# NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations (Ex 5.2)

The CBSE Syllabus for 2022–2023 includes Class 11 Math Chapter 5: Complex Numbers and Quadratic Equations. Class 11’s Chapter 5  Exercise 5.2, primarily focuses on two areas. The NCERT Solutions Class 11 Maths Chapter 5 Exercise 5.2 and the topics covered in Chapter 5. The topics discussed in Complex Numbers and Quadratic Equations are as follows:

X and Y in the Cartesian plane are analogous to the real and imaginary components of a complex number in an Argand diagram. The complex plane, also known as the Argand plane, is the region of an Argand diagram.

Complex numbers can be represented in polar form, hence the term “polar representation of a complex number.”

In order to succeed in Class 11 Mathematics, students should develop a dedicated study plan. The CBSE Mathematics Class 11 syllabus has specific requirements for each chapter. In order to gain an understanding of the topics covered in Mathematics, students are required to read and understand the syllabus.

Students are aware that practising Math is the best way to excel at it. Class 11 students can easily avail the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 by logging in to their registered accounts on the Extramarks mobile application and the Extramarks website. Extramarks provides live interactive sessions, engaging modules, live classes, workshops, and test series, which makes it so distinct from every other study material provider. The Class 11 Maths NCERT Solutions Chapter 5 Exercise 5.2 are specially crafted keeping in mind the importance that the chapter holds in the final examinations.

Extramarks has also designed Question Banks for NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 for a better overview of the crucial concepts and formulas in the chapter.

## NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations (Ex 5.2) Exercise 5.2

According to Extramarks’ teachers, the greatest approach to getting better at mathematics is to practise it. That is why students at Extramarks adhere to the “Learn-Practice-Test” pedagogy.Students will gain the understanding they require to ace the board test by resolving the textual problems using the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 provided by Extramarks. Students can avail all the answers of the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 by downloading the PDF notes available on the Extramarks website and mobile application.

It is undeniable that Mathematics is a high-scoring subject, and with the right practice, one can score well in it. As part of the study plan, students are required to solve past years’ question papers as well as sample question papers on a regular basis. During the preparation of Chapter 5, it is necessary to solve exercise questions and revise concepts.

### NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.2

A fascinating and crucial area of Mathematics is Complex Numbers and Quadratic Equations. In the IIT, JEE and other exams, at least 1 to 3 problems from this chapter appear every year. Many later chapters, including those on functions and coordinate geometry, will make use of the ideas in this one.Examinees can refer to the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 by Extramarks to get a better understanding of the question paper pattern.

Students are strongly advised to begin by understanding basic concepts like the meaning of a complex number, the integral powers of iota, and different complex number formats. Then move on to algebra with complex numbers. The triangle, the Argand plane, the modulus, and the argument of a complex number are all basic ideas. They should also review the ideas from the previously solved problems once more before doing the same with the quadratic equation. Moreover, they can observe the problems solved in class for such topics as complex numbers and quadratic equations.

In order to study for the tests, students are expected to refer to Extramarks’ Complex Numbers and Quadratic Equations NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 and download the PDF version of the solutions.

Extramarks, along with NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 also provides basic notes for Class 11 Maths Chapter 5 Exercise 5.2.

### NCERT Solutions for Class 11 Maths Chapters

The NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 presented by Extramarks, have been solved by the top Math instructors with years of expertise in CBSE Board education. The NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 have been created to help students as they work through the challenging problems found in the CBSE textbook. The NCERT Maths Class 11 Solutions cover all 16 of the CBSE-mandated chapters. Using NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 can help students prepare for their exams because all the necessary concepts have been covered in one place.

Students taking the Class 11 Mathematics examinations will benefit greatly from these NCERT Solutions for Class 11 Maths Chapter 5—Exercise 5.2.

Using NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 can help students prepare for their exams because all the chapter content is available in one place.

These NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 will be of great help to students taking the Class 11 Math exam. Students can more clearly understand the chapter as a whole with the help of the in-depth solutions provided here.

