ∈ Q}
i.e., R = {(x, y): y = x – 2 for x = 5, 6, 7)}

(ii) Roster form is R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7} = P,
Range of R = {3, 4, 5} = Q.

Ex 2.2 Class 11 Maths Question 5.
Let A = {1,2,3,4,6}. Let R be the relation on A defined by {(a, b): a, b ∈ A, b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R?.
Solution.
Given A = {1, 2, 3, 4, 6}
Given relation is R = {(a, b):a,b ∈ A, b is exactly divisible by a}
(i) Roster form of R = {(1,1), (1,2), (1,3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6) (4, 4), (6, 6)}.
(ii) Domain of R = {1, 2, 3, 4, 6} = A.
(iii) Range of R = {1, 2, 3, 4, 6} = A.

Ex 2.2 Class 11 Maths Question 6.
Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
Solution.
Given relation is R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5)}
= {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5,10)}
∴Domain of R = {0,1, 2, 3, 4, 5} and
Range of R = {5, 6, 7, 8, 9, 10}.

Ex 2.2 Class 11 Maths Question 7.
Write the relation R = {(x, x³): x is a prime number less than 10} in roster form.
Solution.
Given relation is R = {(x, x³): x is a prime number less than 10)
= {(x, x³): x ∈ {2, 3, 5, 7}}
= {(2, 23), (3, 33), (5, 53), (7, 73)}
= {(2, 8), (3, 27), (5, 125), (7, 343)}.

Ex 2.2 Class 11 Maths Question 8.
Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Solution.
Given A = {x, y, z} and B = {1, 2}
∴ n(A) = 3 & n(B) = 2
Since n(A x B) = n(A) x n(B)
∴ n(A x B) = 3 x 2 = 6
Number of relations from A to B is equal to the number of subsets of A x B.
Since A x B contains 6 elements.
⇒ Number of subsets of A x B = 2⁶ = 64
So, there are 64 relations from A to B.

Ex 2.2 Class 11 Maths Question 9.
Let R be the relation on Z defined by R = {{a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.
Solution.
Given relation is R = {(a, b): a, b ∈ Z, a – b is an integer}
If a, b ∈ Z, then a- b ∈ Z ⇒ Every ordered pair of integers is contained in R.
R = {(a, b) :a,b ∈ Z}
So, Range of R = Domain of R = Z.

We hope the NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2, drop a comment below and we will get back to you at the earliest.

Chapter 1 - Relations and functions EX 2.3 NCERT Solutions

Ex 2.3 Class 11 Maths Question 1.
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {{2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
(iii) {(1, 3), (1, 5), (2, 5)}.
Solution.
(i) We have a relation R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)} Since 2, 5, 8, 11, 14, 17 are the elements of domain of R having their unique images.
∴ The given relation is a function.
Hence domain = {2, 5, 8, 11, 14, 17) and Range = {1}.

(ii) We have a relation
R = {(2,1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Since 2, 4, 6, 8, 10, 12, 14 are the elements of domain of R having their unique images.
∴ The given relation is a function.
Hence domain = {2, 4, 6, 8, 10, 12, 14} and Range = {1, 2, 3, 4, 5, 6, 7}.

(iii) We have a relation R = {(1, 3), (1, 5), (2, 5)}
Since the distinct ordered pairs (1, 3) and (1, 5) have same first element i.e., 1 does not have a unique image under R.
∴ It is not a function.

Ex 2.3 Class 11 Maths Question 2.
Find the domain and range of the following real functions:
(i) f(x) = $-\left| x \right|$
(ii) f(x) = $\sqrt { 9-{ x }^{ 2 } }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3 1

Ex 2.3 Class 11 Maths Question 3.
A function f is defined by f (x) = 2x – 5. Write down the values of
(i) f (0)
(ii) f (7)
(iii) f (-3)
Solution.
We are given f (x) = 2x – 5
(i) f (0) = 2(0) – 5 = 0- 5 = -5
(ii) f (7) = 2(7) – 5 = 14- 5 = 9
(iii) f (-3) = 2(-3) – 5 = -6 – 5 = -11.

Ex 2.3 Class 11 Maths Question 4.
The function T which maps temperature in degree Celsius into temperature in degree by
$t(C)=\frac { 9C }{ 5 } +32$
Find
(i) t (0)
(ii) t (28)
(iii) t (-10)
(iv) The value of C, when t (C = 212
Solution.
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3 2
Ex 2.3 Class 11 Maths Question 5.
Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x>0.
(ii) f(x)=x²+ 2, x is a real number.
(iii) f (x) = x, x is a real number.
Solution.
(i) Given f (x) = 2 – 3x, x ∈ R, x > 0
∵ x > 0 ⇒ -3x < 0 ⇒ 2 – 3x < 2 + 0 ⇒ f (x) < 2
∴ The range of f (x) is (-2).

(ii) Given f (x) = x² + 2, x is a real number
We know x²≥ 0 ⇒ x² + 2 ≥ 0 + 2
⇒ x² + 2 > 2 ∴ f (x) ≥ 2
∴ The range of f (x) is [2, ∞).

(iii) Given f (x) = x, x is a real number.
Let y =f (x) = x ⇒ y = x
∴ Range of f (x) = Domain of f (x)
∴ Range of f (x) is R.

We hope the NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3, drop a comment below and we will get back to you at the earliest.

