As part of systematic reforms, the National Curriculum Framework 2005 recommends that subjects like mathematics and English could be examined at two different levels – standard and high level. At the same time, they want to use the topic for business, data interpretation, finance, graphical representation, etc. etc. CBSE Class 11 applied maths algebra comprises 10 marks.
The objectives of introducing this new module are:
- Students will develop the skills of logical reasoning and apply the same while solving problems.
- A sense is developed from representing data, data interpreting, data analyzing and data organizing, and meaningful inferences are made from that in the real-life problem.
- A connection is developed between mathematics and other disciplines.
CBSE Class 11 Applied Maths Algebra 2020-21 | Topics and Questions
Following are the algebra topics in applied maths in CBSE class 11 along with some of the crucial questions:
Name of the topics |
1. Sets |
2. Types of sets |
3. Venn diagram |
4. De Morgan’s Law |
5. Problem solving using Venn Diagram |
6. Relations and types of relations |
7. Introduction of sequences, series |
8. Arithmetic and Geometric progression |
9. Relationship between AM and GM |
10. Basic concepts of Permutations and Combinations |
11. Permutations, Circular Permutations, Permutations with restrictions |
12. Combinations with standard results |
Following are the algebra topics in applied maths in CBSE class 11 along with some of the important questions:
Question 1:
Write the following sets in the roaster form.
(i) A = {x | x is a positive integer less than 10 and 2x – 1 is an odd number}
(ii) C = {x : x2 + 7x – 8 = 0, x ∈ R}
Question 2:
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}
Question 1:
Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}. Write the set (A U B)’.
Question 2:
Let U = {1, 2, 3, 4, 5, 6, 7}, A = {2, 4, 6}, B = {3, 5} and C = {1, 2, 4, 7}, find
(i) A′ ∪ (B ∩ C′)
(ii) (B – A) ∪ (A – C)
Question 1:
In a class of 50 students, 10 take Guitar lessons and 20 take singing classes, and 4 take both. Find the number of students who don’t take either Guitar or singing lessons.
Question 1:
Prove De Morgan’s Laws by Venn Diagram
(i) (A∪B)’= A’∩ B’
- Problem-solving using Venn diagram
Same as Venn diagram question
- Relations and types of relations
Question 1:
Write the range of a Signup function.
Question 2:
The Cartesian product A × A has 9 elements, among which are found (–1, 0) and (0,1). Find the set A and the remaining aspects of A × A.
- Introduction of Sequences, Series
Question 1:
The sums of n terms of two arithmetic progressions are in the ratio 5n+4: 9n+6. Find the ratio of their 18th terms.
Question 2:
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
- Arithmetic and Geometric progression
Question 1
Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Question 2
What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?
- 897
- 1,64,850
- 1,64,749
- 1,49,700
- Relationship between AM and GM
Question 1
Find the AM, GM, and HM between 12 and 30
- Basic concepts of Permutations and Combinations
Question 1:
Find the 3-digit numbers that can be formed from the given digits: 1, 2, 3, 4 and 5, assuming that
- a) digits can be repeated.
- b) digits are not allowed to be repeated.
Question 2:
A coin is tossed 6 times, and the outcomes are noted. How many possible results can be there?
- Permutations, Circular Permutations, Permutations with restrictions
Question 1:
From a team of 6 students, in how many ways can we choose a captain and vice-captain assuming one person cannot hold more than one position?
Question 2:
Find the number of ways in which 10 beads can be arranged to form a necklace.
Question 3:
Find the number of ways in which four girls and three boys can arrange themselves in a row so that none of the boys is together? How is this arrangement different from that in a circular way?
- Combinations with standard results
Question 1:
How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word – DAUGHTER?
Question 2:
Find the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
You are also advised to know some of the CBSE class 11 applied maths algebraic expressions. Algebra is basically the study of unknown quantities. Some of the topics of algebraic expressions and formulas are:
Algebraic identities
Various equality equations are consisting of different variables in algebraic identities.
- a) Linear Equations in One Variable: A linear equation in one variable has the maximum of one variable present in the order 1. It is depicted in the form of ax + b = 0, where x is represented as the variable.
- b) Linear Equations in Two Variables: A linear equation in two variables consists of the utmost two variables present in order 2. The equation is depicted in the form: ax2 + bx + c = 0. The two variables are quite important because your coursebook has a lot of questions based on it. So, you need to stay focussed on important algebra formulas to find the solution.
