NCERT Solutions for Exercise 1.3 Class 12 Maths Chapter 1 - Relations and Functions

# NCERT Solutions for Exercise 1.3 Class 12 Maths Chapter 1 - Relations and Functions

Edited By Ramraj Saini | Updated on Dec 03, 2023 01:28 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.3

NCERT Solutions for Exercise 1.3 Class 12 Maths Chapter 1 Relations and Functions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT Solutions for Class 12 Maths chapter 1 exercise 1.3 introduces a few more concepts of relations and functions which includes inverse function, imposition of one function on other eg. fog(x) or gof(x) etc. Exercise 1.3 Class 12 Maths will help students to grasp the basic concepts related to finding the inverse of a function. Practicing this exercise is of utmost importance because most of the questions related to finding the inverse of a function are asked hence students can go through NCERT Solutions for Class 12 Maths chapter 1 exercise 1.3 to score well in CBSE class 12 board exam. In competitive exams also like JEE main ,some questions can be asked from class 12 ex 1.3. Concepts related to inverse functions discussed in Class 12th Maths chapter 1 exercise 1.3 can be useful for NEET physics syllabus, as the problems in physics use the concepts of functions.

12th class Maths exercise 1.3 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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## NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions: Exercise 1.3

Answer:

Given : $f : \{1, 3, 4\}\rightarrow \{1, 2, 5\}$ and $g : \{1, 2, 5\} \rightarrow \{1, 3\}$

$f = \{(1, 2), (3, 5), (4, 1)\}$ and $g = \{(1, 3), (2, 3), (5, 1)\}$

$gof(1) = g(f(1))=g(2) = 3$ $\left [ f(1)=2 \, and\, g(2)=3 \right ]$

$gof(3) = g(f(3))=g(5) = 1$ $\left [ f(3)=5 \, and\, g(5)=1 \right ]$

$gof(4) = g(f(4))=g(1) = 3$ $\left [ f(4)=1 \, and\, g(1)=3 \right ]$

Hence, $gof$ = $\left \{ (1,3),(3,1),(4,3) \right \}$

Answer:

To prove : $\\(f + g) o h = foh + goh$

$((f + g) o h)(x)$

$=(f + g) ( h(x) )$

$=f ( h(x) ) +g(h(x))$

$=(f o h)(x) +(goh)(x)$

$=\left \{ (f o h) +(goh) \right \}(x)$ $x\forall R$

Hence, $\\(f + g) o h = foh + goh$

To prove: $(f \cdot g) o h = (foh) \cdot (goh)$

$((f . g) o h)(x)$

$=(f . g) ( h(x) )$

$=f ( h(x) ) . g(h(x))$

$=(f o h)(x) . (goh)(x)$

$=\left \{ (f o h) .(goh) \right \}(x)$ $x\forall R$

$\therefore$ $(f \cdot g) o h = (foh) \cdot (goh)$

Hence, $(f \cdot g) o h = (foh) \cdot (goh)$

Answer:

$f (x) = | x |$ and $g(x) = \left | 5x-2 \right |$

$gof$ $= g(f(x))$

$= g( | x |)$

$= |5 | x |-2|$

$fog$ $= f(g(x))$

$=f( \left | 5x-2 \right |)$

$=\left \| 5x-2 \right \|$

$=\left | 5x-2 \right |$

Question:3(ii) Find gof and fog, if

Answer:

The solution is as follows

(ii) $f (x) = 8x^{3}$ and $g(x) = x^{\frac{1}{3}}$

$gof$ $= g(f(x))$

$= g( 8x^{3})$

$= ( 8x^{3})^{\frac{1}{3}}$

$=2x$

$fog$ $= f(g(x))$

$=f(x^{\frac{1}{3}} )$

$=8((x^{\frac{1}{3}} )^{3})$

$=8x$

Answer:

