NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.2) Exercise 5.2

NCERT Maths Class 12 Exercise 5.2 Solutions - Free PDF Download

The Class 12 Maths NCERT Solutions, chapter 5, exercise 5.2 are the key to the success of the students. The ex 5.2 class 12 maths NCERT solutions guide the students in the learning and the revision procedure. These solutions act as a guide for the students throughout their study course. The student of class 12 can get in-depth knowledge of the chapter by referring to the class 12 maths exercise 5.2 solutions. The NCERT Maths Solution Class 12 Chapter 5 Exercise 5.2 is designed as per the latest CBSE syllabus of Exercise 5.2 class 12 maths. The solutions are designed keeping in mind all the topics that are given in exercise 5.2, class 12 maths. These solutions explain the different concepts of Continuity of a function and differentiability. The exercise 5.2 class 12 maths NCERT solutions are prepared by the best teachers, who have years of experience. Thus, it guarantees the success of the students. 

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Access NCERT Solutions for Class 12 Mathematics Chapter 5 – Continuity and Differentiability part-1

Access NCERT Solutions for Class 12 Mathematics Chapter 5 – Continuity and Differentiability

1. Differentiate the function with respect to $x$. $\sin \left( {{x^2} + 5} \right)$

Ans: Let $f(x) = \sin \left( {{x^2} + 5} \right),u(x) = {x^2} + 5$, and $v(t) = \sin t$

Then, $(vou)(x) = v(u(x)) = v\left( {{x^2} + 5} \right)$

$ = \tan \left( {{x^2} + 5} \right) = f(x)$

Thus, $f$ is a composite of two functions.

Put $t = u(x) = {x^2} + 5$

Then, we obtain $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos \left( {{x^2} + 5} \right)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2} + 5} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}(5) = 2x + 0 = 2x$

Therefore, by chain rule. $\dfrac{{df}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos \left( {{x^2} + 5} \right) \times 2x$

$ = 2x\cos \left( {{x^2} + 5} \right)$


2. Differentiate the functions with respect of ${\text{x}}$. $\cos (\sin x)$

Ans: Let $f(x) = \cos (\sin x),u(x) = \sin x$, and $v(t) = \cos t$

Then, $(\operatorname{vou} )(x) = v(u(x)) = v(\sin x) = \cos (\sin x) = f(x)$

Thus, $f$ is a composite function of two functions.

Put $t = u(x) = \sin x$

$\therefore \dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}[\cos t] =  - \sin t =  - \sin (\sin x)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sin x) = \cos x$

By chain rule, $\dfrac{{df}}{{dx}},\dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} =  - \sin (\sin x) \cdot \cos x$

$ =  - \cos x\sin (\sin x)$


3. Differentiate the functions with respect of ${\text{x}}$.

 $\sin (ax + b)$

Ans: Let $f(x) = \sin (ax + b),u(x) = ax + b$, and $v(t) = \sin t$

Then, $(vou)(x) = v(u(x)) = v(ax + b) = \sin (ax + b) = f(x)$

Thus, $f$ is a composite function of two functions $u$ and $v$.

Put $t = u(x) = ax + b$

Therefore, $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos (ax + b)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(ax + b) = \dfrac{d}{{dx}}(ax) + \dfrac{d}{{dx}}(b) = a + 0 = a$

Hence, by chain rule, we obtain

$\dfrac{{df}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos (ax + b) \cdot a$

$ = a\cos (ax + b)$


4. Differentiate the functions with respect of ${\text{x}}$. 

$\sec (\tan (\sqrt x ))$

Ans: Let $f(x) = \sec (\tan (\sqrt x )),u(x) = \sqrt x ,v(t) = \tan t$, and $w(s) = \sec s$

Then, $(vou)(x) = w[v(u(x))] = w[v(\sqrt x )] = w(\tan \sqrt x ) = \sec (\tan \sqrt x ) = f(x)$

Thus, $f$ is a composite function of three functions, $u, v$ and $w$. Put $s = v(t) = \tan t$ and $t = u(x) = \sqrt x $

$[s = \tan t]$

Then, $\dfrac{{dw}}{{ds}} = \dfrac{d}{{ds}}(\sec s) = \sec s\tan s$

$ = \sec (\tan t) \cdot \tan (\tan t)$

$ = \sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x )\quad [t = \sqrt x ]$