A few important chapters for the Class 11 Mathematics Examinations are:

• Sets
• Relations and Functions
• Trigonometric Functions
• Principal of Mathematical Induction
• Complex Numbers and Quadratic Equations
• Linear Inequalities
• Binomial Theorem and
• Permutations and Combinations

All the topics covered in the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 have accurate, simple solutions in the NCERT Solutions, which were created by Extramarks’ subject experts. In order to assist students in all classes, NCERT solutions for classes 1 to 12 at Extramarks provide a thorough explanation of the answers to NCERT questions in PDF format. This helps students find the best approach for accurately answering the textbook questions.  Students can quickly improve their speed and learn problem-solving techniques by practising textbook solutions.Classes 1, 2, 3, and 4 in primary grades, as well as classes 6, 7, 8, 9, 10, and 11 in secondary and higher education, are covered by our NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 for NCERT Books.

Class-wise list of  NCERT Solutions available on Extramarks:

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10
• NCERT Solutions Class 11
• NCERT Solutions Class 12

Q.1 Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

$\begin{array}{l}1.\text{ }\mathrm{z}=-1-\mathrm{i}\sqrt{3}\\ 2.\text{ z}=-\sqrt{3}+\text{i}\end{array}$

Ans.

1.

$\begin{array}{l}\mathrm{z}=-1-\mathrm{i}\sqrt{3}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}\left(\mathrm{cos\theta }+\mathrm{isin\theta }\right),\text{we get}\\ \text{rcos}\mathrm{\theta }=-\text{1 and rsin}\mathrm{\theta }=-\sqrt{3}\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)={\left(-1\right)}^{2}+{\left(-\sqrt{3}\right)}^{2}\\ \text{\hspace{0.17em}}{\mathrm{r}}^{2}=1+3\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=2\\ \mathrm{So},\text{​ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}modulus}=\text{2}\end{array}$

$\begin{array}{l}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\frac{-1}{2}\text{and sin}\mathrm{\theta }=-\frac{\sqrt{3}}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=-\mathrm{cos}\frac{\mathrm{\pi }}{3}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=-\mathrm{sin}\frac{\mathrm{\pi }}{3}\\ \mathrm{Since},\text{sin}\mathrm{\theta }\text{and cos}\mathrm{\theta }\text{are negative in III quadrant.}\\ \mathrm{So},\text{}\mathrm{\theta }=\text{\hspace{0.17em}}-\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)=-\frac{2\mathrm{\pi }}{3}\\ \mathrm{Thus},\text{modulus and argument of}\left(-1-\mathrm{i}\sqrt{3}\right)\text{is 2 and}-\frac{2\mathrm{\pi }}{3}\text{}\\ \text{respectively.}\end{array}$

2.

$\begin{array}{l}\mathrm{z}=-\text{\hspace{0.17em}}\sqrt{3}+\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}\left(\mathrm{cos\theta }+\mathrm{isin\theta }\right),\text{we get}\\ \text{rcos}\mathrm{\theta }=-\text{\hspace{0.17em}}\sqrt{3}\text{and rsin}\mathrm{\theta }=1\\ \mathrm{Squarring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)={\left(-\text{\hspace{0.17em}}\sqrt{3}\right)}^{2}+{\left(1\right)}^{2}\\ \text{\hspace{0.17em}}{\mathrm{r}}^{2}=3+1=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=2\\ \mathrm{So},\text{​ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}modulus}=\text{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\frac{-\text{\hspace{0.17em}}\sqrt{3}}{2}\text{and sin}\mathrm{\theta }=\frac{1}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=-\mathrm{cos}\frac{\mathrm{\pi }}{6}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=\mathrm{sin}\frac{\mathrm{\pi }}{6}\\ \mathrm{Since},\text{sin}\mathrm{\theta }\text{is positive and cos}\mathrm{\theta }\text{is negative in II quadrant.}\end{array}$

$\begin{array}{l}\mathrm{So},\text{}\mathrm{\theta }=\text{\hspace{0.17em}}\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)=\frac{5\mathrm{\pi }}{6}\\ \mathrm{Thus},\text{modulus and argument of}\left(-\text{\hspace{0.17em}}\sqrt{3}+\mathrm{i}\right)\text{is 2 and}\frac{5\mathrm{\pi }}{6}\\ \text{respectively.}\end{array}$

Q.2 Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

3. 1 – i
4. – 1 + i
5. – 1 – i
6. – 3
7.

$\sqrt{3}+\mathrm{i}$

8. i

Ans.

