NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5 - Continuity and Differentiability

Q: Can i get CBSE Class 10 exam pattern ?

Click on the link to get CBSE Class 10 Exam Pattern .

Komal MiglaniUpdated on 24 Apr 2025, 09:49 AM IST

Continuity means a function does not jump or disappear, while Differentiability means the function does not stumble and keeps going without any sharp or awkward turns. Understanding how functions change is not just about finding their slopes, but we can go one step further and find how those slopes change to get a better look at how the functions behave. This is where the second-order derivative plays an important role in calculus, it helps us to determine the curvature of the function. In exercise 5.7 of the chapter Continuity and Differentiability, we will learn about the concept of the second-order derivative, which can tell us about how the first-order derivative, i.e. the rate of change itself, is changing. This article on the NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5 - Continuity and Differentiability provides detailed solutions for the problems given in the exercise, so that students can clear their doubts and get a clear understanding of the method and logic behind these solutions. For syllabus, notes, and PDF, refer to this link: NCERT.

Class 12 Maths Chapter 5 Exercise 5.7 Solutions: Download PDF

Download PDF

Continuity and Differentiability Exercise: 5.7

Question:1 Find the second order derivatives of the functions given in Exercises 1 to 10.

$x^2 + 3x+ 2$

Answer:

Given function is
$y=x^2 + 3x+ 2$
Now, differentiation w.r.t. x
$\frac{dy}{dx}= 2x+3$
Now, second order derivative
$\frac{d^2y}{dx^2}= 2$
Therefore, the second order derivative is $\frac{d^2y}{dx^2}= 2$

Question:2 Find the second order derivatives of the functions given in Exercises 1 to 10.

$x ^{20}$

Answer:

Given function is
$y=x ^{20}$
Now, differentiation w.r.t. x
$\frac{dy}{dx}= 20x^{19}$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= 20.19x^{18}= 380x^{18}$
Therefore, second-order derivative is $\frac{d^2y}{dx^2}= 380x^{18}$

Question:3 Find the second order derivatives of the functions given in Exercises 1 to 10.

$x \cos x$

Answer:

Given function is
$y = x \cos x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}= \cos x + x(-\sin x ) = \cos x-x\sin x$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= -\sin x-(\sin x+x\cos x) = -2\sin x - x\sin x$
Therefore, the second-order derivative is $\frac{d^2y}{dx^2}= -2\sin x - x\sin x$

Question:4 Find the second order derivatives of the functions given in Exercises 1 to 10.

$\log x$

Answer:

Given function is
$y=\log x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{1}{x}$
Now, second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{x^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2}= \frac{-1}{x^2}$

Question:5 Find the second order derivatives of the functions given in Exercises 1 to 10.

$x ^3 \log x$

Answer:

Given function is
$y=x^3\log x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=3x^2.\log x+x^3.\frac{1}{x}= 3x^2.\log x+ x^2$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= 6x.\log x+3x^2.\frac{1}{x}+2x=6x.\log x+3x+2x = x(6.\log x+5)$
Therefore, the second-order derivative is $\frac{d^2y}{dx^2} = x(6.\log x+5)$

Question:6 Find the second order derivatives of the functions given in Exercises 1 to 10.

$e ^x \sin5 x$

Answer:

Given function is
$y= e^x\sin 5x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=e^x.\sin 5x +e^x.5\cos 5x = e^x(\sin5x+5\cos5x)$
Now, second order derivative is
$\frac{d^2y}{dx^2}= e^x(\sin5x+5\cos5x)+e^x(5\cos5x+5.(-5\sin5x))$
$= e^x(\sin5x+5\cos5x)+e^x(5\cos5x-25\sin5x)=e^x(10\cos5x-24\sin5x)$
$=2e^x(5\cos5x-12\sin5x)$
Therefore, second order derivative is $\frac{dy}{dx}=2e^x(5\cos5x-12\sin5x)$

Question:7 Find the second order derivatives of the functions given in Exercises 1 to 10.

