NCERT solutions of chapter 5-Continuity and Differentiability

Class 12 maths NCERT solutions of chapter 5- Continuity and Differentiability are published here for helping class 12 students in their preparation of forthcoming examinations and CBSE board exams of 2020-21. All the questions of chapter 5 – Continuity and Differentiability are solved by an expert teacher of CBSE maths as per the norms of the CBSE board. The NCERT solutions of chapter 5-Continuity and Differentiability of class 12 maths will help you a complete understanding of the value of the function and limits at a particular point the function is defined for. You will learn in chapter 5-Continuity and Differentiability of class 12 maths textbook that a particular function is differentiable or not means the derivatives of the function exist or not at the given point.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

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Class 12 maths NCERT solutions of Exercise 5.1 chapter 5-Continuity and Differentiability

Q1. Prove that the function f(x) = 5x – 3 is continuous at x= o, x = -3 and x = 5.

Ans. We are given the function f(x) = 5x – 3

The value of f(x) at x = 0

f(0) = 5 × 0 – 3 = -3

The continuety at x = 0

$\boldsymbol{\lim_{x\rightarrow 0}f\left ( x \right )=\lim_{x\rightarrow 0}\left ( 5x-3 \right )=5\times 0-3=-3}$

$\boldsymbol{\therefore \lim_{x\rightarrow 0}f\left ( x \right )=f\left ( 0 \right )}$

So, f(x) is continuous at x = 0

at x = -3, f(-3) = 5 × -3 – 3 = -15 -3 = -18

$\boldsymbol{\lim_{x\rightarrow -3}f\left ( x \right )=\lim_{x\rightarrow -3}\left ( 5x -3 \right )= 5\times-3-3=-15-3=-18 }$

$\boldsymbol{\therefore \lim_{x\rightarrow -3}f\left ( x \right ) =f\left ( -3 \right )}$

Therefore f(x) is continuous at x = -3

The value of f(x) at x = 5

f(0) = 5 × 5 – 3 = 22

The continuety at x = 5

$\boldsymbol{ \lim_{x\rightarrow 5}f\left ( x \right ) = \lim_{x\rightarrow 5}\left ( 5x-3 \right )=5\times 5-3=25-3=22}$

$\boldsymbol{\therefore \lim_{x\rightarrow 5}f\left ( x \right ) = x\left ( 5 \right )}$

Therefore f(x) is continuous at x = 5

Q2. Examine the continuety of the function f(x) = 2x² – 1, at x = 3.

Ans. The value of f(x) at x = 3

f(3) = 2×3² – 1 = 2 × 9 – 1 = 18 – 1 = 17

$\boldsymbol{ \lim_{x\rightarrow 3}f\left ( x \right ) = \lim_{x\rightarrow 3}\left ( 2x^{2}-1 \right )=2\times 3^{2}-1=18-1=17}$

$\boldsymbol{ \therefore \lim_{x\rightarrow 3}f\left ( x \right ) = f\left ( 3 \right )}$

Therefore the given function is continuous at x = 3

Q3.Examine the following function for continuty.

$\boldsymbol{\left ( a \right )\: f\left ( x \right )=\left ( x-5 \right )\: \: \: \: \left ( b \right )f\left ( x \right )=\frac{1}{x-5},x\neq 5}$

$\boldsymbol{\left ( c \right )f\left ( x \right )=\frac{x^{2}-25}{x+5}, x\neq -5\: \: \left ( d \right )f\left ( x \right )=\left | x -5 \right |}$

Ans. (a) Given function f(x) = x – 5, we know f is defined at every real number k

The value of the function at k

f(k) = k – 5

The continuity of f(x) at k

$\boldsymbol{\lim_{x\rightarrow 5}f\left ( x \right )=\lim_{x\rightarrow 5}\left ( x-5 \right )=k-5=f\left ( k \right )}$

Therefore f(x) is continuous at every real number, so f(x) is continuous

$\boldsymbol{The\: given \: function \: is\: f\left ( x \right )=\frac{1}{x-5}, x\neq 5}$

The value of f(x) for any real number k ≠ 5

$\boldsymbol{f\left ( k \right )=\frac{1}{k-5}}$

The continuity of the function for a real number k ≠ 5

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( \frac{1}{x-5} \right )=\frac{1}{k-5}}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

Therefore f(x) is continuous for any real number x ≠ 5 or of the domain of the function

$\boldsymbol{\left ( c \right )The \: given \: function\: is\: f\left ( x \right )=\frac{x^{2}-25}{x+5},x\neq -5}$