3.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}}\mathrm{z}=1-\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}\left(\mathrm{cos\theta }+\mathrm{isin\theta }\right),\text{we get}\\ \text{rcos}\mathrm{\theta }=1\text{and rsin}\mathrm{\theta }=-1\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)={\left(1\right)}^{2}+{\left(-1\right)}^{2}\\ \text{\hspace{0.17em}}{\mathrm{r}}^{2}=1+1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=\sqrt{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\frac{1}{\sqrt{2}}\text{and sin}\mathrm{\theta }=\frac{-1}{\sqrt{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\mathrm{cos}\frac{\mathrm{\pi }}{4}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=-\mathrm{sin}\frac{\mathrm{\pi }}{4}\\ \mathrm{Since},\text{cos}\mathrm{\theta }\text{is positive and sin}\mathrm{\theta }\text{is negative in IV quadrant.}\\ \mathrm{So},\text{}\mathrm{\theta }=\text{\hspace{0.17em}}-\frac{\mathrm{\pi }}{4}\left[\because -\mathrm{\pi }<\mathrm{\theta }\le \mathrm{\pi }\right]\\ \mathrm{Thus},\text{polar form of z}=\sqrt{2}\left\{\mathrm{cos}\left(-\frac{\mathrm{\pi }}{4}\right)+\mathrm{isin}\left(-\frac{\mathrm{\pi }}{4}\right)\right\}.\end{array}$

4.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}}\mathrm{z}=-1+\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}\left(\mathrm{cos\theta }+\mathrm{isin\theta }\right),\text{we get}\\ \text{rcos}\mathrm{\theta }=-1\text{and rsin}\mathrm{\theta }=1\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)={\left(-1\right)}^{2}+{\left(1\right)}^{2}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\end{array}$

$\begin{array}{l}{\mathrm{r}}^{2}=1+1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=\sqrt{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\frac{-1}{\sqrt{2}}\text{and sin}\mathrm{\theta }=\frac{1}{\sqrt{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=-\mathrm{cos}\frac{\mathrm{\pi }}{4}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=\mathrm{sin}\frac{\mathrm{\pi }}{4}\\ \mathrm{Since},\text{sin\hspace{0.17em}}\mathrm{\theta }\text{is positive and cos\hspace{0.17em}}\mathrm{\theta }\text{is negative in II quadrant.}\\ \mathrm{So},\text{}\mathrm{\theta }=\text{\hspace{0.17em}}\mathrm{\pi }-\frac{\mathrm{\pi }}{4}=\frac{3\mathrm{\pi }}{4}\\ \mathrm{Thus},\text{polar form of z}=\sqrt{2}\left\{\mathrm{cos}\left(\frac{3\mathrm{\pi }}{4}\right)+\mathrm{isin}\left(\frac{3\mathrm{\pi }}{4}\right)\right\}.\end{array}$

5.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}}\mathrm{z}=-1-\text{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}\left(\mathrm{cos\theta }+\mathrm{isin\theta }\right),\text{we get}\\ \text{rcos}\mathrm{\theta }=-1\text{and rsin}\mathrm{\theta }=-1\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)={\left(-1\right)}^{2}+{\left(-1\right)}^{2}\\ \text{\hspace{0.17em}}{\mathrm{r}}^{2}=1+1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=\sqrt{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\frac{-1}{\sqrt{2}}\text{and sin}\mathrm{\theta }=\frac{-1}{\sqrt{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=-\mathrm{cos}\frac{\mathrm{\pi }}{4}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=-\mathrm{sin}\frac{\mathrm{\pi }}{4}\\ \mathrm{Since},\text{sin\hspace{0.17em}}\mathrm{\theta }\text{and cos\hspace{0.17em}}\mathrm{\theta }\text{is negative in III quadrant.}\end{array}$

$\begin{array}{l}\mathrm{So},\text{}\mathrm{\theta }=\text{\hspace{0.17em}}-\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)=-\frac{3\mathrm{\pi }}{4}\left[\because -\mathrm{\pi }<\mathrm{\theta }\le \mathrm{\pi }\right]\\ \mathrm{Thus},\text{polar form of z}=\sqrt{2}\left\{\mathrm{cos}\left(-\frac{3\mathrm{\pi }}{4}\right)+\mathrm{isin}\left(-\frac{3\mathrm{\pi }}{4}\right)\right\}.\end{array}$