$e ^{6x}\cos 3x$

Answer:

Given function is
$y= e^{6x}\cos 3x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=6e^{6x}.\cos 3x +e^{6x}.(-3\sin 3x)= e^{6x}(6\cos 3x-3\sin 3x)$
Now, second order derivative is
$\frac{d^2y}{dx^2}= 6e^{6x}(6\cos3x-3\sin3x)+e^{6x}(6.(-3\sin3x)-3.3\cos3x)$
$= 6e^{6x}(6\cos3x-3\sin3x)-e^{6x}(18\sin3x+9\cos3x)$
$e^{6x}(27\cos3x-36\sin3x) = 9e^{6x}(3\cos3x-4\sin3x)$
Therefore, second order derivative is $\frac{dy}{dx} = 9e^{6x}(3\cos3x-4\sin3x)$

Question:8 Find the second order derivatives of the functions given in Exercises 1 to 10.

$\tan ^{-1} x$

Answer:

Given function is
$y = \tan^{-1}x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\tan^{-1}x)}{dx}=\frac{1}{1+x^2}$
Now, second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{(1+x^2)^2}.2x = \frac{-2x}{(1+x^2)^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-2x}{(1+x^2)^2}$

Question:9 Find the second order derivatives of the functions given in Exercises 1 to 10.

$\log (\log x )$

Answer:

Given function is
$y = \log(\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\log(\log x))}{dx}=\frac{1}{\log x}.\frac{1}{x}= \frac{1}{x\log x}$
Now, second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{(x\log x)^2}.(1.\log x+x.\frac{1}{x}) = \frac{-(\log x+1)}{(x\log x)^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-(\log x+1)}{(x\log x)^2}$

Question:10 Find the second order derivatives of the functions given in Exercises 1 to 10.

$\sin (\log x )$

Answer:

Given function is
$y = \sin(\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\sin(\log x))}{dx}=\cos (\log x).\frac{1}{x}= \frac{\cos (\log x)}{x}$
Now, second order derivative is
Using Quotient rule
$\frac{d^2y}{dx^2}=\frac{-\sin(\log x)\frac{1}{x}.x-\cos(\log x).1}{x^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}$

Question:11 If $y = 5 \cos x - 3 \sin x$ prove that $\frac{d^2y}{dx^2}+y = 0$

Answer:

Given function is
$y = 5 \cos x - 3 \sin x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(5\cos x-3\sin x)}{dx}=-5\sin x-3\cos x$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(-5\sin x-3\cos x)}{dx^2}=-5\cos x+3\sin x$
Now,
$\frac{d^2y}{dx^2}+y=-5\cos x+3\sin x+5\cos x-3\sin x = 0$
Hence proved

Question:12 If $y = \cos ^{-1} x$ Find $\frac{d ^2 y }{dx^2 }$ in terms of y alone.

Answer:

Given function is
$y = \cos ^{-1} x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d( \cos ^{-1} x)}{dx}=\frac{-1}{\sqrt{1-x^2}}$
Now, second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{\sqrt{1-x^2}})}{dx^2}=\frac{-(-1)}{(\sqrt{1-x^2})^2}.(-2x) = \frac{-2x}{1-x^2}$ -(i)
Now, we want $\frac{d^2y}{dx^2}$ in terms of y
$y = \cos ^{-1} x$
$x = \cos y$
Now, put the value of x in (i)

$\frac{d^2y}{dx^2} = \frac{-2\cos y}{1 - \cos^2 y} = \frac{-2\cos y}{\sin^2 y} = -2\cot y \, \operatorname{cosec} y$
$\frac{d^2y}{dx^2} = \frac{-2\cos y}{1 - \cos^2 y} = \frac{-2\cos y}{\sin^2 y} = -2\cot y\, \mathrm{cosec}\, y$
$\left(\because\ 1 - \cos^2 x = \sin^2 x,\ \frac{\cos x}{\sin x} = \cot x,\ \text{and}\ \frac{1}{\sin x} = \mathrm{cosec}\, x \right)$
Therefore, answer is $\frac{d^2y}{dx^2} = -2 \cot y \, \mathrm{cosec}\, y$

Question:13 If $y = 3 \cos (\log x) + 4 \sin (\log x)$, show that $x^2 y_2 + xy_1 + y = 0$

Answer:

