Chapter 5 : Continuity and Differentiability - Ncert Solutions for Class 12-science Maths CBSE

Your CBSE Class 12 syllabus for Maths consists of topics such as linear programming, vector quantities, determinants, etc. which lay the foundation for further education in science, engineering, management, etc. Studying differential equations will be useful for exploring subjects such as Physics, Biology, Chemistry, etc. where your knowledge can be applied for scientific investigations.

On TopperLearning, you can find study resources such as sample papers, mock tests, Class 12 Maths NCERT solutions and more. These learning materials can help you understand concepts such as differentiation of functions, direction cosines, integrals, and more. Also, you can practise the Maths problems by going through the solutions given by our experts.

Maths is considered as one of the most difficult subjects in CBSE Class 12 Science. Our Maths experts simplify complex Maths problems by assisting you with the right methods to solve problems and score full marks. You may still have doubts while referring to the Maths revision notes or Maths NCERT solutions. Solve those doubts by asking an expert through the “Undoubt” feature on the student dashboard.

Read  more

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.1

Solution 1

The given function is f(x) = 5x - 3

At x = 0, f(0) = 5 × 0 - 3 = -3

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

 
Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity.
 
Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

The given function is f(x) = x2 - sinx + 5

It is evident that f is defined at x = ∏

Solution 21

Solution 22

Therefore, cosecant is continuous except at x = np, n begin mathsize 12px style element of end styleZ
Therefore, cotangent is continuous except at x = np, n element ofZ
Solution 23

Solution 24

 
begin mathsize 12px style negative straight x squared less or equal than straight x squared sin 1 over straight x less or equal than straight x squared end style
Solution 25

Solution 26

Solution 27

Solution 28

 

The given function f is continuous at x = begin mathsize 12px style straight pi end style, if f is defined at x = begin mathsize 12px style straight pi end style and if the value

of f at x = begin mathsize 12px style straight pi end style equals the limit of f at x = begin mathsize 12px style straight pi end style

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

 
Solution 34

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.2

Solution 1

Then, (v ο u)(x) = v(u(x)) = v(x2 + 5) = sin (x2 + 5) = f(x)

 

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.4

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.5

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.6

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.7

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.8

Solution 1

 

Rolle's theorem states that there is a point c element of (-4, -2) such that f'(c) = 0

Solution 2

 

then there exists some c begin mathsize 12px style element of end style(a, b) such that f'(c) = 0

Let n be an integer such that n begin mathsize 12px style element of end style (5, 9).

Let n be an integer such that n begin mathsize 12px style element of end style (-2, 2).

Solution 3

Therefore, by the Mean Value Theorem, there exists c begin mathsize 12px style element of end style (-5, 5) such that

 

Solution 4

 

Mean Value Theorem states that there is a point c begin mathsize 12px style element of end style(1, 4) such that f'(c) = 1

Solution 5

Solution 6


Chapter 5 - Continuity and Differentiability Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

 

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

 

where sin x > cosx 

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

begin mathsize 12px style fraction numerator open square brackets 1 plus open parentheses begin display style dy over dx end style close parentheses squared close square brackets to the power of begin display style 3 over 2 end style end exponent over denominator fraction numerator straight d squared straight y over denominator dx squared end fraction end fraction equals fraction numerator begin display style open square brackets 1 plus begin display style open parentheses straight x minus straight a close parentheses squared over open parentheses straight y minus straight b close parentheses squared end style close square brackets to the power of 3 over 2 end exponent end style over denominator negative open square brackets begin display style fraction numerator open parentheses straight y minus straight b close parentheses squared plus open parentheses straight x minus straight a close parentheses squared over denominator open parentheses straight y minus straight b close parentheses cubed end fraction end style close square brackets end fraction equals fraction numerator begin display style open square brackets fraction numerator begin display style open parentheses straight y minus straight b close parentheses squared plus open parentheses straight x minus straight a close parentheses squared end style over denominator begin display style open parentheses straight y minus straight b close parentheses squared end style end fraction close square brackets to the power of 3 over 2 end exponent end style over denominator negative open square brackets fraction numerator open parentheses straight y minus straight b close parentheses squared plus open parentheses straight x minus straight a close parentheses squared over denominator open parentheses straight y minus straight b close parentheses cubed end fraction close square brackets end fraction end style
begin mathsize 12px style equals fraction numerator open square brackets begin display style straight c squared over open parentheses straight y minus straight b close parentheses squared end style close square brackets over denominator begin display style fraction numerator negative straight c squared over denominator open parentheses straight y minus straight b close parentheses cubed end fraction end style end fraction equals fraction numerator begin display style straight c cubed over open parentheses straight y minus straight b close parentheses cubed end style over denominator begin display style fraction numerator negative straight c squared over denominator open parentheses straight y minus straight b close parentheses cubed end fraction end style end fraction equals negative straight c space which space is space constant space and space is space independent space of space straight a space and space straight b end style
 
Hence, proved.
Solution 16

begin mathsize 12px style rightwards double arrow sin open parentheses straight a plus straight y minus straight y close parentheses dy over dx equals cos squared open parentheses straight a plus straight y close parentheses
rightwards double arrow dy over dx equals fraction numerator cos squared open parentheses straight a plus straight y close parentheses over denominator sina end fraction end style

 

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

 Yes.

Consider the function f(x)=|x-1|+|x-2|

Since we know that the modulus function is continuous everywhere, so there sum is also continuous

Therefore, function f is continuous everywhere

Now, let us check the differentiability of f(x) at x=1,2

At x=1

LHD =

 [Take x=1-h, h>0 such that h0 as x1-]

  

Now,

RHD =

 [Take x=1+h, h>0 such that h0 as x1+]

≠ LHD

Therefore, f is not differentiable at x=1.

 

At x=2

LHD =

 [Take x=2-h, h>0 such that h0 as x2-]

  

Now,

RHD =

 [Take x=2+h, h>0 such that h0 as x2+]

≠ LHD

Therefore, f is not differentiable at x=2.

Hence, f is not differentiable at exactly two points.

Solution 22

Solution 23

Loading...

Why CBSE Class 12 Science Maths solutions are important?

Maths is a subject which requires practising a variety of problems to understand concepts clearly. By solving as many problems as you can, you’ll be able to train your brain in thinking the logical way to solve maths problems. For practising problems, study materials such as sample papers, previous year papers, and NCERT solutions are needed.

Some of the best Maths experts work with us to give you the best solutions for Maths textbook questions and sample paper questions. Chapter-wise NCERT solutions for Class 12 Science Maths can be easily accessible on TopperLearning. Use these solutions to practise problems based on concepts such as direction ratios, probability, area between lines, inverse trigonometric functions, and more.

To prepare for your Maths exam, you need to attempt solving different kinds of Maths questions. One of the best ways to assess your problem-solving abilities is to attempt solving previous year papers with a set timer. Our Maths solutions will come in handy to help you with checking your answers and thus, improving your learning experience. So, to score more marks in your Class 12 board exams, use our Maths solutions that will enable you with the appropriate preparation.