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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## Continuity and Differentiability Exercise: 5.8

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 $f (x) = x^2 + 2x - 8$

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)
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 $f (x) = [x]$ , $x \ \epsilon \ [5,9]$

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

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

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

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

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 $f (x) = x^2 - 4x - 3$

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

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)
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)
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.

## 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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## Subject Wise NCERT Exampler Solutions

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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?

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.

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 ?

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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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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.

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.

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.

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.

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.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 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.

With increase of temperature, which of these changes?

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

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

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