$\dfrac{{ds}}{{dt}} = \dfrac{d}{{dt}}(\tan t) = {\sec ^2}t = {\sec ^2}\sqrt x $

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sqrt x ) = \dfrac{d}{{dx}}\left( {{x^{\frac{1}{2}}}} \right)$

$ = \dfrac{1}{2} \cdot {x^{\frac{1}{2} - 1}} = \dfrac{1}{{2\sqrt x }}$

Hence, by chain rule, we obtain $\dfrac{{dt}}{{dx}} = \dfrac{{dw}}{{ds}} \cdot \dfrac{{ds}}{{dt}} \cdot \dfrac{{dt}}{{dx}}$

$ = \sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x ) \times {\sec ^2}\sqrt x  \times \dfrac{1}{{2\sqrt x }}$

$ = \dfrac{1}{{2\sqrt x }}{\sec ^2}\sqrt x (\tan \sqrt x )\tan (\tan \sqrt x )$

$ = \dfrac{{{{\sec }^2}\sqrt x \sec (\tan \sqrt x )\tan (\tan \sqrt x )}}{{2\sqrt x }}$


5. Differentiate the functions with respect of ${\text{X}}$. $\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}$

Ans: The given function is $f(x) = \dfrac{{\sin (ax + b)}}{{\cos (cx + d)}} = \dfrac{{g(x)}}{{h(x)}},$where $g(x) = \sin (ax + b)$ and

$h(x) = \cos (cx + d)$

$\therefore f = \dfrac{{{g^\prime }h - g{h^\prime }}}{{{h^2}}}$

Consider $g(x) = \sin (ax + b)$

Let $u(x) = ax + b,v(t) = \sin t$

Then $(vou)(x) = v(u(x)) = v(ax + b) = \sin (ax + b) = g(x)$

$\therefore g$ is a composite function of two functions, $u$ and $v$.

Put $t = u(x) = ax + b$

$\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos (ax + b)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(ax + b) = \dfrac{d}{{dx}}(ax) + \dfrac{d}{{dx}}(b) = a + 0 = a$

Therefore, by chain rule, we obtain ${g^\prime } = \dfrac{{dg}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos (ax + b) \cdot a = a\cos (ax + b)$

Consider $h(x) = \cos (cx + d)$

Let $p(x) = cx + d,q(y) = \cos y$

Then, $(qop)(x) = q(p(x)) = q(cx + d) = \cos (cx + d) = h(x)$

$\therefore h$ is a composite function of two functions, $p$ and $q$.

Put $y = p(x) = cx + d$

$\dfrac{{dq}}{{dy}} = \dfrac{d}{{dy}}(\cos y) =  - \sin y =  - \sin (cx + d)$

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(cx + d) = \dfrac{d}{{dx}}(cx) + \dfrac{d}{{dx}}(d) = c$

Therefore, by chain rule, we obtain ${h^\prime } = \dfrac{{dh}}{{dx}} = \dfrac{{dq}}{{dy}} \cdot \dfrac{{dy}}{{dx}} =  - \sin (cx + d){\text{xc}} =  - c\sin (cx + d)$

$\therefore {f^\prime } = \dfrac{{a\cos (ax + b) \cdot \cos (cx + d) - \sin (ax + b)\{  - c\sin cx + d\} }}{{{{[\cos (cx + d)]}^2}}}$

$ = \dfrac{{a\cos (ax + b)}}{{\cos (cx + d)}} + c\sin (ax + b) \cdot \dfrac{{\sin (cx + d)}}{{\cos (cx + d)}} \times \dfrac{1}{{\cos (cx + d)}}$

$ = a\cos (ax + b)\sec (cx + d) + c\sin (ax + b)\tan (cx + d)\sec (cx + d)$


6. Differentiate the function with respect to ${\text{x}}$. 

$\cos {x^3} \cdot {\sin ^2}\left( {{x^5}} \right)$

Ans: $\cos {x^3} \cdot {\sin ^2}\left( {{x^5}} \right)$

$\dfrac{d}{{dx}}\left[ {\cos {x^3} \cdot {{\sin }^2}\left( {{x^5}} \right)} \right] = {\sin ^2}\left( {{x^5}} \right){\text{x}}\dfrac{d}{{dx}}\left( {\cos {x^3}} \right) + \cos {x^3}{\text{x}}\dfrac{d}{{dx}}\left[ {{{\sin }^2}\left( {{x^5}} \right)} \right]$