Given function is
$y = 3 \cos (\log x) + 4 \sin (\log x)$
Now, differentiation w.r.t. x
$y_1=\frac{dy}{dx}=\frac{d( 3 \cos (\log x) + 4 \sin (\log x))}{dx}=-3\sin(\log x).\frac{1}{x}+4\cos (\log x).\frac{1}{x}$
$=\frac{4\cos (\log x)-3\sin(\log x)}{x}$ -(i)
Now, second order derivative is
By using the Quotient rule
$y_2 = \frac{d^2y}{dx^2} = \frac{d^2\left(\frac{4\cos(\log x) - 3\sin(\log x)}{x}\right)}{dx^2}$
$= \frac{\left(-4\sin(\log x) \cdot \frac{1}{x} - 3\cos(\log x) \cdot \frac{1}{x}\right) \cdot x - 1 \cdot \left(4\cos(\log x) - 3\sin(\log x)\right)}{x^2}$
$= \frac{-\sin(\log x) + 7\cos(\log x)}{x^2} \ \text{-(ii)}$
Now, from equation (i) and (ii) we will get $y_1 \ and \ y_2$
Now, we need to show
$x^2 y_2 + xy_1 + y = 0$
Put the value of $y_1 \ and \ y_2$ from equation (i) and (ii)
$x^2\left( \frac{-\sin(\log x) + 7\cos(\log x)}{x^2} \right) + x\left( \frac{4\cos(\log x) - 3\sin(\log x)}{x} \right) + 3\cos(\log x) + 4\sin(\log x)$
$-\sin(\log x) - 7\cos(\log x) + 4\cos(\log x) - 3\sin(\log x) + 3\cos(\log x) + 4\sin(\log x)$
$=0$
Hence proved

Question:14 If $y = A e ^{mx} + Be ^{nx}$ , show that $\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0$

Answer:

Given function is
$y = A e ^{mx} + Be ^{nx}$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(A e ^{mx} + Be ^{nx})}{dx}=mAe^{mx}+nBe^{nx}$ -(i)
Now, second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(mAe^{mx}+nBe^{nx})}{dx^2}= m^2Ae^{mx}+n^2Be^{nx}$ -(ii)
Now, we need to show
$\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0$
Put the value of $\frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}$ from equation (i) and (ii)
$m^2Ae^{mx}+n^2Be^{nx}-(m+n)(mAe^{mx}+nBx^{nx}) +mn(Ae^{mx}+Be^{nx})$
$m^2Ae^{mx}+n^2Be^{nx}-m^2Ae^{mx}-mnBx^{nx}-mnAe^{mx} -n^2Be^{nx}+mnAe^{mx}$$+mnBe^{nx}$
$=0$
Hence proved

Question:15 If $y = 500 e ^{7x} + 600 e ^{- 7x }$ , show that $\frac{d^2 y}{dx ^2} = 49 y$
Answer:

Given function is
$y = 500 e ^{7x} + 600 e ^{- 7x }$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(500 e ^{7x} + 600 e ^{- 7x })}{dx}=7.500e^{7x}-7.600e^{-7x} =3500e^{7x}-4200e^{-7x}$ -(i)
Now, second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(3500e^{7x}-4200e^{-7x})}{dx^2}$
$= 7.3500e^{7x}-(-7).4200e^{-7x}= 24500e^{7x}+29400e^{-7x}$ -(ii)
Now, we need to show
$\frac{d^2 y}{dx ^2} = 49 y$
Put the value of $\frac{d^2y}{dx^2}$ from equation (ii)
$24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})$
$= 24500e^{7x}+29400e^{-7x}$
Hence, L.H.S. = R.H.S.
Hence proved

Question:16 If $e ^y (x+1) = 1$ show that $\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2$

Answer:

Given function is
$e ^y (x+1) = 1$
We can rewrite it as
$e^y = \frac{1}{x+1}$
Now, differentiation w.r.t. x
$\frac{d(e^y)}{dx}=\frac{d(\frac{1}{x+1})}{dx}\\ e^y.\frac{dy}{dx}= \frac{-1}{(x+1)^2}\\ \frac{1}{x+1}.\frac{dy}{dx}= \frac{-1}{(x+1)^2} \ \ \ \ \ \ \ \ \ (\because e^y = \frac{1}{x+1})\\ \frac{dy}{dx}= \frac{-1}{x+1}$ -(i)
Now, second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{x+1})}{dx^2}=\frac{-(-1)}{(x+1)^2} = \frac{1}{(x+1)^2}$ -(ii)
Now, we need to show
$\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2$
Put value of $\frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}$ from equation (i) and (ii)
$\frac{1}{(x+1)^2}=\left ( \frac{-1}{x+1} \right )^2$
$=\frac{1}{(x+1)^2}$
Hence, L.H.S. = R.H.S.
Hence proved