$f(x) = \frac{4x + 3}{6x - 4}, x \neq \frac{2}{3}$

$fof (x) = x$

$(fof) (x) = f(f(x))$

$=f( \frac{4x + 3}{6x - 4})$

$=\frac{4( \frac{4x + 3}{6x - 4}) +3}{6( \frac{4x + 3}{6x - 4}) -4}$

$= \frac{16x+12+18x-12}{24x+1824x+16}$

$= \frac{34x}{34}$

$\therefore fof(x) = x$ , for all $x \neq \frac{2}{3}$

$\Rightarrow fof=Ix$

Hence,the given function $f$ is invertible and the inverse of $f$ is $f$ itself.

(i) $f : \{1, 2, 3, 4\} \rightarrow\{10\}$

with $f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}$

Answer:

(i) $f : \{1, 2, 3, 4\} \rightarrow\{10\}$ with
$f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}$

From the given definition,we have:

$f\left ( 1 \right )=f\left ( 2 \right )=f\left ( 3 \right )=f(4)=10$

$\therefore$ f is not one-one.

Hence, f do not have an inverse function.

(ii) $g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\}$ with
$g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$

Answer:

(ii) $g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\}$ with
$g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$

From the definition, we can conclude :

$g(5)=g(7)=4$

$\therefore$ g is not one-one.

Hence, function g does not have inverse function.

Question:5(iii) State with reason whether following functions have inverse

(iii) $h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\}$ with
$h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$

Answer:

(iii) $h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\}$ with
$h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$

From the definition, we can see the set $\left \{ 2,3,4,5 \right \}$ have distant values under h.

$\therefore$ h is one-one.

For every element y of set $\left \{ 7,9,11,13 \right \}$ ,there exists an element x in $\left \{ 2,3,4,5 \right \}$ such that $h(x)=y$

$\therefore$ h is onto

Thus, h is one-one and onto so h has an inverse function.

Answer:

$f : [-1, 1] \rightarrow R$

$f(x) = \frac{x}{(x + 2)}$

One -one:

$f(x)=f(y)$

$\frac{x}{x+2}=\frac{y}{y+2}$

$x(y+2)=y(x+2)$

$xy+2x=xy+2y$

$2x=2y$

$x=y$

$\therefore$ f is one-one.

It is clear that $f : [-1, 1] \rightarrow Range f$ is onto.

Thus,f is one-one and onto so inverse of f exists.

Let g be inverse function of f in $Range f\rightarrow [-1, 1]$

$g: Range f\rightarrow [-1, 1]$

let y be an arbitrary element of range f

Since, $f : [-1, 1] \rightarrow R$ is onto, so

$y=f(x)$ for $x \in \left [ -1,1 \right ]$

$y=\frac{x}{x+2}$

$xy+2y=x$

$2y=x-xy$

$2y=x(1-y)$

$x = \frac{2y}{1-y}$ , $y\neq 1$

$g(y) = \frac{2y}{1-y}$

$f^{-1}=\frac{2y}{1-y},y\neq 1$

Answer:

$f : R \rightarrow R$ is given by $f (x) = 4x + 3$

One-one :

Let $f(x)=f(y)$

$4x + 3 = 4y+3$

$4x=4y$

$x=y$

$\therefore$ f is one-one function.

Onto:

$y=4x+3\, \, \, , y \in R$

$\Rightarrow x=\frac{y-3}{4} \in R$

So, for $y \in R$ there is $x=\frac{y-3}{4} \in R$ ,such that

$f(x)=f(\frac{y-3}{4})=4(\frac{y-3}{4})+3$

$= y-3+3$

$= y$

$\therefore$ f is onto.

Thus, f is one-one and onto so $f^{-1}$ exists.