$ = {\sin ^2}\left( {{x^5}} \right){\text{x}}\left( { - \sin {x^3}} \right){\text{x}}\dfrac{d}{{dx}}\left( {{x^3}} \right) + \cos {x^3} + 2\sin \left( {{x^5}} \right) \cdot \dfrac{d}{{dx}}\left[ {\sin {x^5}} \right]$

The given function is

$ = \sin {x^3}{\sin ^2}\left( {{x^5}} \right) \times 3{x^2} + 2\sin {x^5}\cos {x^3} \cdot \cos {x^5} \times \dfrac{d}{{dx}}\left( {{x^5}} \right)$

$ = 3{x^2}\sin {x^3} \cdot {\sin ^3}\left( {{x^5}} \right) + 2\sin {x^5}\cos {x^5}\cos {x^3} \cdot x5{x^4}$

$ = 10{x^4}\sin {x^5}\cos {x^5}\cos {x^3} - 3{x^2}\sin {x^3}{\sin ^2}\left( {{x^5}} \right)$


7. Differentiate the functions with respect to ${\text{x}}$. $2\sqrt {\cot \left( {{x^2}} \right)} $

Ans: $\dfrac{d}{{dx}}\left[ {2\sqrt {\cot \left( {{x^2}} \right)} } \right]$

$ = 2 \cdot \dfrac{1}{{2\sqrt {\cot \left( {{x^2}} \right)} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {\cot \left( {{x^2}} \right)} \right]$

$ = \sqrt {\dfrac{{\sin \left( {{x^2}} \right)}}{{\cos \left( {{x^2}} \right)}}} {\text{x}} - {\operatorname{cosec} ^2}\left( {{x^2}} \right) \times \dfrac{d}{{dx}}\left( {{x^2}} \right)$

$ = \sqrt {\dfrac{{\sin \left( {{x^2}} \right)}}{{\cos \left( {{x^2}} \right)}}}  \times \dfrac{1}{{{{\sin }^2}\left( {{x^2}} \right)}}x(2x)$

$ = \dfrac{{ - 2x}}{{\sqrt {\cos {x^2}\sqrt {\sin {x^2}\sin {x^2}} } }}$

$ = \dfrac{{ - 2\sqrt 2 x}}{{\sqrt {2\sin {x^2}\cos {x^2}} \sin {x^2}}}$

$ = \dfrac{{ - 2\sqrt 2 x}}{{\sin {x^2}\sqrt {\sin 2{x^2}} }}$


8. Differentiate the functions with respect to $x$ $\cos (\sqrt x )$

Ans: Let $f(x) = \cos (\sqrt x )$

Also, let $u(x) = \sqrt x $

And, $v(t) = \cos t$

Then, $(vou)(x) = v(u(x))$

$ = v(\sqrt x )$

$ = \cos \sqrt x $

$ = f(x)$

Clearly, $f$ is a composite function of two functions, $u$ and $v$, such that $t = u(x) = \sqrt x $

Then,

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sqrt x ) = \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right)$

$\dfrac{1}{2}{x^{ - \dfrac{1}{2}}} = \dfrac{1}{{2\sqrt x }}$

And, $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\cos t) =  - \sin t =  - \sin \sqrt x $

By using chain rule, we obtain $\dfrac{{dt}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}}$

$ =  - \sin (\sqrt x ) \cdot \dfrac{1}{{2\sqrt x }}$

$ =  - \dfrac{1}{{2\sqrt x }}\sin (\sqrt x )$

$ =  - \dfrac{{\sin (\sqrt x )}}{{2\sqrt x }}$


9. Prove that the function $f$ given by $f(x) = |x - 1|,x \in {\mathbf{R}}$ is not differentiable at $x = 1$.

Ans: The given function is $f(x) = |x - 1|,x \in {\mathbf{R}}$

It is known that a function $f$ is differentiable at a point $x = c$ in its domain if both $\mathop {\lim }\limits_{k \to {0^ - }} \dfrac{{f(c + h) - f(c)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(c + h) - f(c)}}{h}$ are finite and equal.