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Class 11th Chapter -1 Sets | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter 1 Sets NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Sets Class 11 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Sets NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter 1 SETS | NCERT MATHS SOLUTION |

Chapter 1 - Sets EX 1.1 NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1

Ex 1.1 Class 11 Maths Question 1.
Which of the following are sets? Justify your answer.
(i) The collection of all the months of a year beginning with the letter J.
(ii) The collection of ten most talented writers of India.
(iii) A team of eleven best-cricket batsmen of the world.
(iv) The collection of all boys in your class.
(v) The collection of all natural numbers less than 100.
(vi) A collection of novels written by the writer Munshi Prem Chand.
(vii) The collection of all even integers.
(viii) The collection of questions in this Chapter,
(ix) A collection of most dangerous animals of world.
Solution.
(i) The collection of all months of a year beginning with J is {J anuary, June, July}, which is well defined and hence it forms a set.
(ii) The collection of most talented writers of India is not well defined because opinions about ‘most talented writers’ vary from person to person and hence it does not form a set.
(iii) A team of eleven best-cricket batsmen of the world is not well defined because opinions about ‘best-cricket batsmen’ vary from person to person and hence it does not form a set.
(iv) The collection of all boys in your class is well defined and hence it forms a set.
(v) The collection of all natural numbers less than 100 is {1, 2, 3, 4,…………, 99}, which is well
defined and hence it forms a set.
(vi) A collection of novels written by the writer Munshi Prem Chand is well defined and hence it forms a set.
(vii) The collection of all even integers is {…………..,-4, -2, 0, 2, 4,……….. } which is well defined and hence it forms a set.
(viii) The collection of questions in this chapter is well defined and hence it forms a set.
(ix) A collection of most dangerous animals of the world is not well defined because opinions about ‘most dangerous animals’ vary from person to person and hence it does not form a set.

Ex 1.1 Class 11 Maths Question 2.
Let A = {1, 2, 3, 4, 5, 6). Insert the appropriate symbol ∈ or ∉ in the blank spaces:
(i) 5…A
(ii) 8…A
(iii) 0…A
(iv) 4…A
(v) 2…A
(vi) 10…A
Solution.
(i) Since 5 is the element of A. ∴ 5 ∉ A.
(ii) As 8 is not the element of A. ∴ 8 ∉ A
(iii) As 0 is not the element of A ∴ 0 ∈ A.
(iv) 4 is the element of A ∴ 4 ∈ A.
(v) 2 is the element of A ∴ 2 ∈ A.
(vi) 10 is not the element of A ∴ 10 ∉ A.

Ex 1.1 Class 11 Maths Question 3.
Write the following sets in roster form:
(i) A = {x: x is an integer and -3 < x < 7}
(ii) B = {x: x is a natural number less than 6}
(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}
(v) E = The set of all letters in the word TRIGONOMETRY
(vi) F = The set of all letters in the word
Solution.
(i) Integers lying between -3 and 7 are -2, -1, 0, 1, 2, ……….. , 6
∴ A = {-2,-1, 6}.

(ii) Natural numbers less than 6 are 1, 2, 3, 4, 5.
∴ B = 11, 2, 3, 4, 5}

(iii) Two digit natural numbers such that the sum of its digits is 8 are 17, 26, 35, 44, 53, 62, 71, 80.
∴ C= (17, 26, 35, 44, 53, 62, 71,80}

(iv) Prime number divisors of 60 are 2, 3, 5.
∴ D = (2, 3, 5}

(v) Word TRIGONOMETRY is formed by using the letters T, R, I, G, O, N, M, E, Y.
∴ E = (T, R, I, G, N, O, M, E, Y}

(vi) Word BETTER is formed by using the letters B, E, T, R
∴ F = (B, E, T, R}

Ex 1.1 Class 11 Maths Question 4.
Write the following sets in the set-builder form:
(i) {3, 6, 9, 12}
(ii) {2, 4, 8, 16, 32}
(iii) {5, 25, 125, 625}
(iv) {2,4,6,…}
(v) {1,4,9,……..,100}
Solution.
(i) Let A = (3, 6, 9, 12}
All elements of the set are natural numbers that are multiples of 3.
∴ A = (x : x = 3n, n∈N and 1 ≤ n ≤4}

(ii) Let B = (2, 4, 8, 16, 32} = (2¹, 2², 2³, 2⁴, 2⁵}
∴ B = {x : x = 2ⁿ, n ∈ N and 1 ≤ n ≤ 5}

(iii) Let C = (5, 25, 125, 625} = (5¹, 5², 5³, 5⁴}
∴ C = {x : x = 5ⁿ, n ∈ N and 1 ≤ n ≤ 4}

(iv) Let D = (2, 4, 6,……………..}
All elements of the set are even natural numbers.
∴ D = (x: x is an even natural number)

(v) Let E = {1,4,9,……….,100}
All elements of the set are perfect squares.
∴ E = {x: x = n², n ∈ N and 1 ≤ n ≤ 10}

Ex 1.1 Class 11 Maths Question 5.
List all the elements of the following sets:
(i) A = {x: x is an odd natural number}
(ii) B = {x: x is an integer, $-\frac { 1 }{ 2 }$ < x < $\frac { 9 }{ 2 }$ }
(iii) C = {x: x is an integer, x² ≤ 4}
(iv) D = {x: x is a letter in the word “LOYAL”}
(v) E = {x: x is a month of a year not having 31 days}
(vi) F = {x : x is a consonant in the English alphabet which precedes k}.
Solution.
(i) A = {x: x is an odd natural number}
∴ A = {1, 3, 5, 7,……………}

(ii) B = {x: x is an integer, $-\frac { 1 }{ 2 }$ < x < $\frac { 9 }{ 2 }$ }
∴ B = { 0, 1, 2, 3, 4}