6.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}}\mathrm{z}=-3+0\text{\hspace{0.17em}}\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}\left(\mathrm{cos\theta }+\mathrm{isin\theta }\right),\text{we get}\\ \text{rcos}\mathrm{\theta }=-3\text{and rsin}\mathrm{\theta }=0\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)={\left(-3\right)}^{2}+{\left(0\right)}^{2}\\ \text{\hspace{0.17em}}{\mathrm{r}}^{2}=9\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=3\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\frac{-3}{3}=-\text{1 and sin}\mathrm{\theta }=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=-\mathrm{cos}0\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=\mathrm{sin}0\\ \mathrm{Since},\text{sin\hspace{0.17em}}\mathrm{\theta }\text{is positive and cos\hspace{0.17em}}\mathrm{\theta }\text{is negative in II quadrant.}\\ \mathrm{So},\text{}\mathrm{\theta }=\text{\hspace{0.17em}}\left(\mathrm{\pi }-0\right)=\mathrm{\pi }\\ \mathrm{Thus},\text{polar form of z}=3\left\{\mathrm{cos\pi }+\mathrm{isin\pi }\right\}.\end{array}$

7.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}}\mathrm{z}=\sqrt{3}+\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}\left(\mathrm{cos\theta }+\mathrm{isin\theta }\right),\text{we get}\\ \text{rcos}\mathrm{\theta }=\sqrt{3}\text{and rsin}\mathrm{\theta }=1\\ \mathrm{Squaring}\text{and adding, we get}\end{array}$

$\begin{array}{l}{\text{r}}^{\text{2}}\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)={\left(\sqrt{3}\right)}^{2}+{\left(1\right)}^{2}\\ \text{\hspace{0.17em}}{\mathrm{r}}^{2}=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=2\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\frac{\sqrt{3}}{2}\text{and sin}\mathrm{\theta }=\frac{1}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\mathrm{cos}\frac{\mathrm{\pi }}{6}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=\mathrm{sin}\frac{\mathrm{\pi }}{6}\\ \mathrm{So},\text{}\mathrm{\theta }=\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{\pi }}{6}\\ \mathrm{Thus},\text{polar form of z}=2\left\{\mathrm{cos}\frac{\mathrm{\pi }}{6}+\mathrm{isin}\frac{\mathrm{\pi }}{6}\right\}.\end{array}$

8.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}}\mathrm{z}=0+\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}\left(\mathrm{cos\theta }+\mathrm{isin\theta }\right),\text{we get}\\ \text{rcos}\mathrm{\theta }=0\text{and rsin}\mathrm{\theta }=1\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)={\left(0\right)}^{2}+{\left(1\right)}^{2}\\ \text{\hspace{0.17em}}{\mathrm{r}}^{2}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=1\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=0\text{and sin}\mathrm{\theta }=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cos\theta }=\mathrm{cos}\frac{\mathrm{\pi }}{2}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=\mathrm{sin}\frac{\mathrm{\pi }}{2}\\ \mathrm{So},\text{}\mathrm{\theta }=\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{\pi }}{2}\end{array}$

$\mathrm{Thus},\text{polar form of z}=1\left\{\mathrm{cos}\frac{\mathrm{\pi }}{2}+\mathrm{isin}\frac{\mathrm{\pi }}{2}\right\}.$

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### 1. Are the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 provided by Extramarks beneficial?

Yes, the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 provided by Extramarks are very vital for exam preparations as they give an in-depth knowledge of the complex concepts of the chapter.

### 2. How can examinees avail the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2?

Students can access the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 are available for download in PDF format or accessible online on the Extramarks website and the Extramarks mobile application.

### 3. Does Extramarks provide live interactive sessions?

Yes, Extramarks does provide Live Interactive Sessions for students to help them clear their doubts by the “Question and Answer” method. Moreover, many assessments are taken from time to time so that students can get an idea of their level of preparation.

### 4. How to prepare for examinations using the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2?

The NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 have very detailed answers to all the queries that a student may have while learning the chapter. Students are advised to first get a grasp of the entire chapter including notes, PDF notes, and Chapter – wise solutions and practice extensively with the NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.2 to gain better results.