Some basic identities to note are:
- The combination of literal numbers obeys every basic rule of addition, subtraction, multiplication and division.
- x × y = xy; such as 5 × a = 5a = a × 5.
- a × a × a × … 9 more times = a12
- If a number is x8, then x is the base and 8 is the exponent.
- A constant is a symbol with a fixed numerical value.
- Law of exponent
The degrees and powers in any mathematical expressions are known as exponent. Some of the laws of exponent are:
- a0 = 1
- a-m = 1/am
- (am)n = amn
- am / an = am-n
- am x bm = (ab)m
- am / bm = (a/b)m
- (a/b)-m =(b/a)m
- (1)n = 1 for infinite values of n
- Quadratic equations
The linear equations in two variables are known as quadratic equations.
The roots of the equation ax2 + bx + c = 0 (where a ≠ 0) can be given as:
−b±b2−4ac√2a
Below given are some important points about the equation as a part of important algebra formulas:
- Δ = b2 − 4ac is also known as a discriminant.
- For roots;
- Δ > 0 happens when the roots are real and distinct
- For real and coincident roots, Δ = 0
- Δ < 0 happens in the case when the roots are non-real
- If α and β are the two roots of the equation ax2 + bx + c then,
α + β = (-b / a) and α × β = (c / a).
- If the roots of a quadratic equation are α and β, the equation will be
(x − α)(x − β) = 0.
The general algebra formulas can be given as:
- n is a natural number: an – bn = (a – b)(an-1 + an-2b+…+ bn-2a + bn-1)
- If n is even: (n = 2k), an + bn = (a – b)(an-1 + an-2b +…+ bn-2a + bn-1)
- n is odd: (n = 2k + 1), an + bn = (a + b)(an-1 – an-2b +an-3b2…- bn-2a + bn-1)
- General square Formula: (a + b + c + …)2 = a2 + b2 + c2 + … + 2(ab + ac + bc + ….)
- List of important formulas
* (a + b)2 = a2 + 2ab + b2
* (a – b)2 = a2 – 2ab + b2
* (a + b) (a – b) = a2 -b2
* (x + a) (x + b) = x2 + (a + b) x + ab
* (x + a) (x – b) = x2 + (a – b) x – ab
* (x – a) (x + b) = x2 + (b – a) x – ab
* (x – a) (x – b) = x2 – (a + b) x + ab
* (a + b)3 = a3 + b3 + 3ab (a + b)
* (a – b)3 = a3 – b3 – 3ab (a – b)
* (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
* (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4
* (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2xz
* (x + y – z)2 = x2 + y2 + z2 + 2xy – 2yz – 2xz
* (x – y + z)2 = x2 + y2 + z2 – 2xy – 2yz + 2xz
* (x – y – z)2 = x2 + y2 + z2 – 2xy + 2yz – 2xz
* x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz -xz)
* x2 + y2 = 12
* [(x + y)2 + (x – y)2]
* (x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
* x3 + y3 = (x + y) (x2 – xy + y2)
* x3 – y3 = (x – y) (x2 + xy + y2)
* x2 + y2 + z2 – xy – yz – z [(x – y)2 + (y – z)2 + (z – x)2]
Topics included in CBSE Class 11 Applied Maths Algebra
Below mentioned topics are included in the CBSE Class 11 applied maths Algebra:
- Sets
- Types of sets
- Venn diagram
- De Morgan's laws
- Problem-solving using Venn diagram
- Relations and types of relations
- Introduction of Sequences, Series
- Arithmetic and Geometric progression
- Relationship between AM and GM
- Basic concepts of Permutations and Combinations
- Permutations, Circular Permutations, Permutations with restrictions
- Combinations with standard results
NCERT Solution for Class 11 Applied Math Algebra PDF
NCERT Solutions for CBSE Class 11 Applied mathematics Algebra is an essential resource for preparing for the class 11 exam. It will help you to clear all your doubts with their solved question paper samples, tips, notes, and summary.
Students are asked to practice the NCERT solutions for applied maths algebra part in CBSE class 11 syllabus pdf regularly as it will help them score good marks in the competitive exams. You will be able to solve sample question papers chapter wise that will help you to know all the details of the topics.