(iii) C = {x: x is an integer, x² ≤ 4}
x² ≤ 4⇒ -2 ≤ x ≤ 2
∴ C = {-2, -1, 0, 1, 2}

(iv) D = {x: x is a letter in the word “LOYAL”}
∴ D = {L, O, Y, A}

(v) E = {x: x is a month of a year not having 31 days}
∴ E = {February, April, June, September, November}

(vi) F = {x: x is a consonant in the English alphabet which precedes k}
∴ F = {b, c, d, f, g, h, j}

Ex 1.1 Class 11 Maths Question 6.
Match each of the set on the left in the roster form with the same set on the right described in set-builder form:
i. {1,2, 3,6}   (a) {x: x is a prime number and a divisor of 6}
ii. {2,3}     (b) {x : x is an odd natural number less than 10}
iii. {M, A, T, H, E,I,C, S}   (c) {x: x is natural number and divisor of 6}

iv. {1, 3, 5, 7, 9} (d) {x : x is a letter of the word MATHEMATICS}.

Solution.
(i) → (c),
(ii) → (a),
(iii) → (d),
(iv) → (b).
The sets which are in set-builder form can be written in roster form as follows:
(a) {x : x is a prime number and a divisor of 6} = {2, 3}
(b) {x: x is an odd natural number less than 10} = {1, 3, 5, 7, 9}
(c) {x : x is natural number and divisor of 6} = {1, 2, 3, 6}
(d) {x: x is a letter of the word MATHEMATICS} = {M, A, T, H, E, I, C, S}

We hope the NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1, drop a comment below and we will get back to you at the earliest.

Chapter-1 Sets Ex. 1.2 NCERT Solution

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2

Ex 1.2 Class 11 Maths Question 1.
Which of the following are examples of the null set?
(i) Set of odd natural numbers divisible by 2
(ii) Set of even prime numbers
(iii) {x: x is a natural number, x ≤ 5 and x > 7}
(iv) {y: y is a point common to any two parallel lines}
Solution.
(i) Set of odd natural numbers divisible by 2 is a null set because odd natural numbers are not divisible by 2.
(ii) Set of even prime numbers is {2} which is not a null set.
(iii) {x: x is a natural number, x < 5 and x >7} is a null set because there is no natural number which satisfies x < 5 and x > 7 simultaneously,
(iv) [y: y is a point common to any two parallel lines) is a null set because two parallel lines
do not have any common point.

Ex 1.2 Class 11 Maths Question 2.
Which of the following sets are finite or infinite?
(i) The set of months of a year
(ii) {1,2,3,…}
(iii) {1,2,3, …,99,100}
(iv) The set of positive integers greater than 100
(v) The set of prime numbers less than 99
Solution.
(i) The set of months of a year is finite set because there are 12 months in a year.
(ii) {1, 2, 3, …} is an infinite set because there are infinite elements in the set.
(iii) {1, 2, 3, …, 99, 100) is a finite set because the set contains finite number of elements.
(iv) The set of positive integers greater than 100 is an infinite set because there are infinite
number of positive integers greater than 100.
(v) The set of prime numbers less than 99 is a finite set because the set contains finite number of elements.

Ex 1.2 Class 11 Maths Question 3.
State whether each of the following set is finite or infinite:
(i) The set of lines which are parallel to the x-axis
(ii) The set of letters in the English alphabet
(iii) The set of numbers which are multiple of 5
(iv) The set of animals living on the earth
(v) The set of circles passing through the origin (0,0)
Solution.
(i) The set of lines which are parallel to the x-axis is an infinite set because we can draw infinite number of lines parallel to x-axis.
(ii) The set of letters in the English alphabet is a finite set because there are 26 letters in the English alphabet.
(iii) The set of numbers which are multiple of 5 is an infinite set because there are infinite multiples of 5.
(iv) The set of animals living on the earth is a finite set because the number of animals living on the earth is very large but finite.
(v) The set of circles passing through the origin (0, 0) is an infinite set because we can draw infinite number of circles passing through origin of different radii.

Ex 1.2 Class 11 Maths Question 4.
In the following, state whether A = B or not:
(i) A = {a, b, c, d};B = {d, c, b, a}
(ii) A = {4, 8, 12, 16};B = {8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10}
B = {x : x is positive even integer and x≤ 10}
(iv) A = {x: x isa multiple of 10}
B = {10, 15, 20, 25, 30,…}
Solution.
(i) A = {a, b, c, d} and B = {d, c, b, a} are equal sets because order of elements does not changes a set.
∴ A = B = [a, b, c, d}.

(ii) A = {4, 8, 12, 16} and B = {8, 4, 16, 18} are not equal sets because 12 ∈ A but 12 ∉ B and 18 ∉ B but 18 ∉ A.

(iii) A = {2, 4, 6, 8,10} and B = {x: x is a positive even integer and x ≤ 10) which can be written in roster form as B = (2, 4, 6, 8, 10) are equal sets.
∴ A = B = {2, 4, 6, 8,10).

(iv) A = {x: x is a multiple of 10) can be written in roster form as A = {10, 20, 30, 40,…….. } and
B – {10, 15, 20, 25, 30, ………..} are not equal sets because 15 ∈ B but 15 ∉ A.