Question:17 If $y = (\tan^{-1} x)^2$ show that $(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$

Answer:

Given function is
$y = (\tan^{-1} x)^2$
Now, differentiation w.r.t. x
$y_1=\frac{dy}{dx}=\frac{d((\tan^{-1}x)^2)}{dx}= 2.\tan^{-1}x.\frac{1}{1+x^2}= \frac{2\tan^{-1}x}{1+x^2}$ -(i)
Now, the second-order derivative is
By using the quotient rule
$y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{2\tan^{-1}x}{1+x^2})}{dx^2}=\frac{2.\frac{1}{1+x^2}.(1+x^2)-2\tan^{-1}x(2x)}{(1+x^2)^2}=\frac{2-4x\tan^{-1}x}{(1+x^2)^2}$ -(ii)
Now, we need to show
$(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$
Put the value from equation (i) and (ii)
$(x^2+1)^2.\frac{2-4x\tan^{-1}x}{(1+x^2)^2}+2x(x^2+1).\frac{2\tan^{-1}x}{x^2+1}\\ \Rightarrow 2-4x\tan^{-1}x+4x\tan^{-1}x = 2$
Hence, L.H.S. = R.H.S.
Hence proved

Topics covered in Chapter 5, Continuity and Differentiability: Exercise 5.7

The main topics covered in Chapter 5 of continuity and differentiability, exercises 5.7 are:

Second-order derivative: Understanding how second-order derivatives work and how to evaluate second-order derivatives. For example the second-order derivatives of $y=f(x)$ can be written as $\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})$.
Applications of second-order derivatives: There are many applications of second-order derivatives, like finding the maxima and minima of a function, determining the curvature of a graph, etc.

Aakash Repeater Courses

Take Aakash iACST and get instant scholarship on coaching programs.

Apply

Related eBooks & Sample Papers

CBSE Class 12 Sample Paper English Elective 2025-26 with Solution

68+ Downloads

CBSE Class 12 Sample Paper English Core 2025-26 with Solution

66+ Downloads

NCERT Solutions Subject Wise

Below are some useful links for subject-wise NCERT solutions for class 12.

NCERT Exemplar Solutions Subject Wise

Here are some links to subject-wise solutions for the NCERT exemplar class 12.

Frequently Asked Questions (FAQs)

Q: If y = c is a function where c is a constant then find dy/dx ?

y = c

dy/dx = 0

Q: If y = c is a function where c is a constant then find the second order derivative of y ?

y = c

dy/dx = 0

d(dy/dx)/dx = 0

Q: Find the first derivative of y = x ?

Given y = x

dy/dx = 1

Q: Find the second order derivative of y = x ?

Given y = x

dy/dx = 1

d(dy/dx)/dx = 0

Q: What is the second order derivative of y = e^x ?

y = e^x

dy/dx = e^x

d(dy/dx)/dx = e^x

d^(2)y/dx^2 = e^x

Q: Can i get CBSE Class 10 exam pattern ?

Click on the link to get CBSE Class 10 Exam Pattern.

Q: Can I get detailed syllabus for CBSE Class 10 ?

Click here to get Syllabus for CBSE Class 10

Q: Can I get detailed syllabus for CBSE Class 10 Maths ?

Click on the given link to get Syllabus for CBSE Class 10 Maths.