The value of f(x) for any real number k ≠ -5
$\boldsymbol{ f\left ( k \right )=\frac{k^{2}-25}{k+5}=\frac{k^{2}-5^{2}}{k+5}=\frac{\left ( k+5 \right )\left ( k-5 \right )}{k+5}=k-5}$
The continuity of the function for any real number k ≠ -5
$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( \frac{x^{2}-25}{x+5} \right )=\lim_{x\rightarrow k}\frac{x^{2}-5^{2}}{x+5}=\lim_{x\rightarrow k}\frac{\left ( x+5 \right )\left ( x-5 \right )}{x+5}=\lim_{x\rightarrow k}\left ( x-5 \right )}$
$\boldsymbol{=k-5=f\left ( k \right )}$

$\boldsymbol{\left ( d \right )\: f\left ( x \right )=\left | x-5 \right |}$

The given function is $\boldsymbol{f\left ( x \right )=\left | x-5 \right |}$

The given function f(x) is a modulus function

$\boldsymbol{f\left ( x \right )=\left | x-5 \right |=\left\{\begin{matrix}5-x \: if\:x < 5 & \\ x-5 \: if \: x\geq 5& \end{matrix}\right.}$

The given function is defined for all real numbers

Let k be a real number such that k< 5

Then f(x) = 5 – x

The value of f(x) at x = k

f(k) = 5 – k

The continuity of f(x) at x = k

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( 5-x \right )=5-k}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

Therefore f(x) is continuous at all real numbers less than 5.

In the case when k = 5

f(x) = x – 5

The value of f(x) at x =k

f(k) = k -5

The continuity of f(x) at x = k

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( x-5 \right )=k-5}$

Q4.Prove that the f(x) =xⁿis continuous at x = n where n is a positive integer.

Ans. The given function is f(x) =xⁿ

The value of f(x) at x = n

f(n) = nⁿ

The continuty of f(x) at x = n

$\boldsymbol{\lim_{x\rightarrow n}f\left ( x \right )=\lim_{x\rightarrow n}x^{n}=n^{n}}$

$\boldsymbol{\lim_{x\rightarrow n}f\left ( x \right )=f\left ( x \right )}$

Therefore the given function is continuous at x = n

Q5. Is the given function f defined by

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}x, if\: x\leq 1 & \\ 5,if\: x>1& \end{matrix}\right.}$

continuous at x = 0 at x = 1 at x = 2.

Ans. The given function is

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}x, if\: x\leq 1 & \\ 5,if\: x>1& \end{matrix}\right.}$

The function is defined for x< 1

f (x) = x

So, its value at x = 0 is f(0) = 0

The continuty of f(x) at x= 0

$\boldsymbol{\lim_{x\rightarrow 0}f\left ( x \right )=\lim_{x\rightarrow 0}x=0}$

$\boldsymbol{\therefore \lim_{x\rightarrow 0}f\left ( x \right )=f\left ( x \right )}$

Therefore f(x) is continuous at x = 0

The function is defined for x = 1

f (x) = x

The value of f(x) at x = 1

f(1) = 1

The continuty of the function f(x) at x = 1 since f(x) can be approached from LHS and RHS of 1, so calculating LHS limit and RHS limit of the f(x) at x = 1

LHS limit of f(x) at x = 1

$\boldsymbol{ \lim_{x\rightarrow 1^{-}}f\left ( x \right )=\lim_{x\rightarrow 1^{-}}x=1}$

RHS limit of f(x) at x = 1

$\boldsymbol{ \lim_{x\rightarrow 1^{+}}f\left ( x \right )=\lim_{x\rightarrow 1^{+}}x=5}$

$\boldsymbol{ \lim_{x\rightarrow 1^{-}}f\left ( x \right )\neq \lim_{x\rightarrow 1^{+}}f\left ( x \right )}$

LHS limit ≠ RHS limit

Therefore f(x) is not continuous at x = 1

For x >1 the function is defined f(x) = 5

The value of f(x) at x = 2 is f(2) = 5

$\boldsymbol{\lim_{x\rightarrow 2}f\left ( x \right )=\lim_{x\rightarrow 2}5=5}$

$\boldsymbol{\lim_{x\rightarrow 2}f\left ( x \right )=f\left ( x \right )}$

Therefore f(x) is continuous at x = 2

Q6. Find all the points of discontinuty of f , where f is defined by

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}2x+3 \: if\: x\leq 2 & & \\ 2x-3 \: if \: x> 2 & & \end{matrix}\right.}$

Ans. The given function f(x) is defined as follows

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}2x+3 \: if\: x\leq 2 & & \\ 2x-3 \: if \: x> 2 & & \end{matrix}\right.}$

It is clear that the function is valid for x ≤ 2 and x > 2, so is defined for all values of real numbers