Ex 1.2 Class 11 Maths Question 5.
Are the following pair of sets equal ? Give reasons.
(i) A = {2, 3}, B={x: x is solution of x² + 5x + 6 = 0}
(ii) A = {x: x is a letter in the word FOLLOW}
B = {y: y is a letter in the word WOLF}
Solution.
(i) A = (2, 3} and B = {x: x is a solution of x² + 5x + 6 = 0}
Now, x² + 5x + 6 = 0 ⇒ x² + 3x + 2x + 6 = 0 ⇒ (x + 3)(x + 2) = 0 ⇒ x = -3, -2
∴ B = {-2, -3}
Hence, A and B are not equal sets.

(ii) A = {x : x is a letter in the word FOLLOW} = {F, O, L, W}
B = {y: y is a letter in the word WOLF}
= {W, O, L, F}
Hence, A = B = {F, O, L, W}.

Ex 1.2 Class 11 Maths Question 6
From the sets given below, select equal sets:
A = {2, 4, 8, 12),
B = {1, 2, 3, 4},
C = {4, 8, 12, 14},
D ={3,1,4,2},
E ={-1, 1},
F ={0, a},
G ={1, -1},
H ={0, 1}
Solution.
From the given sets, we see that sets B and D have same elements and also sets E and G have same elements.
∴ B = D = {1 ,2, 3, 4} and E = G = {-1, 1}.

We hope the NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2, drop a comment below and we will get back to you at the earliest.

Chapter-1 Sets Ex. 1.3 NCERT Solution

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3

Ex 1.3 Class 11 Maths Question 1.
Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:
(i) {2, 3, 4} …{1, 2, 3, 4, 5}
(ii) {a, b, c}… {b, c, d}
(iii) {x: x is a student of Class XI of your school} … {x: x student of your school}
(iv) {x : x is a circle in the plane}… {x: x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane}… {x : x is a rectangle in the plane}
(vi) {x: x is an equilateral triangle in a plane} … {x: x is a triangle in the same plane}
(vii) {x: x is an even natural number}… {x: x is an integer}
Solution.
(i) {2, 3, 4} ⊂ {11, 2, 3, 4, 5}
(ii) [a, b, c) ⊄ {{b, c, d}
(iii) {x : x is a student of Class XI of your school} ⊂ {x : x student of your school}
(iv) {x : x is a circle in the plane} ⊄ {x : x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane} ⊄ {x : x is a rectangle in the plane}
(vi) {x : x is an equilateral triangle in a plane} ⊂ {x : x is a triangle in the same plane}
(vii) {x: x is an even natural number} ⊂ {x: x is an integer}

Ex 1.3 Class 11 Maths Question 2.
Examine whether the following statements are true or false:
(i) {a, b} ⊄{b, c, a}
(ii) {a, e} ⊂ {x : x is a vowel in the English alphabet}
(iii) {1, 2, 3} ⊂ {1, 3, 5}
(iv) {a} ⊂ {a, b, c}
(v) {a} ∈ la, b, c}
(vi) {x: x is an even natural number less than 6} ⊂ {x: x is a natural number which divides 36}
Solution.

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3 2

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3 3

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3 4

Ex 1.3 Class 11 Maths Question 3.
Let A = {1, 2, {3, 4}, 5}. Which of the following statements are incorrect and why?

Solution.

Ex 1.3 Class 11 Maths Question 4.
Write down all the subsets of the following sets
(i) {a}
(ii) {a,b}
(iii) {1,2,3}
(iv) φ
Solution.
(i) Number of elements in given set = 1
Number of subsets of given set = 2¹ = 2
∴ Subsets of given set are φ , {a}.

(ii) Number of elements in given set = 2
Number of subsets of given set = 2¹² = 4
∴ Subsets of given set are φ, {a}, {b}, {a, b}.

(iii) Number of elements in given set = 3
Number of subsets of given set = 2³ = 8
Subsets of given set are φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}.

(iv) Number of elements in given set = 0
Number of subsets of given set = 2⁰= 1
∴ Subset of given set is φ.

Ex 1.3 Class 11 Maths Question 5.
How many elements has P(A), if A = φ?
Solution.
Number of elements in set A = 0
Number of subset of set A = 2⁰ = 1
Hence, number of elements of P(A) is 1.

Ex 1.3 Class 11 Maths Question 6.
Write the following as intervals:
(i) {x: x ∈ R, -4 < x ≤ 6}
(ii) {x: x ∈ R, -12 < x < -10}
(iii) {x: x ∈ R, 0 ≤ x < 7}
(iv) {x: x ∈ R, 3 ≤ x ≤ 4}
Solution.
(i)Let A = {x: x ∈ R, -4 < x ≤ 6}
It can be written in the form of interval as (-4, 6)
(ii) Let A= {x: x ∈ R, -12 < x < -10}
It can be written in the form of interval as (-12, -10)
(iii) Let A = {x: x ∈ R, 0 ≤ x < 7}
It can be written in the form of interval as (0, 7).
(iv) Let A = {x: x ∈ R, 3 ≤ x ≤ 4}
It can be written in the form of interval as (3,4).

Ex 1.3 Class 11 Maths Question 7.
Write the following intervals in set-builder form:
(i) (-3,0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [-23, 5)
Solution.
(i) The interval (-3, 0) can be written in set-builder form as {x : x ∈ R,-3 < x < 0}.
(ii) The interval [6, 12] can be written in set-builder form as {x : x ∈ R, 6 ≤ x ≤ 12}.
(iii) The interval (6, 12] can be written in set-builder form as {x : x ∈ R, 6 < x ≤ 12}
(iv) The interval [-23,5) can be written in set-builder form as {x : x ∈ R, -23 ≤ x < 5}

Ex 1.3 Class 11 Maths Question 8.
What universal set(s) would you propose for each of the following:
(i) The set of right triangles.
(ii) The set of isosceles triangles.
Solution.
(i) Right triangle is a type of triangle. So the set of triangles contain all types of triangles.
∴ U = {x : x is a triangle in a plane}

(ii) Isosceles triangle is a type of triangle. So the set of triangles contain all types of triangles.
∴ U = }x : x is a triangle in a plane}

Ex 1.3 Class 11 Maths Question 9.
Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set(s) for all the three sets A, B and C
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) φ
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) {1, 2, 3, 4, 5, 6, 7, 8}
Solution.