Let, $g:R\rightarrow R$ by $g(x)=\frac{y-3}{4}$

Now,

$(gof)(x)= g(f(x))= g(4x+3)$

$=\frac{(4x+3)-3}{4}$

$=\frac{4x}{4}$

$=x$

$(fog)(x)= f(g(x))= f(\frac{y-3}{4})$

$= 4\times \frac{y-3}{4}+3$

$= y-3+3$

$= y$

$(gof)(x)= x$ and $(fog)(x)= y$

Hence, function f is invertible and inverse of f is $g(y)=\frac{y-3}{4}$ .

Answer:

It is given that
$f : R^+ \rightarrow [4,\infty)$ , $f(x) = x^2+4$ and

Now, Let f(x) = f(y)

⇒ x 2 + 4 = y 2 + 4

⇒ x 2 = y 2

⇒ x = y

⇒ f is one-one function.

Now, for y $\epsilon$ [4, ∞), let y = x 2 + 4.

⇒ x 2 = y -4 ≥ 0

⇒ for any y $\epsilon$ R, there exists x = $\epsilon$ R such that

= y -4 + 4 = y.

⇒ f is onto function.

Therefore, f is one–one and onto function, so f-1 exists.

Now, let us define g: [4, ∞) → R+ by,

g(y) =

Now, gof(x) = g(f(x)) = g(x 2 + 4) =

And, fog(y) = f(g(y)) = =

Therefore, gof = gof = I R .

Therefore, f is invertible and the inverse of f is given by

f-1(y) = g(y) =

Answer:

$f : R_+ \rightarrow [- 5, \infty)$

$f (x) = 9x^2 + 6x - 5$

One- one:

Let $f(x)=f(y)$ $for \, \, x,y\in R$

$9x^{2}+6x-5=9y^{2}+6y-5$

$9x^{2}+6x=9y^{2}+6y$

$\Rightarrow$ $9(x^{2}-y^{2})=6(y-x)$

$9(x+y)(x-y)+6(x-y)= 0$

$(x-y)(9(x+y)+6)=0$

Since, x and y are positive.

$(9(x+y)+6)> 0$

$\therefore x=y$

$\therefore$ f is one-one.

Onto:

Let for $y \in [-5,\infty)$ , $y=9x^{2}+6x-5$

$\Rightarrow$ $y=(3x+1)^{2}-1-5$

$\Rightarrow$ $y=(3x+1)^{2}-6$

$\Rightarrow$ $y+6=(3x+1)^{2}$

$(3x+1)=\sqrt{y+6}$

$x = \frac{\sqrt{y+6}-1}{3}$

$\therefore$ f is onto and range is $y \in [-5,\infty)$ .

Since f is one-one and onto so it is invertible.

Let $g : [-5,\infty)\rightarrow R_+$ by $g(y) = \frac{\sqrt{y+6}-1}{3}$

$(gof)(x)=g(f(x))=g(9x^{2}+6x)-5=g((3x+1)^{2}-6)\\=\sqrt{{(3x+1)^{2}}-6+6} -1$

$(gof)(x)=\frac{3x+1-1}{3}=\frac{3x}{3}= x$

$(fog)(x)=f(g(x))=f(\frac{\sqrt{y+6}-1}{3})$

$=[3(\frac{\sqrt{y+6}-1}{3})+1]^{2}-6$

$=(\sqrt{y+6})^{2}-6$

$=y+6-6$

$=y$

$\therefore gof=fog=I_R$

Hence, $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )$

Answer:

Let $f : X \rightarrow Y$ be an invertible function

Also, suppose f has two inverse $g_1 and g_2$

For $y \in Y$ , we have

$fog_1(y) = I_y(y)=fog_2(y)$

$\Rightarrow$ $f(g_1(y))=f(g_2(y))$

$\Rightarrow$ $g_1(y)=g_2(y)$ [f is invertible implies f is one - one]

$\Rightarrow$ $g_1=g_2$ [g is one-one]

Thus,f has a unique inverse.