Let there is a real number k, so there arises 3 cases as follows

First case, when k < 2

The value of f(x) at x = k

f(k) = 2k + 3

The continuty of f(x) at x = k

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( 2x+3 \right )=2k+3}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

So, f(x) is continuous at x = k

Second case, when x = k

The value of f(x) at x = k

f(k) = 2k + 3

Checking the LHS and RHS limit of the f(x) at k = 2

LHS limit

$\boldsymbol{ \: \lim_{x\rightarrow 2^{-}}f\left ( x \right )=\lim_{x\rightarrow 2^{-}}\left ( 2x+3 \right )=2\times 2+3=7}$

RHS limit

$\boldsymbol{ \: \lim_{x\rightarrow 2^{+}}f\left ( x \right )=\lim_{x\rightarrow 2^{+}}\left ( 2x-3 \right )=2\times 2-3=1}$

LHS limit ≠ RHS

Therefore the function f(x) is discontinuous at x = 2

The third case , when k > 2

The value of f(x) at x = k

f(k) = 2k + 3

The continuty of f(x) at x = k

$\boldsymbol{ \: \lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( 2x-3 \right )=2\times k-3=2k-3}$

$\boldsymbol{ \therefore \: \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

So, f(x) is continuous at x = k

Therefore f(x) is discontinuous at x = 2

Q7. Find all the points of discontinuity of f, where f is defined by

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\left | x \right |+3,\: if\: x\leq -3 & \\-2x,if -3< x< 3 \\6x+2,\: if\: x\geq 3 & \end{matrix}\right.}$

Ans. The given function f is defined as follows

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\left | x \right |+3,\: if\: x\leq -3 & \\-2x,if -3< x< 3 \\6x+2,\: if\: x\geq 3 & \end{matrix}\right.}$

As it is seen that the function f(x) is defined for all values of real numbers

Let k is a point on the real number line, then, there arise following cases

First case when k < 3

The value of f(x) at x = k

$\boldsymbol{f\left ( k \right )= \left | k \right |+1}$

For k < 3

$\boldsymbol{f\left ( k \right )=\left | k \right |+3=-k+3}$

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}\left ( \left | x \right |+3 \right )=\lim_{x\rightarrow k}-x+3=-k+3}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

So, f(x) is continuous at all points when x < -3.

Second case when k = -3

The value of f(x) at k = -3

$\boldsymbol{f\left ( k \right )=\left | k \right |+3=-k+3}$

f(-3) = -(-3) + 3 = 6

LHS limit of f(x) at x = -3

$\boldsymbol{\lim_{x\rightarrow -3^{-}}f\left ( x \right )=\lim_{x\rightarrow -3^{-}}-x+3=-\left ( -3 \right )+3=6}$

RHS limit of f(x) at x = -3

$\boldsymbol{\lim_{x\rightarrow -3^{+}}f\left ( x \right )=\lim_{x\rightarrow -3^{+}}-2x=-2\times -3=6}$

f(x) = LHS limit = RHS limit

So, the f(x) is continuous at x = -3

Third case when -3 < x < 3

Let – 3 < k < 3

The value of f(x) at x = k

f(k) = -2k

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}-2x=-2k}$

Thus

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

So, f(x) is continuous at -3 < x < 3

Fourth case when k > 3

f(x) = 6x + 2

The value of f(x) at x = k

f(k) = 6 × k + 2 = 6k + 2

The limit of the f(x) at x = k

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

So, f(x) is continuous at k > 3

Fifth case when k = 3

LHS limit at k = 3

$\boldsymbol{\lim_{x\rightarrow 3^{-}}f\left ( x \right )=\lim_{x\rightarrow 3^{-}}-2x=-2\times 3=-6}$

RHS limit at k = 3

$\boldsymbol{\lim_{x\rightarrow 3^{+}}f\left ( x \right )=\lim_{x\rightarrow 3^{+}}6x+2=6\times 3+2=20}$

RHS limit ≠ LHS limit

So, f(x) is not continuous at k = 3

Q8. Find all the points of discontinuity of f, where f is defined by

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\frac{\left | x \right |}{x}\: if \: x\neq 0 & \\ & \\ 0\: if\: x=0& \end{matrix}\right.}$

Ans. The given function is as follows

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\frac{\left | x \right |}{x}\: if \: x\neq 0 & \\ & \\ 0\: if\: x=0& \end{matrix}\right.}$