(i) {0, 1, 2, 3, 4, 5, 6} is not a universal set for A, B, C because 8 ∈ C but 8 is not a member of {0, 1, 2, 3, 4, 5, 6}.
(ii) φ is a set which contains no element. So it is not a universal set for A, B, C.
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is a universal set for A, B, C because all members of A, B, C are present in {0,1 , 2, 3, 4, 5, 6, 7, 8, 9, 10).
(iv) (1, 2, 3, 4, 5, 6, 7, 8) is not a universal set for A, B, C because 0 ∈ C but 0 is not a member of {1, 2, 3, 4, 5, 6, 7, 8)

We hope the NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3, drop a comment below and we will get back to you at the earliest.

Chapter-1 Sets Ex. 1.4 NCERT Solution

Ex 1.4 Class 11 Maths Question 1.
Find the union of each of the following pairs of sets:
(i) X = {1 ,3, 5}, Y= {1, 2, 3}
(ii) A = {a, e, i, o, u}, B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6} B = (x:x is a natural number and 6 <x< 10}
(v) A = {1, 2, 3}, B = φ
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4 1

Ex 1.4 Class 11 Maths Question 2.
Let A = {a, b}, B = {a, b, c}. Is A ⊂ B ? What is A ∪B?
Solution.
Here A = {a, b} and B = {a, b, c}. All elements of set A are present in set B.
∴ A ⊂ B. Now, A ∪ B = {a, b, c) = B.

Ex 1.4 Class 11 Maths Question 3.
If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Solution.
Here A and B are two sets such that A ⊂ B.
Take A = {1, 2} and B = {1, 2, 3}.
A ∪ B = {1, 2, 3) = B.

Ex 1.4 Class 11 Maths Question 4.
If A = {11, 2, 3, 4}, B = {3, 4, 5, 6}, C={5, 6, 7, 8} and D = {7, 8, 9, 10}; find
(i) A ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) B ∪ O
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vii) B ∪ C ∪ D
Solution.
Here A = {11, 2, 3, 4}, B = {3, 4, 5, 6}, C={5, 6, 7, 8} and D = {7, 8, 9, 10}
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4 2

Ex 1.4 Class 11 Maths Question 5.
Find the intersection of each pair of sets of .
(i) X = {1 ,3, 5}, Y= {1, 2, 3}
(ii) A = {a, e, i, o, u}, B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6} B = (x:x is a natural number and 6 <x< 10}
(v) A = {1, 2, 3}, B = φ
Solution.
(i) Here X = {1, 3, 5} and Y = {1, 2, 3}
∴ X ∩ Y= {1,3}

(ii) Here A = {a, e, i, o, u} and B = {a, b, c}
∴ A ∩ B = {a}

(iii) Here A = {x: x is a natural number and multiple of 3} = {3, 6, 9,12,….} and B = {x: x is a natural number less than 6}
= {1, 2, 3, 4, 5} ∴ A ∩ B = {3}

(iv) Here A = {x: x is a natural number and 1 < x < 6} ={2, 3, 4, 5, 6} and B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9} ∴ A ∩ B = φ

(v) Here A = {1, 2, 3) and B = φ
∴ A ∩ B = φ

Ex 1.4 Class 11 Maths Question 6.
If A = (3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find
(i) A ∩ B
(ii) B ∩ C
(iii) A ∩ C ∩ D
(iv) A ∩ C
(v) B ∩ D
(vi) A ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪ D)
(ix) (A ∪ B) ∩ (B ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)
Solution.
Here A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4 3

Ex 1.4 Class 11 Maths Question 7.
If A = {x: x is a natural number), B = {x: x is an even natural number}, C={x : x is an odd natural number} and D = {x: x is a prime number}, find
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Solution.
Here A = {x: x is a natural number}
= (1, 2, 3, 4, 5, …….}
B = {x: x is an even natural number}
= 12, 4, 6,………}
C = {x: x is an odd natural number}
= {1, 3, 5, 7,………}
and D = {x: x is a prime number}
= {2, 3, 5, 7,….}

(i) A ∩ B = {x: x is a natural number} ∩ {x: x is an even natural number}
= {x: x is an even natural number} = B.

(ii) A ∩ C = {x: x is a natural number} ∩ {x: x is an odd natural number}
= {x: x is an odd natural number} = C.

(iii) A ∩ D = {x: x is a natural number} ∩ {x: x is a prime number}
= {x: x is a prime number} = D.

(iv) B ∩ C = {x: x is an even natural number} ∩{x: x is an odd natural number} = φ .

(v) B ∩ D = [x: x is an even natural number} ∩ {x: x is a prime number} = {2}.

(vi) C ∩ D = {x: x is an odd natural number} ∩ {x: x is a prime number} = {x: x is an odd prime number}.