Answer:

It is given that
$f : \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}$

$f(1) = a, f(2) = b \ and \ f(3) = c$

Now,, lets define a function g :
$\left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}$ such that

$g(a) = 1, g(b) = 2 \ and \ g(c) = 3$
Now,

$(fog)(a) = f(g(a)) = f(1) = a$
Similarly,

$(fog)(b) = f(g(b)) = f(2) = b$

$(fog)(c) = f(g(c)) = f(3) = c$

And

$(gof)(1) = g(f(1)) = g(a) = 1$

$(gof)(2) = g(f(2)) = g(b) = 2$

$(gof)(3) = g(f(3)) = g(c) = 3$

Hence, $gof = I_X$ and $fog = I_Y$ , where $X = \left \{ 1,2,3 \right \}$ and $Y = \left \{ a,b,c \right \}$

Therefore, the inverse of f exists and $f^{-1} = g$

Now,
$f^{-1} : \left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}$ is given by

$f^{-1}(a) = 1, f^{-1}(b) = 2 \ and \ f^{-1}(c) = 3$

Now, we need to find the inverse of $f^{-1}$ ,

Therefore, lets define $h: \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}$ such that

$h(1) = a, h(2) = b \ and \ h(3) = c$

Now,

$(goh)(1) = g(h(1)) = g(a) = 1$

$(goh)(2) = g(h(2)) = g(b) = 2$

$(goh)(3) = g(h(3)) = g(c) = 3$

Similarly,

$(hog)(a) = h(g(a)) = h(1) = a$

$(hog)(b) = h(g(b)) = h(2) = b$

$(hog)(c) = h(g(c)) = h(3) = c$

Hence, $goh = I_X$ and $hog = I_Y$ , where $X = \left \{ 1,2,3 \right \}$ and $Y = \left \{ a,b,c \right \}$

Therefore, inverse of $g^{-1} = (f^{-1})^{-1}$ exists and $g^{-1} = (f^{-1})^{-1} = h$

$\Rightarrow h = f$

Therefore, $(f^{-1})^{-1} = f$

Hence proved

Answer:

$f : X \rightarrow Y$

To prove: $(f^{-1})^{-1} = f$

Let $f:X\rightarrow Y$ be a invertible function.

Then there is $g:Y\rightarrow X$ such that $gof =I_x$ and $fog=I_y$

Also, $f^{-1}= g$

$gof =I_x$ and $fog=I_y$

$\Rightarrow$ $f^{-1}of = I_x$ and $fof^{-1} = I_y$

Hence, $f^{-1}:Y\rightarrow X$ is invertible function and f is inverse of $f^{-1}$ .

i.e. $(f^{-1})^{-1} = f$

(A) $x^{\frac{1}{3}}$

(B) $x^3$

(C) $x$

(D) $(3 - x^3)$

Answer:

$f(x) = (3 - x^3)^{\frac{1}{3}}$

$fof(x)$ $=f(f(x))=f((3-x^{3})^{\frac{1}{3}})$

$=[3- ((3-x^{3})^{\frac{1}{3}})^{3}]^{\frac{1}{3}}$

$=([3- ((3-x^{3})]^{\frac{1}{3}})$

$= (x^{3})^{\frac{1}{3}}$

$=x$

Thus, $fof(x)$ is x.

Hence, option c is correct answer.

(A) $g(y) = \frac{3y}{3 -4y}$

(B) $g(y) = \frac{4y}{4 -3y}$

(C) $g(y) = \frac{4y}{3 -4y}$

(D) $g(y) = \frac{3y}{3 -4y}$

Answer:

$f: R - \left\{-\frac{4}{3}\right\} \rightarrow R$

$f(x) = \frac{4x}{3x + 4}$

Let f inverse $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$

Let y be the element of range f.

