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NCERT Solutions for Exercise 5.8 Class 12 Maths Chapter 5 - Continuity and Differentiability

NCERT Solutions for Exercise 5.8 Class 12 Maths Chapter 5 - Continuity and Differentiability

Edited By Ramraj Saini | Updated on Dec 03, 2023 05:16 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.8

NCERT Solutions for Exercise 5.8 Class 12 Maths Chapter 5 Continuity and Differentiability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In the previous exercises of this Class 12 NCERT syllabuschapter, you have already learned about the first-order derivatives and second-order derivatives. In NCERT solutions for Class 12 Maths chapter 5 exercise 5.8, you will learn about the two important theorems called Rolle's theorem and the Mean value theorem. These theorems are used to prove the inequality of derivatives, study the properties of the derivatives. The proof of these theorems also given in the Class 12 Maths ch 5 ex 5.8. You can go through the proof to get in-depth knowledge of these theorems.

There are some examples given in the NCERT book before the Class 12th Maths chapter 5 exercise 5.8. You should be thorough with the exercise 5.8 Class 12 Maths as one question from this exercise is generally asked in the board exams. 12th class Maths exercise 5.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Assess NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.8

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Continuity and Differentiability Exercise: 5.8

Question:1 Verify Rolle’s theorem for the functionf (x) = x^2 + 2x - 8, x \epsilon [- 4, 2].

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a c \ \epsilon \ (x,y) such that f^{'}(c)= 0
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
f (x) = x^2 + 2x - 8
Now, being a polynomial function, f (x) = x^2 + 2x - 8 is both continuous in [-4,2] and differentiable in (-4,2)
Now,
f (-4) = (-4)^2 + 2(-4) - 8= 16-8-8=16-16=0
Similalrly,
f (2) = (2)^2 + 2(2) - 8= 4+4-8=8-8=0
Therefore, value of f (-4) = f(2)=0 and value of f(x) at -4 and 2 are equal
Now,
According to roll's theorem their is point c , c \ \epsilon (-4,2) such that f^{'}(c)=0
Now,
f^{'}(x)=2x+2\\ f^{'}(c)=2c+2\\ f^{'}(c)=0\\ 2c+2=0\\ c = -1
And c = -1 \ \epsilon \ (-4,2)
Hence, Rolle's theorem is verified for the given function f (x) = x^2 + 2x - 8

Question:2 (1) Examine if Rolle’s theorem is applicable to any of the following functions. Can
you say some thing about the converse of Rolle’s theorem from these example?
f (x) = [x] \: \: for \: \: x \epsilon [ 5,9]

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a c \ \epsilon \ (x,y) such that f^{'}(c)= 0
If all these conditions are satisfied then we can verify Rolle's theorem
Given function is
f (x) = [x]
It is clear that Given function f (x) = [x] is not continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n \ \epsilon \ [5,9]
\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h}= -\infty
( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)
Now,
R.H.L. at x = n , n \ \epsilon \ [5,9]
\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0
( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, the function is not differential in (5,9)
Hence, Rolle's theorem is not applicable for given function f (x) = [x] , x \ \epsilon \ [5,9]

Question:2 (2) Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?

f (x) = [x] \: \:for \: \: x \epsilon [ -2,2]

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a c \ \epsilon \ (x,y) such that f^{'}(c)= 0
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
f (x) = [x]
It is clear that Given function f (x) = [x] is not continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n \ \epsilon \ [-2,2]
\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h}= -\infty
( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)
Now,
R.H.L. at x = n , n \ \epsilon \ [-2,2]
\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0
( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (-2,2)
Hence, Rolle's theorem is not applicable for given function f (x) = [x] , x \ \epsilon \ [-2,2]

Question:2 (3) Examine if Rolle’s theorem is applicable to any of the following functions. Can
you say some thing about the converse of Rolle’s theorem from these example?
f (x) = x^2 - 1 \: \:for \: \: x \epsilon [ 1,2]

Answer:

According to Rolle's theorem function must be
a ) continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then there exist a c \ \epsilon \ (x,y) such that f^{'}(c)= 0
If all these conditions are satisfied then we can verify Rolle's theorem
Given function is
f (x) = x^2-1
Now, being a polynomial , function f (x) = x^2-1 is continuous in [1,2] and differentiable in(1,2)
Now,
f(1)=1^2-1 = 1-1 = 0
And
f(2)=2^2-1 = 4-1 = 3
Therefore, f(1)\neq f(2)
Therefore, All conditions are not satisfied
Hence, Rolle's theorem is not applicable for given function f (x) = [x] , x \ \epsilon \ [-2,2]