Ex 1.4 Class 11 Maths Question 8.
Which of the following pairs of sets are disjoint?
(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6}
(ii) {a, e, i, o, u] and {c, d, e, f}
(iii) {x: x is an even integer} and {x: x is an odd integer}
Solution.
(i) Let A = {1,2,3,4}
and B = {x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}
∴ A ∩ B = {1,2,3,4} n {4,5, 6} = {4}
Hence A and B are not disjoint sets.

(ii) Let A = {a, e, i, o, u} and B = {c, d, e, f}
∴ A ∩ B = {e}
Hence A and B are not disjoint sets.

(iii) Let A = {x : x is an even integer} and B = {x: x is an odd integer}
∴ A ∩ B = φ. Hence A and B are disjoint sets.

Ex 1.4 Class 11 Maths Question 9.
If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}; find
(i) A – B
(ii) A – C
(iii) A – D
(iv) B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(x) D – B
(xi) C – D
(xii)D – C
Solution.
Here A = {3, 6, 9, 12, 15, 18, 21},
B = {4, 8, 12, 16, 20},
C ={2, 4, 6, 8, 10, 12, 14, 16},
D = {5, 10, 15, 20}
(i) A – B = {3, 6, 9, 12, 15, 18, 21} – {4, 8,12,16, 20} = {3, 6, 9,15,18, 21}
(ii) A – C = {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16} = {3, 9, 15, 18, 21}
(iii) A – D = {3, 6, 9, 12, 15, 18, 21} – {5,10,15, 20} = {3, 6, 9, 12, 18, 21}
(iv) B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21} = {4, 8,16, 20}
(v) C – A = {2,4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21} = {2, 4, 8, 10, 14, 16}
(vi) D – A = {5, 10, 15, 20} – {3, 6, 9, 12, 15, 18, 21} = {5, 10, 20}
(vii) B – C={4, 8, 12, 16, 20} – {2, 4, 6, 8, 10, 12, 14, 16} = {20}
(viii) B – D = {4, 8, 12, 16, 20} – {5, 10, 15, 20} = {4, 8, 12, 16}
(ix) C – B = {2,4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20} = {2, 6, 10, 14}
(x) D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20} = {5, 10, 15}
(xi) C – D = {2, 4, 6, 8, 10, 12, 14, 16} – {5, 10, 15, 20} = {2, 4, 6, 8, 12, 14, 16}
(xii) D – C={5, 10, 15, 20} – {2, 4, 6, 8, 10, 12, 14, 16} = {5, 15, 20}

Ex 1.4 Class 11 Maths Question 10.
If X= {a, b, c, d} and Y={f, b, d, g}, find
(i) X – Y
(ii) Y – X
(iii) X ∩ Y
Solution.
Here X = {a, b, c, d} and Y = {f, b, d, g}
(i) X – Y = {a, b, c, d} – {f, b, d, g} = {a, c}
(ii) Y – X = {f, b, d, g} – {a, b, c, d} = {f, g}
(iii) X ∩ Y = {a, b, c, d} ∩ {f, b, d, g} = {b, d}

Ex 1.4 Class 11 Maths Question 11.
If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
Solution.
We know that set of real numbers contain rational and irrational numbers. So R – Q = set of irrational numbers.

Ex 1.4 Class 11 Maths Question 12.
State whether each of the following statement is true or false. Justify your answer.
(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
(ii) {a, e, i, o, u} and {a, b, c, d} are disjoint sets.
(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4 5

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Chapter-1 Sets Ex. 1.5 NCERT Solution

Ex 1.5 Class 11 Maths Question 1.
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = { 1, 2, 3, 4}, B = (2,4,6,8} and C = {3,4,5,6}. Find
(i) A’
(ii) B’
(iii) (A ∪ C)’
(iv) (A ∪B)’
(v) (A’)’
(vi) (B – C)’
Solution.
Here U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C ={3, 4, 5, 6}
(i) A’=U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}

(ii) B’=U – B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}

(iii) A ∪ C = {1, 2, 3, 4} ∪ {3, 4, 5, 6}
= (1, 2, 3, 4, 5, 6}
(A∪C)’=U-(A∪C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6}
= {7, 8, 9}

(iv) A ∪ B = {1, 2, 3,4} ∪ {2, 4, 6, 8}
= {1, 2, 3, 4, 6, 8}
(A∪B)’ = U – (A∪B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8}
= {5, 7, 9}

(v) We know that A’ = {5, 6, 7, 8, 9}
(A’)’ =U – A’
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9}
= {1, 2, 3, 4}

(vi) B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {2, 8}
(B-C)’=U – (B-C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8}
= {1, 3, 4, 5, 6, 7, 9}.

Ex 1.5 Class 11 Maths Question 2.
If U = {a,b, c, d, e, f, g, h}, find the complements of the following sets:
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}
Solution.
(i) A’ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c}
= {d, e,f, g, h}

(ii) B’ = U – B = {a, b, c, d, e,f, g, h} – {d, e, f, g}
= {a, b, c, h}

(iii) C’ = U – C = {a, b, c, d, e, f, g, h} – {a, c, e, g}
= {b, d, f, h}

(iv) D’ = U – D = {a, b, c, d, e, f, g, h} – {f, g, h, a}
= {b, c, d, e}.

