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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
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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.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

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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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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
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    • Practice Regularly: Consistent practice is key to mastering chemistry.

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  4. Focus on NEET 2025 Preparation:

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  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

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Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024
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.

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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
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12.89×10−3 kg

 
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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)

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

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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
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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.
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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.
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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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