Articles

NCERT Exemplar Class 12 Chemistry Solutions chapter 16 Chemistry in Everyday life

3 days ago

NCERT Resources

NCERT Exemplar Class 12 Chemistry Solutions - Chapter Wise Problems with Solutions

3 days ago

NCERT Resources

NCERT Exemplar Class 12 Chemistry Solutions chapter 13 Amines

3 days ago

NCERT Resources

NCERT Exemplar Class 12 Chemistry Solutions Chapter 15 Polymers

3 days ago

NCERT Resources

Upcoming School Exams

National Means Cum-Merit Scholarship

Ongoing Dates

NMMS Application Date

25 Jul'25 - 30 Sep'25 (Online)

Assam High School Leaving Certificate Examination

Ongoing Dates

Assam HSLC Application Date

1 Sep'25 - 4 Oct'25 (Online)

Telangana Open School Society Intermediate Examination

Ongoing Dates

TOSS Intermediate Late Fee Application Date

8 Sep'25 - 20 Sep'25 (Online)

Admit Card

Result

Certifications By Top Providers

Introduction to Accounting-Part 1 Basics of Financial Statements

Via Indian Institute of Management Bangalore

Basic 3D Animation using Blender

Via Indian Institute of Technology Bombay

People Management with Impact

Via Indian Institute of Management Bangalore

Essentials of Program Strategy and Evaluation

Via Stanford University, Stanford

Preparing for the AP English Language and Composition Exam

Via Tennessee Board of Regents

Learn to Speak Korean 1

Via Yonsei University, Seoul

1074 courses

817 courses

394 courses

273 courses

Harvard University, Cambridge

97 courses

University of Michigan, Ann Arbor

83 courses

IT & Software

Teaching and Academics

Programming and Development

Health, Fitness and Medicine

Business and Management

Engineering and Architecture

General Management

Financial Management

Environmental Studies

History

Psychology

Artificial Intelligence

Explore Top Universities Across Globe

University of Essex, Colchester

Wivenhoe Park Colchester CO4 3SQ

University College London, London

Gower Street, London, WC1E 6BT

The University of Edinburgh, Edinburgh

Old College, South Bridge, Edinburgh, Post Code EH8 9YL

University of Delaware, Newark

Newark, DE 19716

University of Nottingham, Nottingham

University Park, Nottingham NG7 2RD

Bristol Baptist College, Bristol

The Promenade, Clifton Down, Bristol BS8 3NJ

Cost of Living in South Korea for Indian Students 2025

11 min ⋆ May 21, 2025 11:05 AM IST

Cheapest Universities in London For International Students 2025

12 min ⋆ Sep 09, 2025 11:09 AM IST

Study MBBS in Kyrgyzstan 2025: Fees, Top Medical Colleges

15 min ⋆ Apr 07, 2025 05:04 AM IST

MBBS in Uzbekistan 2025: Fees, Top Medical Colleges & Admission Process

16 min ⋆ Sep 09, 2025 10:09 AM IST

Study in Dubai 2025-26: Cost, Admission Process, Eligibility, Visa, Scholarship

13 min ⋆ Sep 09, 2025 09:09 AM IST

GMAT 2025 Exam: Dates, Registration, Fees, Eligibility, Syllabus, Pattern, Result

28 min ⋆ Sep 09, 2025 09:09 AM IST

Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?

Can i improve my 12 result this year in PCM

Hello,

If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5 - Continuity and Differentiability

Class 12 Maths Chapter 5 Exercise 5.7 Solutions: Download PDF

Continuity and Differentiability Exercise: 5.7

Topics covered in Chapter 5, Continuity and Differentiability: Exercise 5.7

Aakash Repeater Courses

Related eBooks & Sample Papers

NCERT Solutions Subject Wise

NCERT Exemplar Solutions Subject Wise

Questions related to CBSE Class 12th

Popular CBSE Class 12th Questions

Study Resources, Applications and Opportunities

Aakash Repeater Courses

NEET Most scoring concepts

NEET Previous 10 Year Questions

JEE Main Important Physics formulas

JEE Main Important Chemistry formulas

JEE Main high scoring chapters and topics

CBSE Class 12th Notifications

Latest updates

Never miss an CBSE Class 12th update

NCERT Solutions

NCERT Exemplars

NCERT Books

NCERT Notes

CBSE

CBSE Class 10

CBSE Class 12th

CISCE Board 10th

IMO

IEO

NSO

NMMS

NVS

JNVST

Schools By Medium

By Ownership

By City

By State

Download Careers360 App

All this at the convenience of your phone

Scan and download the app