Ex 1.5 Class 11 Maths Question 3.
Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x: x is an even natural number}
(ii) {x: x is an odd natural number}
(iii) {x: x is a positive multiple of 3}
(iv) {x: x is a prime number}
(v) {x: x is a natural number divisible by 3 and 5}
(vi) {x: x is a perfect square}
(vii) {x: x is a perfect cube}
(viii) {x: x + 5 = 8}
(ix) (x: 2x + 5 = 9)
(x) {x: x ≥ 7}
(xi) {x: x ∈ W and 2x + 1 > 10}
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 1

Ex 1.5 Class 11 Maths Question 4.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that
(i) (A ∪ B)’ = A’∩B’
(ii) (A ∩ B)’ = A’∪B’
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 4

Ex 1.5 Class 11 Maths Question 5.
Draw appropriate Venn diagram for each of the following:
(i) (A ∪ B)’
(ii) A’∩B’
(iii) (A ∩ B)’
(iv) A’ ∪ B’
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 6

Ex 1.5 Class 11 Maths Question 6.
Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A’?
Solution.
Here U = {x : x is a triangle}
A = {x: x is a triangle and has at least one angle different from 60°}
∴ A’ = U – A = {x : x is a triangle} – {x : x is a triangle and has atleast one angle different from 60°}
= {x : x is a triangle and has all angles equal to 60°)
= Set of all equilateral triangles.

Ex 1.5 Class 11 Maths Question 7.
Fill in the blanks to make each of the following a true statement:
(i) A ∪ A’ = …….
(ii) φ’ ∩ A = .…….
(iii) A ∩ A’ = …….
(iv) U’ ∩ A = .…….
Solution.
(i) A ∪ A’= U
(ii) φ’ ∩ A = U ∩ A = A
(iii) A ∩ A’ = φ
(iv) U’ ∩ A = φ ∩ A = φ

We hope the NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5, drop a comment below and we will get back to you at the earliest.

Chapter-1 Sets Ex. 1.6 NCERT Solution

Ex 1.6 Class 11 Maths Question 1.
If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n (X ∪ Y) = 38, find n(X ∩ Y).
Solution.
Here n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38
We know that
n(X ∪ Y) = n(X) + n(Y) -n(X ∩ Y)
⇒ 38 = 17 + 23 – n(X ∩ Y)
∴ n(X ∩ Y) = 40 – 38 = 2.

Ex 1.6 Class 11 Maths Question 2.
If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩ Y have?
Solution.
Here n(X ∪ Y) = 18. n(X) = 8 and n(Y) = 15
We know that
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 18 = 8 +15 – n(X ∩ Y)
∴ n(X ∩ Y) = 23-18 = 5.

Ex 1.6 Class 11 Maths Question 3.
In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?
Solution.
Let H be the set of people speaking Hindi and E be the set of people speaking English.
∴ n(H) = 250, n(E) = 200 and n(H ∪ E) = 400,
We know that
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
400 = 250 + 200 – n(H ∩ E)
∴ n(H ∩ E) = 450 – 400 = 50.

Ex 1.6 Class 11 Maths Question 4.
If 5 and Tare two sets such that 5 has 21 elements, T has 32 elements, and S ∩T has 11 elements, how many elements does S ∪ T have?
Solution.
Here n(S) = 21, n(T) = 32 and n(S ∩T) = 11
We know that
n(S ∪ T) = n(S) + n(T) – n(S ∩ T) n(S ∪ T)
= 21 + 32 – 11 = 42.

Ex 1.6 Class 11 Maths Question 5.
If X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements, and X ∩ Y has 10 elements, how many elements does X have?
Solution.
Here n{X) = 40, n(X ∪ Y) = 60 and n(X ∩ Y) = 10
We know that
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 60 = 40 + n(Y) – 10
∴ n(Y) = 60 – 30 = 30.

Ex 1.6 Class 11 Maths Question 6.
In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea?
Solution.
Let C be the set of persons who like coffee and T be the set of persons who like tea.
∴ n(C) = 37, n(T) = 52 and n(C ∪ T) = 70
We know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 70 = 37 + 52 – n(C ∩ T)
∴ n(C ∩ T) = 89 – 70 = 19.

Ex 1.6 Class 11 Maths Question 7.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Solution.
Let C be the set of people who like cricket and T be the set of people who like tennis. Here n(Q) = 40, n(C ∩ T) = 10 and n(C ∪ T) = 65 .
We know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 65 = 40 + n(T) -10
⇒ n{T} = 65 – 30 = 35
∴ Number of people who like tennis = 35 j Now number of people who like tennis only and not cricket
=n(T – C) = n(T) – n(C ∩ T) = 35 – 10 = 25.

Ex 1.6 Class 11 Maths Question 8.
In a committee, 50 people speak French, 20 f speak Spanish and 10 speak both Spanish and
French. How many speak at least one of these two languages?
Solution.
Let F be the set of people who speak French and S be the set of people who speak Spanish.
Here n(F) = 50, n(S) = 20 and n(F ∩ S) = 10
We know that
n(F ∪ S) = n(F) + n(S) – n(F ∩ S)
n(F ∪ S) = 50 + 20 -10 = 60
∴ Number of people who speak at least one of these two languages = 60

We hope the NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6, drop a comment below and we will get back to you at the earliest.