Then there is $x \in R - \left\{-\frac{4}{3}\right\}$ such that

$y=f(x)$

$y=\frac{4x}{3x+4}$

$y(3x+4)=4x$

$3xy+4y=4x$

$3xy-4x+4y=0$

$x(3y-4)+4y=0$

$x= \frac{-4y}{3y-4}$

$x= \frac{4y}{4-3y}$

Now , define $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$ as $g(y)= \frac{4y}{4-3y}$

$gof(x)= g(f(x))= g(\frac{4x}{3x+4})$

$= \frac{4(\frac{4x}{3x+4})}{4-3(\frac{4x}{3x+4})}$

$=\frac{16x}{12x+16-12x}$

$=\frac{16x}{16}$

$=x$

$fog(y)=f(g(y))=f(\frac{4y}{4-3y})$

$= \frac{4(\frac{4y}{4-3y})}{3(\frac{4y}{4-3y}) + 4}$

$=\frac{16y}{12y+16-12y}=\frac{16y}{16}$

$=y$

Hence, g is inverse of f and $f^{-1}=g$

The inverse of f is given by $g(y)= \frac{4y}{4-3y}$ .

The correct option is B.

## More About NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3

The NCERT class 12 maths chapter Relations and Functions consists of a total of 5 exercises including miscellaneous exercise. Exercise 1.3 Class 12 Maths covers solutions to 14 main questions and their sub-questions. Most questions are related to finding out the inverse of a function. Hence NCERT Solutions for Class 12 Maths chapter 1 exercise 1.3 can be referred by students in case of any doubt.

## Benefits of NCERT Solutions for class 12 maths chapter 1 exercise 1.3

• The Class 12th maths chapter 1 exercise is provided above in detail which is solved by subject matter experts according to the NCERT syllabus .
• Students are recommended to practice Exercise 1.3 Class 12 Maths to prepare for topics like inverse functions etc., direct questions are asked in Board exams.
• These Class 12 Maths NCERT book chapter 1 exercise 1.3 solutions can be referred by students to revise just before the exam.
• NCERT Solutions for Class 12 Maths chapter 1 exercise 1.3 can be used to prepare inverse function topics of physics also.
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## Key Features Of NCERT Solutions for Exercise 1.3 Class 12 Maths Chapter 1

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 1.3 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 1.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 1.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 1.3 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 1.3 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 1.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

### Also see-

NCERT Solutions Subject Wise

Subject wise NCERT Exemplar solutions

Happy learning!!!

### Frequently Asked Question (FAQs)

1. Which concepts are covered in Exercise 1.3 Class 12 Maths?

Concepts related to inverse function, imposition of one function on other eg. fog(x) or gof(x) etc., are discussed in the Exercise 1.3 Class 12 Maths

2. What are the important topics in chapter relations and functions ?

Definitions of relations and functions, types of relations, types of functions, composition of functions, invertible function and binary operations are the important topics in this chapter.

3. What is the weightage of the chapter relations and functions for CBSE board exam ?

Two chapters 'relation and function' and 'inverse trigonometry' combined has 10 % weightage in the CBSE final board exam.

4. How are the NCERT solutions helpful in the board exam ?

As CBSE board exam paper is designed entirely based on NCERT textbooks and most of the questions in CBSE board exam are directly asked from NCERT textbook, students must know the NCERT very well to perform well in the exam. NCERT solutions are not only important when you are stuck while solving the problems but students will get how to answer in the board exam in order to get good marks in the board exam.

5. What are relations in Class 12 Maths?

In maths, relation defines the relationship between sets of values of ordered pairs

6. What are some types of relations discussed in Exercise 1.3 Class 12 Maths

Questions related to inverse function, imposition of one function on other eg. fog(x) or gof(x) etc are discussed in the Exercise 1.3 Class 12 Maths

7. How many questions are covered in Exercise 1.3 Class 12 Maths ?

There are 14 questions in Exercise 1.3 Class 12 Maths NCERT syllabus.

8. How many exercises are there in the NCERT class 12 maths chapter 1 ?

There are 5 exercises including a miscellaneous exercise in the NCERT  book Class 12 maths chapter 1.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

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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.

Good luck with your studies!

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)

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 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

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

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 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
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