NCERT published students and teachers like textbooks. They have a straightforward way of illustrations and explanations. The format of these textbooks is straightforward and easy. The content of the book and its problem-solving procedures are straightforward; therefore, they are liked by students. The more straightforward solutions do not compromise on the quality of the content.
The syllabus in these books is covered very clearly so that students do not miss out any topic from the curriculum.
Complex Number Class 11 Notes with Examples
Complex Numbers Class 11 is defined when a number can be represented in form p + iq. Here, p and q are real numbers and i=−1−−−√. For a complex number z = p + iq, p is known as the real part, represented by Re z, and q is known as the imaginary part; it is represented by Im z of complex number z.
Concepts of complex numbers class 11
The topics and the subtopics taught in Complex numbers CBSE class 11 applied maths algebra are:
- Introduction
- Complex Numbers
- Algebra of Complex Numbers
- Addition of two complex Numbers
- The difference between two complex Numbers
- Multiplication of two complex Numbers
- Division of two complex Numbers
- Power of i
- The square root of a negative real number
- Identities
- The Modulus and the Conjugate of Complex Numbers
- Argand Plane and Polar Representation
- Polar Representation of Complex Numbers
The notes for class 11 gives a detailed knowledge describing the concepts involved in complex numbers. Some of the examples are:
Example 1:
If 4x + i(3x – y) = 3 + i (– 6), where x and y are real numbers, then find the values of x and y.
Solution: Given,
4x + i (3x – y) = 3 + i (–6) ….(1)
By equating the real and the imaginary parts of equation (1),
4x = 3, 3x – y = –6,
Now, 4x = 3
⇒ x = 3/4
And 3x – y = -6
⇒ y = 3x + 6
Substituting the value of x,
⇒ y = 3(3/4) + 6
⇒ y = 33/4
Therefore, x = 3/4 and y = 33/4.
Example 2:
Express (-√3 + √-2)(2√3 – i) in the form of a + ib.
Solution: We know that i2 = -1
(-√3 + √-2)(2√3 – i) = (-√3 + i√2)(2√3 – i)
= (-√3)(2√3) + (i√3) + i(√2)(2√3) – i2√2
= -6 + i√3(1 + 2√2) + √2
= (-6 + √2) + i√3(1 + 2√2)
This is of the form a + ib, where a = -6 + √2 and b = √3(1 + √2).
Example 3:
Find the multiplicative inverse of 2 – 3i.
Solution: Let z = 2 – 3i
z¯ = 2 + 3i
|z|2 = (2)2 + (-3)2 = + = 13
We know that the multiplicative inverse of z is given by the formula:
z−1=z¯|z|2
= (2 + 3i)/13
= (2/13) + i(3/13)
Alternatively,
Multiplicative inverse of z is:
z-1 = 1/(2 – 3i)
By rationalizing the denominator we get,
= (2 + 3i)/(4 + 9)
= (2 + 3i)/ 13
= (2/13) + i(3/13)
Example 4:
Represent the complex number z = 1 + i√3 in the polar form.
Solution: Given, z = 1 + i√3
Let 1 = r cos θ, √3 = r sin θ
By squaring and adding, we get
r2(cos2θ + sin2θ) = 4
r2 = 4
r = 2 (as r > 0)
Therefore, cos θ = 1/2 and sin θ = √3/2
This is possible when θ = π/3.
Thus, the required polar form is z = 2[cos π/3 + i sin π/3].
Hence, the complex number z = 1 + i√3 is represented as shown in the below figure.
Important Questions with Solutions on CBSE Class 11 Applied Maths Algebra
Question 1:
How many words can be formed each of 2 vowels and 3 consonants from the given the word's letters – DAUGHTER?
Solution:
No. of Vowels in the word – DAUGHTER is 3.
No. of Consonants in the word Daughter is 5.
No of ways to select a vowel = 3c2 = 3!/2!(3 – 2)! = 3
No. of ways to select a consonant = 5c3 = 5!/3!(5 – 3)! = 10
Now you know that the number of combinations of 3 consonants and 2 vowels = 10x 3 = 30
Total number of words = 30 x 5! = 3600 ways.
Question 2:
It is needed to seat 5 boys and 4 girls in a row to get the even places. How many are such arrangements possible?
Solution:
5 boys and 4 girls are to be seated in a row to get the even places.
The 5 boys can be seated in 5! Ways.