Question:3 If f ; [ -5 ,5] \rightarrow R is a differentiable function and if f ' (x) does not vanish
anywhere, then prove that f (-5) \neq f(5)

Answer:

It is given that
f ; [ -5 ,5] \rightarrow R is a differentiable function
Now, f is a differential function. So, f is also a continuous function
We obtain the following results
a ) f is continuous in [-5,5]
b ) f is differentiable in (-5,5)
Then, by Mean value theorem we can say that there exist a c in (-5,5) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
f^{'}(c) = \frac{f(5)-f(-5)}{5-(-5)}\\ f^{'}(c)= \frac{f(5)-f(-5)}{10}\\ 10f^{'}(c)= f(5)-f(-5)
Now, it is given that f ' (x) does not vanish anywhere
Therefore,
10f^{'}(c)\neq 0\\ f(5)-f(-5) \neq 0\\ f(5)\neq f(-5)
Hence proved

Question:4 Verify Mean Value Theorem, if f (x) = x^2 - 4x - 3in the interval [a, b], where
a = 1 and b = 4.

Answer:

Condition for M.V.T.
If f ; [ a ,b] \rightarrow R
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, there exist a c in (a,b) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
It is given that
f (x) = x^2 - 4x - 3 and interval is [1,4]
Now, f is a polynomial function , f (x) = x^2 - 4x - 3 is continuous in[1,4] and differentiable in (1,4)
And
f(1)= 1^2-4(1)-3= 1-7= -6
and
f(4)= 4^2-4(4)-3= 16-16-3= 16-19=-3
Then, by Mean value theorem we can say that their exist a c in (1,4) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
f^{'}(c) = \frac{f(4)-f(1)}{4-1}\\ f^{'}(c)= \frac{-3-(-6)}{3}\\ f^{'}(c)= \frac{3}{3}\\ f^{'}(c)= 1
Now,
f^{'}(x) =2x-4\\ f^{'}(c)-2c-4\\ 1=2c-4\\ 2c=5\\ c=\frac{5}{2}
And c=\frac{5}{2} \ \epsilon \ (1,4)
Hence, mean value theorem is verified for the function f (x) = x^2 - 4x - 3

Question:5 Verify Mean Value Theorem, iff (x) = x^3 - 5x^2- 3xin the interval [a, b], where
a = 1 and b = 3. Find all c \epsilon (1,3) for which f '(c) = 0.

Answer:

Condition for M.V.T.
If f ; [ a ,b] \rightarrow R
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, their exist a c in (a,b) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
It is given that
f (x) = x^3 - 5x^2- 3x and interval is [1,3]
Now, f being a polynomial function , f (x) = x^3 - 5x^2- 3x is continuous in[1,3] and differentiable in (1,3)
And
f(1)= 1^3-5(1)^2-3(1)= 1-5-3=1-8=-7
and
f(3)= 3^3-5(3)^2-3(3)= 27-5.9-9= 18-45=-27
Then, by Mean value theorem we can say that their exist a c in (1,4) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
f^{'}(c) = \frac{f(3)-f(1)}{3-1}\\ f^{'}(c)= \frac{-27-(-7)}{2}\\ f^{'}(c)= \frac{-20}{2}\\ f^{'}(c)= -10
Now,
f^{'}(x) =3x^2-10x-3\\ f^{'}(c)=3c^2-10c-3\\ -10=3c^2-10c-3\\ 3c^2-10c+7=0\\ 3c^2-3c-7c+7=0\\ (c-1)(3c-7)=0\\ c = 1 \ \ \ and \ \ \ c = \frac{7}{3}
And c=1,\frac{7}{3} \ and \ \frac{7}{3}\ \epsilon \ (1,3)
Hence, mean value theorem is varified for following function f (x) = x^3 - 5x^2- 3x and c=\frac{7}{3} is the only point where f '(c) = 0

Question:6 Examine the applicability of Mean Value Theorem for all three functions given in
the above exercise 2.