To check the differentiability of the given function at $x = 1$, Consider the left hand limit of $f$ at $x = 1$ $\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f|I + h - 1||1 - 1|}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{|h| - 0}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - h}}{h}\quad (h < 0 \Rightarrow |h| =  - h)$

$ =  - 1$

Consider the right hand limit of $f$ at $x = 1$

$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f|I + h - 1| - |1 - 1|}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{|h| - 0}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{h}{h}\quad (h > 0 \Rightarrow |h| = h)$

$ = 1$

Since the left and right hand limits of $f$ at $x = 1$ are not equal, $f$ is not differentiable at $x = 1$


10. Prove that the greatest integer function defined by $f = (x) = [x],0 < x < 3$ is not differentiable at $x = 1$ and $x = 2$.

Ans: The given function $f$ is $f = (x) = [x],0 < x < 3$

It is known that a function $f$ is differentiable at a point $x = c$ in its domain if both $\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(c + h) - f(c)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(c + h) - f(c)}}{h}$ are finite and equal.

To check the differentiable of the given function at $x = 1$, consider the left hand limit of $f$ at

$x = 1$

$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{[1 + h] - [1]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{0 - 1}}{h} = \mathop {\lim }\limits_{h \to {0^ - }}  = \dfrac{{ - 1}}{h} = \infty $

Consider the right hand limit of $f$ at $x = 1$ $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{[1 + h][1]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{1 - 1}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$

Since the left and right limits of $f$ at $x = 1$ are not equal, $f$ is not differentiable at $x = 1$

To check the differentiable of the given function at $x = 2$, consider the left hand limit of $f$ at

$x = 2$

$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(2 + h) - f(2)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{[2 + h] - [2]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{1 - 2}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - 1}}{h} = \infty $

Consider the right hand limit of $f$ at $x = 1$ 

$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(2 + h) - f(2)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{[2 + h] - [2]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{1 - 2}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$

Since the left and right hand limits of $f$ at $x = 2$ are not equal, $f$ is not differentiable at $x = 2$


Chapter 5 Continuity and Differentiability Class 12 Maths

The CBSE Class 12 maths exercise 5.2 solutions PDF provides the most trustable and reliable information for the students. The NCERT Solutions of Class 12 maths chapter 5 exercise 5.2 comes with an answer to each and every question given in the class 12 maths chapter 5 exercise 5.2. These solutions guide the students through the learning procedure and help them secure better grades. The solutions tend to decrease the fear of examination from the minds of the students. 

The ex 5.2 class 12 maths NCERT Solutions PDF is easy to understand and is written and crafted in simple language for the quick understanding of the students. Exercise 5.2 class 12 maths NCERT solutions explain all the things about limits and continuity. The Solutions explain every detail of the Class 12 maths ex 5.2. The class 12 maths exercise 5.2 explains what continuity and differentiability are. The solutions come with all the solved questions and answers which provide the students with an overview of the chapter.

 

NCERT Math Solution Class 12 Chapter 5 Exercise 5.2 

The Class 12 Maths Chapter 5.2, explains various concepts of continuity. The Class 12th maths chapter 5 exercise 5.2 teaches the students how to find continuity of a function. These solutions teach various methods to find whether a function is continuous or not. Exercise 5.2 class 12th maths is mostly based on the different rules for addition and subtraction of continuity and differentiability. The exercise explains the various concepts of functions that are continuous and are without any breaks. The class 12 maths ch 5 ex 5.2 explains that continuity is a unique feature of a function that represents the continuity of a function in a graphical form. The graph represents a continuous wave that shows the continuity of the function. The exercise also explains the binary operations that are involved in the continuity of function. 

The maths class 12 ex 5.2 explains in details the conditions that can be used to prove the continuity of a function, and these conditions are as follows:

A function f(x) can be proved to be continuous if the following three conditions are verified.

  • If the value of f(a) is finite.

  • Limx→a f(x) is available.

  • Limx→a f(x) = f(a).

As we can see that the Class 12 maths chapter 5 exercise 5.2 explains all the details about different concepts of continuity. So referring to these solutions can prove to be helpful to the students.