Chapter-1 Sets Mislleneous NCERT Solution

Miscellaneous Exercise Class 11 Maths Question 1.
Decide, among the following sets, which sets are subsets of one and another:
A = {x : x ∈R and x satisfy x² – 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8,…}, D = {6}.
Solution.
Here A = {x : x ∈ R and x satisfies x² – 8x + 12 = 0}
= {x : x ∈ R and (x – 6)(x – 2) = 0} = {2, 6}
B = {2, 4, 6}, C = {2, 4, 6, 8,…….} and D = {6}
Now, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B and D ⊂ C

Miscellaneous Exercise Class 11 Maths Question 2.
In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B,then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 1

Miscellaneous Exercise Class 11 Maths Question 3.
Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B – C.
Solution.
Given that A ∩ B = A ∩C and A ∪ B=A ∪ C
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 2

Miscellaneous Exercise Class 11 Maths Question 4.
Show that the following four conditions are equivalent :
(i) A ⊂ B
(ii) A – B = φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Solution.
(i) ⇒ (ii)
A – B = {x: x∈ A and x∉B]
Since A⊂B
∴ A – B = φ

(ii) ⇒ (iii)
A – B = φ ⇒ A⊂B ⇒ A∪B = B

(iii) ⇒ (iv)
AuB = B ⇒A⊂B ⇒ A∩B = A

(iv) ⇒ (i)
A∩B = A ⇒ A⊂B
Thus (i) ⇔ (ii) ⇔ (iii) ⇔ (iv).

Miscellaneous Exercise Class 11 Maths Question 5.
Show that if A ⊂ B, then C – B ⊂ C – A.
Solution.
Let x ∈ C – B ⇒x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [∵ A ⊂ B]
⇒ x ∈ C – A
Hence C – B ⊂C – A

Miscellaneous Exercise Class 11 Maths Question 6.
Assume that P(A) = P(B). Show that A = B
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 3

Miscellaneous Exercise Class 11 Maths Question 7.
Is it true that for any sets A and B, P(A) ∪ P(B) = P(A ∪ B) Justify your answer.
Solution.
No, it is not true.
Take A = {1, 2} and B = {2,3}
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 4

Miscellaneous Exercise Class 11 Maths Question 8.
Show that for any sets A and B,
A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
Solution.
(A ∩ B) ∪ (A – B) = (A ∩ B) ∪ (A ∩ B’)
= A ∩ (B ∪ B’) (By distributive law)
= A ∩ U = A
Hence, A = (A ∩ B) ∪ (A – B)
Also A ∪ (B – A) = A ∪ (B ∩ A’)
= (A ∪ B) ∩ (A ∪ A’) (By distributive law)
= (A ∪ B) ∩ U= A ∪ B
Hence, A ∪ (B – A) = A ∪ B.

Miscellaneous Exercise Class 11 Maths Question 9.
Using properties of sets, show that
(i) A ∪ (A ∩ B)=A
(ii) A ∩ (A ∪ B) = A.
Solution.
(i) We know that if A ⊂ B, then
A ∪ B = B. Also, A ∩ B ⊂ A
∴ A ∪ (A ∩ B) = A.
(ii) We know that if A ⊂ B,
then A ∩ B = A Also, A ⊂ A ∪ B
∴ A ∩ (A ∪ B) = A.

Miscellaneous Exercise Class 11 Maths Question 10.
Show that A ∩ B = A ∩ C need not imply B = C.
Solution.
Let A = {1, 2, 3, 4}, B = {2, 3, 4, 5, 6}, C = {2, 3, 4, 9,10}.
∴ A ∩ B = [1, 2,3,4} ∩ {2,3,4, 5, 6]
= {2, 3, 4}
A ∩ C = {1, 2, 3, 4} ∩ {2, 3, 4, 9, 10} = {2, 3, 4}
Now we have A ∩ B = A ∩ C. But B ≠ C.

Miscellaneous Exercise Class 11 Maths Question 11.
Let A and B be sets. If A ∩ X=B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints A = A ∩ (A∪X), B = B ∩ (B ∪ X) and use Distributive law)
Solution.
Here
A ∪ X = B ∪ X for some set X
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 5

Miscellaneous Exercise Class 11 Maths Question 12.
Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ
Solution.
Take A = {1, 2}, B = {1, 4} and C = {2, 4}
Now, A ∩ B = {1} ≠ φ, B ∩ C = {4} ≠ φ and
A ∩ C = {2} ≠ φ
But A ∩ B ∩ C = φ.

Miscellaneous Exercise Class 11 Maths Question 13.
In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Solution.
Let T be the set of students who takes tea and C be the set of students who takes coffee. Here, n(T) = 150, n(C) = 225 and n(C ∩ T) = 100
We know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
= 150 + 225 -100 = 275
∴ Number of students taking either tea or coffee = 275
∴ Number of students taking neither tea nor coffee = 600 – 275 = 325.

Miscellaneous Exercise Class 11 Maths Question 14.
In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Solution.
Let H be the set of students who know Hindi and E be the set of students who know English.
Here n(H) = 100, n(E) = 50 and n(H ∩ E) = 25
We know that
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25 = 125

Miscellaneous Exercise Class 11 Maths Question 15.
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper H, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T
and I, 3 read all three newspapers. Find:
(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 6

Miscellaneous Exercise Class 11 Maths Question 16.
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 8

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Class 11th Chapter -4 Principle of Mathematical Induction | NCERT Maths Solution | NCERT Solution | Edugrown

Class 11th Chapter 4 Principle of Mathemarical Induction| NCERT MATHS SOLUTION |

Class 11th Chapter -3 Trigonometric Functions | NCERT Maths Solution | NCERT Solution | Edugrown

Class 11th Chapter 3 Trigonometric Fuctions | NCERT MATHS SOLUTION |

Class 11th Chapter -2 Relations and Functions | NCERT Maths Solution | NCERT Solution | Edugrown

Class 11th Chapter 1 Relations and Functions | NCERT MATHS SOLUTION |

Class 11th Chapter -1 Sets | NCERT Maths Solution | NCERT Solution | Edugrown

Class 11th Chapter 1 SETS | NCERT MATHS SOLUTION |

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3

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