There arises the following cases

For x ≠ 0

$\boldsymbol{f\left ( x \right )=\frac{\left | x \right |}{x}}$

When x < 0, then $\boldsymbol{\left | x \right |=-x}$

$\boldsymbol{f\left ( x \right )=\frac{\left | x \right |}{x}=\frac{-x}{x}=-1}$

When x > 0

$\boldsymbol{f\left ( x \right )=\frac{\left | x \right |}{x}=\frac{x}{x}=1}$

First case, let there is a point on real number line k <0

The value of f(x) at x = k

f(k) = -1

Limit of f(x) at x = k

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}-1=-1}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

So, f(x) is continuous for all values x < 0

Second case, when k > o

Value of f(x) at x = k

f(k) = 1

The limit of f(x) at x = k

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}1=1}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

So, f(x) is continuous for all values x > 0

Third case when x = 0

LHS limit at x = 0

$\boldsymbol{\lim_{x\rightarrow 0^{-}}f\left ( x \right )=\lim_{x\rightarrow 0^{-}}1=-1}$

RHS limit at x = 0

$\boldsymbol{\lim_{x\rightarrow 0^{+}}f\left ( x \right )=\lim_{x\rightarrow 0^{+}}1=1}$

LHS limit ≠ RHS limit

So, f(x) is not continuous at x = 0

Q9.Find all the points of discontinuity of f, where f is defined by

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\frac{x}{\left | x \right |},\: if\: x<0 & & \\ -1,if\: x\geq 0& & \end{matrix}\right.}$

Ans. The given function is f(x) is defined as follows

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}\frac{x}{\left | x \right |},\: if\: x<0 & & \\ -1,if\: x\geq 0& & \end{matrix}\right.}$

If x < 0

$\boldsymbol{f\left ( x \right )=\frac{x}{\left | x \right |}=\frac{x}{-x}=-1}$

Let there is a point k on the real number line such that k < 0

The value of f(x) at x = k

f(k) = -1

The limit of f(x) at x = k

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}-1=-1}$

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

So, f(x) is continuous at x = k

If x ≥ 0, then f(x) = -1

Let there is a point k on real number line such that k > 0

f(k) = -1

The limit at x = k, will be = -1

So, the f(x) is continuous for all values x > 0

If x = 0, then f(x) = -1

Let there is a point k on real number line such that k = 0

f(k) =f(0) =-1

LHS limit

$\boldsymbol{\lim_{x\rightarrow 0^{-}}f\left ( x \right )=\lim_{x\rightarrow 0^{-}}-1=-1}$

RHS limit

$\boldsymbol{\lim_{x\rightarrow 0^{+}}f\left ( x \right )=\lim_{x\rightarrow 0^{+}}-1=-1}$

Since LHS limit = RHS limit = f(k) = -1

So, the function f(x) is continuous at x = 0

Hence the f(x) is continuous at all points ,thus f(x) is not discontinuous at any point.

Q10.Find all the points of discontinuity of f, where f is defined by

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}x+1\: if\: x\geq 1 & \\ x^{2}+1,if\: x<1& \end{matrix}\right.}$

Ans. The given function f(x) is defined as follows

$\boldsymbol{f\left ( x \right )=\left\{\begin{matrix}x+1\: if\: x\geq 1 & \\ x^{2}+1,if\: x<1& \end{matrix}\right.}$

First case when x > 1

f(x) = x + 1

Let there is point on real number line such that k >1

The value of f(x) at x = k

f(k) = k + 1

The limit of f(x) at x = k

$\boldsymbol{\lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}x+1=k+1}$

$\boldsymbol{\therefore \lim_{x\rightarrow k}f\left ( x \right )=f\left ( x \right )}$

So, f(x) is continuous at all values of x > 1

When x = 1, f(x) = x + 1

The value of f(x) at x = 1

f(1) = 1 + 1 = 2

RHS limit of f(x)

$\boldsymbol{ \lim_{x\rightarrow 1^{+}}f\left ( x \right )=\lim_{x\rightarrow 1^{+}}x+1=1+1=2}$

LHS limit of f(x) at x = 1

$\boldsymbol{ \lim_{x\rightarrow 1^{-}}f\left ( x \right )=\lim_{x\rightarrow 1^{-}}x^{2}+1=1^{}+1=2}$

So, f(x) = LHS limit = RHS limit

So, f(x) is continuous at x = 1

If x <1, then f(x) = x² + 1

Let there is a point k on real number line such that k < 1

The value of f(x) at x = k

f(k) = k² + 1

The limit of f(x) at x = k

$\boldsymbol{ \lim_{x\rightarrow k}f\left ( x \right )=\lim_{x\rightarrow k}x^{2}+1=k^{2}+1}$

$\boldsymbol{ \lim_{x\rightarrow k}f\left ( x \right )=f\left ( k \right )}$

So, the f(x) is continuous at all values of x> 1

Hence f(x) is continuous at all real values of x, so f(x) is discontinuous at any point.

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals