Updated On : February 13, 2024

**Reader's Digest: **Ready to learn the ins and outs of CBSE Class 11 Applied Maths Coordinate Geometry 2024? Read this blog to learn the concept, project topics, syllabus, weightage & prep tips!

The world of mathematics continues to evolve, as does its importance in many professions and daily applications. Recognizing this, the board announced the introduction of a new topic, Applied Mathematics.

A significant chapter from the **CBSE class 11 Applied Mathematics** is '*Algebra*', a cornerstone in mathematics. Delving deeper into the weightage, the algebra unit comprises 15 marks out of the 80 in the theory exam.

Here are the points to be discussed in this blog:

**Basics of CBSE Class 11 Applied Maths Coordinate Geometry:**Dive into the foundational concepts and unique approach that sets Applied Maths Coordinate Geometry apart.**Important Topics of CBSE Applied Maths Coordinate Geometry Class 11:**Identify the core topics that deserve your focus and attention to excel in this subject.Uncover students' pitfalls and learn how to avoid them, ensuring a smoother learning experience.*Table of Common Errors to Avoid:***CBSE Class 11 Applied Maths Coordinate Geometry Solved Sample Questions:**Put your knowledge to the test with solved sample questions that showcase the challenges you'll encounter in this exciting field.

Algebra is a branch of mathematics in which letters and symbols represent numbers and quantities in formulas and equations.

It provides methods to work with finite and infinite dimensional vector spaces and linear transformations between these spaces, including systems of linear equations, matrices, determinants, polynomials, and certain abstract structures, such as groups, rings, and fields.

Essentially, algebra provides a powerful tool to analyze and understand various mathematical structures.

**Examples:**

**Linear Equations:**If we're told that the sum of two numbers is 15 and one number is 5 greater than the other, we can use algebra to represent this situation. Let the smaller number be x. The other number is x + 5. Thus, the equation becomes x+x+5=15x+x+5=15. Solving this, we get x=5, so the numbers are 5 and 10.**Polynomials:**Consider the expression 2x² - 3x - 5. This is a polynomial of degree 2. When we set this equal to zero, we get a quadratic equation, which can be solved using various methods such as factoring, using the quadratic formula, or completing the square.

We can use methods like substitution, elimination, or matrix techniques to find the values of x and y that satisfy both equations simultaneously.

Applied mathematics course prepares you to choose algebraic methods as a means of representation and as a problem-solving tool. Algebra mainly focuses on topics like sets, relations, Venn diagrams, the relation between arithmetic and geometric progression, etc.

Go through the table below to learn the detailed** CBSE Class 11 Applied Maths Syllabus **for algebra**.**

Algebra is the study of unknown quantities. Some of the topics of algebraic expressions and formulae are:

Various equality equations consist of different variables in algebraic identities.

**Linear Equations in One Variable:**A linear equation in one variable has the maximum of one variable in order 1. It is depicted as ax + b = 0, where x represents the variable.**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 as ax2 + bx + c = 0. The two variables are crucial because your coursebook has a lot of questions based on it. So, you must stay focused on important algebra formulas to find the solution.

**Some basic identities to note are:**

- The combination of literal numbers obeys every fundamental 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.

The degrees and powers in any mathematical expression are known as exponents. 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*

**Read More: CBSE Class 11 Applied Maths - Mathematical Reasoning**

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*±*b*2−4*ac*√2*a*

Below are some important points about the equation as a part of essential algebra formulas:

- Δ = b2 − 4ac is also known as a discriminant.
- For roots;

Delta; > 0 happens when the roots are real and distinct

For real and coincident roots, Δ = 0

Delta; < 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]

**Read More: CBSE Class 11 Applied Math Books **

Here's a table detailing common errors students make while solving CBSE Class 11 Applied Maths Algebra problems and tips to avoid those errors:

Common Errors |
Tips to Avoid Errors |
---|---|

Mistaking variables: Confusing one variable for another, especially in multi-variable equations or systems of equations. |
Stay Organized: Label each variable clearly, and when solving, isolate each variable systematically. Drawing diagrams or flowcharts can also help visualize relationships. |

Calculation Errors: Small mistakes in basic arithmetic or order of operations can drastically change results. |
Double-check Calculations: Always revisit calculations, especially in complex problems. Using the BODMAS/BIDMAS rule can help maintain the correct order of operations. |

Misunderstanding Terms: Misinterpreting algebraic terms or confusing terms like coefficients and exponents. |
Revisit Basics: Make flashcards for different algebraic terms and concepts. Frequently reviewing these can help cement understanding. |

Improper Factoring: Incorrectly factoring quadratic expressions or other polynomials. |
Practice: The more problems you solve, the more comfortable you'll get with various factoring techniques. Also, look out for patterns and remember factorization formulas. |

Misapplying Formulas: Using incorrect formulas or misapplying them in certain contexts. |
Understand, Don't Memorize: Instead of rote learning, try to understand the logic behind each formula. This not only helps in correct application but also in recalling them. |

Overlooking Solutions: Especially in quadratic equations, missing one of the two possible solutions. |
Methodical Solving: Always ensure you've considered all possibilities for a solution. For quadratic equations, remember that there can be two, one, or no real solutions. |

Errors in Simplification: Not simplifying fractions or radicals to their lowest terms. |
Always Simplify: After solving a problem, always check if your result can be simplified further. This will ensure answers match standard solutions. |

Graphing Mistakes: Plotting points or drawing graphs incorrectly. |
Plot Carefully: Double-check each point's coordinates. When sketching curves, ensure smoothness and check for symmetry when relevant. |

**Read More: CBSE Class 11 Applied Maths Calculus**

To ease your preparation, we have provided important topic-wise questions for class 11 Applied Maths Algebra in the post below.

**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, 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’

**Question 1:**

Write the range of a Signup function.

**Question 2:**

The Cartesian product A × A has 9 elements, found (–1, 0) and (0,1). Find the set A and the remaining aspects of A × A.

**Question 1:** The sum of n terms of two arithmetic progressions is 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.

**Question 1:** Find the sum of divisible integers from 1 to 100 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

**Check Out - CBSE Class 11 Numerical Applications **

**Question 1:** Find the AM, GM, and HM between 12 and 30

**Question 1:**

Find the 3-digit numbers that can be formed from the given digits: 1, 2, 3, 4 and 5, assuming that

- digits can be repeated.
- 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?

**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?

**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.

**Check Out - CBSE Class 11 Commerce Books**

To help you understand the type of questions that can be asked in the exam, we have provided some sample questions for your reference here.

**Question 1:** How many words can be formed each of 2 vowels and 3 consonants from the given 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 such arrangements are 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.

**Check Also: CBSE Class 11 Applied Maths Probability**

**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)th and (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.

**Question 8**: 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}

**Find More: Differences between Pure Maths and Applied Maths**

**Question 9**: 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

**Find More:** **CBSE Class 11 Commerce Syllabus**

NCERT Solutions for CBSE Class 11 Applied Mathematics Algebra is an essential resource for preparing for the exam. You can verify your answers and understand how each problem is solved.

NCERT Applied Maths Algebra solutions provide a straightforward way of illustrating and explanations. The format of these textbooks is straightforward.

Complex Numbers Class 11 is defined when a number can be represented in the 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, represented by Im z of complex number z.

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 give you 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 x and y values.

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.

In this blog, we delved into the world of CBSE Class 11 Applied Maths Coordinate Geometry for 2024. We explored concepts, project topics, syllabus, weightage, and crucial preparation tips.

**Key Takeaways:**

*CBSE's commitment to making mathematics more applied and relatable with the introduction of Class 11 Applied Mathematics.**The substantial weightage of algebra in the syllabus, commanding 15 marks in the theory exam.**The significance of mastering algebraic fundamentals.**Common mistakes students should avoid while solving algebra problems.**A glimpse of solved sample questions to enhance your understanding.*

Frequently Asked Questions

What has applied mathematics algebra for Class 11?

What are the topics included in the CBSE class 11 applied mathematics algebra syllabus?

What is the difference between pure algebra and applied algebra?

What is the importance of De Morgan's Laws in algebra?

How can I represent a complex number in its polar form?

How can students improve their algebra skills?

What are common mistakes students should avoid when solving algebra problems?

What are some key algebraic topics covered in the CBSE Class 11 Applied Mathematics syllabus?

February 13, 2024

**Reader's Digest: **Ready to learn the ins and outs of CBSE Class 11 Applied Maths Coordinate Geometry 2024? Read this blog to learn the concept, project topics, syllabus, weightage & prep tips!

The world of mathematics continues to evolve, as does its importance in many professions and daily applications. Recognizing this, the board announced the introduction of a new topic, Applied Mathematics.

A significant chapter from the **CBSE class 11 Applied Mathematics** is '*Algebra*', a cornerstone in mathematics. Delving deeper into the weightage, the algebra unit comprises 15 marks out of the 80 in the theory exam.

Here are the points to be discussed in this blog:

**Basics of CBSE Class 11 Applied Maths Coordinate Geometry:**Dive into the foundational concepts and unique approach that sets Applied Maths Coordinate Geometry apart.**Important Topics of CBSE Applied Maths Coordinate Geometry Class 11:**Identify the core topics that deserve your focus and attention to excel in this subject.Uncover students' pitfalls and learn how to avoid them, ensuring a smoother learning experience.*Table of Common Errors to Avoid:***CBSE Class 11 Applied Maths Coordinate Geometry Solved Sample Questions:**Put your knowledge to the test with solved sample questions that showcase the challenges you'll encounter in this exciting field.

Algebra is a branch of mathematics in which letters and symbols represent numbers and quantities in formulas and equations.

It provides methods to work with finite and infinite dimensional vector spaces and linear transformations between these spaces, including systems of linear equations, matrices, determinants, polynomials, and certain abstract structures, such as groups, rings, and fields.

Essentially, algebra provides a powerful tool to analyze and understand various mathematical structures.

**Examples:**

**Linear Equations:**If we're told that the sum of two numbers is 15 and one number is 5 greater than the other, we can use algebra to represent this situation. Let the smaller number be x. The other number is x + 5. Thus, the equation becomes x+x+5=15x+x+5=15. Solving this, we get x=5, so the numbers are 5 and 10.**Polynomials:**Consider the expression 2x² - 3x - 5. This is a polynomial of degree 2. When we set this equal to zero, we get a quadratic equation, which can be solved using various methods such as factoring, using the quadratic formula, or completing the square.

We can use methods like substitution, elimination, or matrix techniques to find the values of x and y that satisfy both equations simultaneously.

Applied mathematics course prepares you to choose algebraic methods as a means of representation and as a problem-solving tool. Algebra mainly focuses on topics like sets, relations, Venn diagrams, the relation between arithmetic and geometric progression, etc.

Go through the table below to learn the detailed** CBSE Class 11 Applied Maths Syllabus **for algebra**.**

Algebra is the study of unknown quantities. Some of the topics of algebraic expressions and formulae are:

Various equality equations consist of different variables in algebraic identities.

**Linear Equations in One Variable:**A linear equation in one variable has the maximum of one variable in order 1. It is depicted as ax + b = 0, where x represents the variable.**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 as ax2 + bx + c = 0. The two variables are crucial because your coursebook has a lot of questions based on it. So, you must stay focused on important algebra formulas to find the solution.

**Some basic identities to note are:**

- The combination of literal numbers obeys every fundamental 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.

The degrees and powers in any mathematical expression are known as exponents. 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*

**Read More: CBSE Class 11 Applied Maths - Mathematical Reasoning**

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*±*b*2−4*ac*√2*a*

Below are some important points about the equation as a part of essential algebra formulas:

- Δ = b2 − 4ac is also known as a discriminant.
- For roots;

Delta; > 0 happens when the roots are real and distinct

For real and coincident roots, Δ = 0

Delta; < 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]

**Read More: CBSE Class 11 Applied Math Books **

Here's a table detailing common errors students make while solving CBSE Class 11 Applied Maths Algebra problems and tips to avoid those errors:

Common Errors |
Tips to Avoid Errors |
---|---|

Mistaking variables: Confusing one variable for another, especially in multi-variable equations or systems of equations. |
Stay Organized: Label each variable clearly, and when solving, isolate each variable systematically. Drawing diagrams or flowcharts can also help visualize relationships. |