Answer:

According to Mean value theorem function
f:[a,b]\rightarrow R must be
a ) continuous in given closed interval say [a,b]
b ) differentiable in given open interval say (a,b)
Then their exist a c \ \epsilon \ (x,y) such that
f^{'}(c)= \frac{f(b)-f(a)}{b-a}
If all these conditions are satisfies then we can verify mean value theorem
Given function is
f (x) = [x]
It is clear that Given function f (x) = [x] is not continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n \ \epsilon \ [5,9]
\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h}= -\infty
( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)
Now,
R.H.L. at x = n , n \ \epsilon \ [5,9]
\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0
( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (5,9)
Hence, Mean value theorem is not applicable for given function f (x) = [x] , x \ \epsilon \ [5,9]

Similaly,
Given function is
f (x) = [x]
It is clear that Given function f (x) = [x] is not continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n \ \epsilon \ [-2,2]
\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h}= -\infty
( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)
Now,
R.H.L. at x = n , n \ \epsilon \ [-2,2]
\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0
( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (-2,2)
Hence, Mean value theorem is not applicable for given function f (x) = [x] , x \ \epsilon \ [-2,2]

Similarly,
Given function is
f (x) = x^2-1
Now, being a polynomial , function f (x) = x^2-1 is continuous in [1,2] and differentiable in(1,2)
Now,
f(1)=1^2-1 = 1-1 = 0
And
f(2)=2^2-1 = 4-1 = 3
Now,
f^{'}(c)= \frac{f(b)-f(a)}{b-a}
f^{'}(c)= \frac{f(2)-f(1)}{2-1}\\ f^{'}(c)=\frac{3-0}{1}\\ f^{'}(c)= 3
Now,
f^{'}(x)= 2x\\ f^{'}(c)=2c\\ 3=2c\\ c=\frac{3}{2}
And c=\frac{3}{2} \ \epsilon \ (1,2)
Therefore, mean value theorem is applicable for the function f (x) = x^2-1


More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.8:-

In Class 12 Maths chapter 5 exercise 5.8 solutions, you will get 6 questions related to verifying the two theorems called the mean value theorem and Rolle's theorem. The three examples given before this exercise are also related to the same also. All the questions in the Class 12th Maths chapter 5 exercise 5.8 are very similar but you must solve all the problems by yourself to get familiar with these types of questions.

Also Read| Continuity and Differentiability Class 12th Chapter 5 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.8:-

  • In Class 12 Maths chapter 5 exercise 5.8 solutions you will get some new ways to approach the problem.
  • NCERT solutions for Class 12 Maths Chapter 5 exercise 5.8 are designed by subject matter experts in a descriptive manner that you can understand very easily.
  • Class 12th Maths chapter 5 exercise 5.8 can be used as revision notes.
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Key Features Of NCERT Solutions for Exercise 5.8 Class 12 Maths Chapter 5

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 5.8 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 5.8, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 5.8 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 5.8 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 5.8 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 5.8 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
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Frequently Asked Questions (FAQs)

1. Do I need to buy NCERT solution book for Class 12 ?

No, you don't need to buy the NCERT solution book for Class 12. NCERT solutions can be easily downloaded from careers360 website. Chapter wise solutions for Class 6 to 10 Mathematics and Science are given. Also solutions to Class 11 and 12 Mathematics, Physics, Chemistry and Biology are given.

2. Can I get NCERT solutions book for Class 12 Maths?

Yes, Click here to get NCERT Solutions for Class 12 Maths.

3. If the function f(x) is not a continuous function on point ‘a’, can it be differentiable at point ‘a’?

No, the f(x) needs to be a continuous function at point ‘a’ to be a differentiable at the given point ‘a’.

4. What is NCERT exemplar book ?

CBSE provides an additional question book for practice which is called NCERT exemplar book.

5. Do CBSE provides NCERT exemplar solutions for free ?

CBSE only provides the NCERT exemplar book which you can solve but it doesn't provide NCERT exemplar solutions.

6. Can you give provide any website that provides NCERT Exemplar Solutions for Class 11 and Class 12 ?

Yes, click on the link to download NCERT Exemplar Solutions for Class 11 and Class 12.

7. Can I get chapter-wise NCERT Exemplar Solutions for Class 12 physics ?

By clicking on the link, you will get Chapter-Wise NCERT Exemplar Solutions for Class 12 physics.

8. What is the use of NCERT Solutions for Class 12 maths ?

NCERT Solutions for Class 12 maths are very useful when you are facing problems while NCERT problems. You can go through these solutions to get conceptual clarity.

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Have a question related to CBSE Class 12th ?
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

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kumardurgesh1802 10 July,2024
I passed my 12th cbse boards in 2023 , and first time appear for jee mains and advance in 2024 so , am i eligible for 2025 jee mains and advance ?

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

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satyamercedes0626 23 July,2024
I've passed my class 12th board exam from cbse in PCB in 2024. Now I'll give mathematics exam from cbse as a additional subject in 2025. However, they'll provide separate marksheet for maths. Will i be eligible for jee?? And how I'll fill the form??

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.


Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.


Hope this resolves your query.

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Abhishek 12 July,2024

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Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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
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