For each of the arrangement, the 4 girls can be seated only at the places which are cross marked to make girls occupy the even places).
B x B x B x B x B
So, the girls can be seated in 4! Ways.
Hence, the possible number of arrangements = 4! × 5! = 24 × 120 = 2880
Question 3:
Find the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Solution:
Take a deck of 52 cards,
To get exactly one king, 5-card combinations have to be made. It should be made in such a way that in each selection of 5 cards, or a deck of 52 cards, there will be 4 kings.
To select 1 king out of 4 kings = 4c1
To select 4 cards out of the remaining 48 cards = 48c4
To get the needed number of 5 card combination = 4c1 x 48c4
= 4x2x 47x 46×45
= 778320 ways.
Question 4:
Find the number of 6 digit numbers that can be formed by using the digits 0, 1, 3, 5, 7, and 9. These digits shall be divisible by 10, and no digit shall be repeated?
Solution:
The number which has a 0 in its unit place is divisible by 10.
If we put 0 in the unit place, _ _ _ _ 0, there will be as many ways to fill 5 vacant places. (1, 3, 5, 7, 9)
The five vacant places can be filled in 5! ways = 120.
Question 5:
Evaluate: 10! – 6!
Solution:
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x1 = 3628800
6! = 6 X 5 x 4 x 3 x 2 x 1 = 720
10! – 6! = 3628800 – 720 = 3628080
Question 6:
Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Solution:
Let a and d be the first term and the common difference of the A.P. respectively. It is known
that the kth term of an A.P. is given by
ak = a +(k -1)d
Therefore, am+n = a +(m+n -1)d
am-n = a +(m-n -1)d
am = a +(m-1)d
Hence, the sum of (m + n)th and (m – n)th terms of an A.P is written as:
am+n+ am-n = a +(m+n -1)d + a +(m-n -1)d
= 2a +(m + n -1+ m – n -1)d
=2a+(2m-2)d
=2a + 2(m-1)d
= 2 [a + (m-1)d]
= 2 am [since am = a +(m-1)d]
Therefore, the sum of (m + n)thand (m – n)th terms of an A.P. is equal to twice the mth term.
Question 7:
Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Solution:
The integers from 1 to 100, which are divisible by 2, are 2, 4, 6 ….. 100.
This forms an A.P. with both the first term and common difference equal to 2.
⇒ 100=2+(n-1)2
⇒ n= 50
Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:
2+4+6+…+100 = (50/2)[2(2)+(50-1)(2)]
= (50/2)(4+98)
= 25(102)
= 2550
The integers from 1 to 100, which are divisible by 5, 10…. 100
This forms an A.P. with both the first term and common difference equal to 5.
Therefore, 100= 5+(n-1)5
⇒5n = 100
⇒ n= 100/5
⇒ n= 20
Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:
5+10+15+…+100= (20/2)[2(5)+(20-1)(5)]
= (20/2)(10+95)
= 10(105)
= 1050
Hence, the integers from 1 to 100, which are divisible by both 2 and 5 are 10, 20, ….. 100.
This also forms an A.P. with both the first term and common difference equal to 10.
Therefore, 100= 10+(n-1)10
⇒10n = 100
⇒ n= 100/10
⇒ n= 10
10+20+…+100= (10/2)[2(10)+(10-1)(10)]
= (10/2)(20+90)
= 5(110)
= 550
Therefore, the required sum is:
= 2550+ 1050 – 550
= 3050
Hence, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.
Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}. Write the set (A U B)’.
Solution:
Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, 0 < x < 10}; B = {2, 3, 5, 7}
U = { 1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {2, 4, 6, 8}
A U B = {2, 3, 4, 5, 6, 7, 8}
(A U B)’ = {1, 9}
Question 8:
In a survey of 600 students in a school, 150 students were drinking Tea and 225 drinking Coffee, and 100 were drinking both Tea and Coffee. Find how many students were drinking neither Tea nor Coffee.
Solution:
Given,
Total number of students = 600
Number of students who were drinking Tea = n(T) = 150
Number of students who were drinking Coffee = n(C) = 225
Number of students who were drinking both Tea and Coffee = n(T ∩ C) = 100
n(T U C) = n(T) + n(C) – n(T ∩ C)
= 150 + 225 -100
= 375 – 100
= 275
Hence, the number of students who are drinking neither Tea nor Coffee = 600 – 275 = 325