Calculation Errors: Small mistakes in basic arithmetic or order of operations can drastically change results. |
Double-check Calculations: Always revisit calculations, especially in complex problems. Using the BODMAS/BIDMAS rule can help maintain the correct order of operations. |

Misunderstanding Terms: Misinterpreting algebraic terms or confusing terms like coefficients and exponents. |
Revisit Basics: Make flashcards for different algebraic terms and concepts. Frequently reviewing these can help cement understanding. |

Improper Factoring: Incorrectly factoring quadratic expressions or other polynomials. |
Practice: The more problems you solve, the more comfortable you'll get with various factoring techniques. Also, look out for patterns and remember factorization formulas. |

Misapplying Formulas: Using incorrect formulas or misapplying them in certain contexts. |
Understand, Don't Memorize: Instead of rote learning, try to understand the logic behind each formula. This not only helps in correct application but also in recalling them. |

Overlooking Solutions: Especially in quadratic equations, missing one of the two possible solutions. |
Methodical Solving: Always ensure you've considered all possibilities for a solution. For quadratic equations, remember that there can be two, one, or no real solutions. |

Errors in Simplification: Not simplifying fractions or radicals to their lowest terms. |
Always Simplify: After solving a problem, always check if your result can be simplified further. This will ensure answers match standard solutions. |

Graphing Mistakes: Plotting points or drawing graphs incorrectly. |
Plot Carefully: Double-check each point's coordinates. When sketching curves, ensure smoothness and check for symmetry when relevant. |

**Read More: CBSE Class 11 Applied Maths Calculus**

To ease your preparation, we have provided important topic-wise questions for class 11 Applied Maths Algebra in the post below.

**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, 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’

**Question 1:**

Write the range of a Signup function.

**Question 2:**

The Cartesian product A × A has 9 elements, found (–1, 0) and (0,1). Find the set A and the remaining aspects of A × A.

**Question 1:** The sum of n terms of two arithmetic progressions is 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.

**Question 1:** Find the sum of divisible integers from 1 to 100 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

**Check Out - CBSE Class 11 Numerical Applications **

**Question 1:** Find the AM, GM, and HM between 12 and 30

**Question 1:**

Find the 3-digit numbers that can be formed from the given digits: 1, 2, 3, 4 and 5, assuming that

- digits can be repeated.
- 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?

**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?

**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.

**Check Out - CBSE Class 11 Commerce Books**

To help you understand the type of questions that can be asked in the exam, we have provided some sample questions for your reference here.

**Question 1:** How many words can be formed each of 2 vowels and 3 consonants from the given 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 such arrangements are 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.

**Check Also: CBSE Class 11 Applied Maths Probability**

**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)th and (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.

**Question 8**: 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}

**Find More: Differences between Pure Maths and Applied Maths**

**Question 9**: 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

**Find More:** **CBSE Class 11 Commerce Syllabus**

NCERT Solutions for CBSE Class 11 Applied Mathematics Algebra is an essential resource for preparing for the exam. You can verify your answers and understand how each problem is solved.

NCERT Applied Maths Algebra solutions provide a straightforward way of illustrating and explanations. The format of these textbooks is straightforward.

Complex Numbers Class 11 is defined when a number can be represented in the 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, represented by Im z of complex number z.

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 give you 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 x and y values.

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.

In this blog, we delved into the world of CBSE Class 11 Applied Maths Coordinate Geometry for 2024. We explored concepts, project topics, syllabus, weightage, and crucial preparation tips.

**Key Takeaways:**

*CBSE's commitment to making mathematics more applied and relatable with the introduction of Class 11 Applied Mathematics.**The substantial weightage of algebra in the syllabus, commanding 15 marks in the theory exam.**The significance of mastering algebraic fundamentals.**Common mistakes students should avoid while solving algebra problems.**A glimpse of solved sample questions to enhance your understanding.*

Frequently Asked Questions

What has applied mathematics algebra for Class 11?

What are the topics included in the CBSE class 11 applied mathematics algebra syllabus?

What is the difference between pure algebra and applied algebra?

What is the importance of De Morgan's Laws in algebra?

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How can students improve their algebra skills?

What are common mistakes students should avoid when solving algebra problems?

What are some key algebraic topics covered in the CBSE Class 11 Applied Mathematics syllabus?