 

NCERT Maths Class 12 Exercise 5.2 Solutions - Weightage of Marks

In this segment of Continuity and differentiability, students can increase their conceptual knowledge about continuity. The students can get an overview of the chapter and can understand the various formulas used to find the continuity of a function. The maths class 12 chapter 5 exercise 5.2 has 10 questions which are mostly short types. The questions are based on differentiating various functions. The questions of class 12 maths ex 5.2 can carry 2-5 marks of weightage in the CBSE Board examination.

 

Benefits of NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.2

The Benefits of CBSE Class 12 Maths exercise 5.2 solutions are multifold. These solutions are beneficial to the students as they provide the solution to different sums given in the exercise. Here we have mentioned the Benefits of these solutions:

  • These Solutions help the students to understand the concepts of continuity and differentiation.

  • The solutions will guide the student throughout the learning procedure and help them secure better grades.

  • The solutions guarantee error-free and quality content as the solutions are designed by experts in Vedantu.

  • The most important part is that these solutions are available for free on the Vedantu site.

FAQs (Frequently Asked Questions)

1. What is Continuity, and what are the criteria to prove the continuity of a function?

Continuity is a unique feature or characteristic of a function under which the function is represented in a graphical manner. The graph of continuity always shows a continuous wave which indicates that a function is continuous. Continuity can be found in many aspects of nature and our surroundings.

The three criteria that prove a function to be continuous are:

If the value of f(a) is finite.

Lim x→a is available.

Lim x→a f(x) = f(a).

2. How are the NCERT Solutions of Class 12 Maths Chapter 5 exercise 5.2 Helpful?

The solutions of class 12 maths chapter 5 are very much helpful to the students. They guide the students and lead them to the path of success. These solutions are structured in simple language for the betterment of the students. The solutions are prepared by our experts and guarantee error-free solutions.

3. How many questions are there in NCERT Solutions Class 12 Maths exercise 5.2?

NCERT Solutions Class 12 Maths Exercise 5.2 consists of ten questions. All the ten questions in Exercise 5.2 are based on the application of Product Rule, Quotient Rule, and Chain Rule. They focus on finding Left Hand Derivatives (LHD) and Right Hand Derivatives (RHD). Only a few of them are difficult, but the vast majority are simple. You must practice them regularly to understand the concepts properly. Finally, this is a useful exercise for honing your skills in revising functions' derivatives.

4. Is Class 12 Maths Chapter 5 important for board exams?

In NCERT Class 12 Maths Chapter 5 you will learn about continuity and differentiability, as well as its more advanced theorems and logarithms. You will be taught a more advanced version of what you learned in your last class. You'll learn about the ideas of continuity, differentiability, and their connections. You'll also learn about inverse trigonometric function differentiation. It belongs to the ‘Calculus’ unit which holds 35 marks in the board exams. You can also get questions worth two-four marks from this chapter. 

5. Where can I download NCERT Solutions for Class 12 Maths Chapter 5 PDF for free?

Students can easily get access to the NCERT Solutions for Class 12 Maths Chapter 5 both in online and offline mode. You can access NCERT Solutions by visiting the website of Vedantu or download the Vedantu app and go through the solutions online. And if you want to access this later offline, then you can download the solution in PDFs free of cost. Once downloaded from Vedantu, you can access them on your device anytime and anywhere. All these steps can be done free of cost. 

6. What are the main topics explained in Class 12 Maths Chapter 5?

Class 12 Maths Chapter 5 ‘Continuity and Differentiability’ is a very important chapter for a 12th grader. This chapter includes some very crucial concepts to understand such as introduction to continuity and differentiability, exponential and logarithmic functions, logarithmic differentiation, derivation of functions in parametric forms, second order derivatives and mean value theorem. To get well-versed with the entire chapter, you should visit Vedantu. The study material provided by Vedantu will help you clear all your concepts. You can go through the NCERT Solutions to revise your basics.

7. What are the benefits of choosing Vedantu for Class 12 Maths Chapter 5?

When it comes to Class 12 Maths Chapter 5, the advantages of using Vedantu are virtually limitless. Vedantu has a team of educators who work together to create high-quality study resources for students. You can benefit from the free study materials provided by Vedantu to its students. You will receive many more benefits if you join the Vedantu family. Live meetings with the teachers can assist you to clear up any confusion you may have. All Vedantu’s NCERT Solutions are prepared by the best teachers in India and are based on the latest syllabus and exam patterns.

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