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

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

Edited By Ramraj Saini | Updated on Dec 04, 2023 11:59 AM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Chapter 1 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 1 class 12 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 miscellaneous exercise discusses concepts related to inverse, injective, onto functions. Also in Class 12 maths chapter 1 miscellaneous exercise some of the topics of binary operations signum function, inverse functions. All in all it can be said that in miscellaneous exercises, diverse questions are asked unlike in earlier exercises of this chapter. Hence class 12 maths chapter 1 miscellaneous exercise is a good source to cover the syllabus in a comprehensive manner. There are a total of 5 exercises in the NCERT solutions for class 12 maths chapter 1 including miscellaneous exercise which is present in NCERT Class 12th book. Also with NCERT questions, students should practice NCERT exemplar and class 12 CBSE previous year questions for better understanding.

Miscellaneous exercise class 12 chapter 1 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 enumerated in NCERT Book together using the link provided below.

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## Access NCERT Solutions for Miscellaneous Exercise Chapter 1 Class 12

NCERT solutions for Class 12 Maths Chapter 1 Relations and Functions: Miscellaneous Exercise

## Question:1 Let $f : R \rightarrow R$ be defined as $f (x) = 10x + 7$ . Find the function $g : R \rightarrow R$ such that $g o f = f o g = I_R$ .

$f : R \rightarrow R$

$f (x) = 10x + 7$

$g : R \rightarrow R$ and $g o f = f o g = I_R$

For one-one:

$f(x)=f(y)$

$10x+7=10y+7$

$10x=10y$

$x=y$

Thus, f is one-one.

For onto:

For $y \in R$ , $y=10x+7$

$x= \frac{y-7}{10} \in R$

Thus, for $y \in R$ , there exists $x= \frac{y-7}{10} \in R$ such that

$f(x) = f(\frac{y-7}{10})=10(\frac{y-7}{10})+7=y-7+7=y$

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Let $g : R \rightarrow R$ as $f(y)=\frac{y-7}{10}$

$gof(x)=g(f(x))= g(10x+7) =\frac{(10x+7)-7}{10}=\frac{10x}{10}=x$

$fog(x)=f(g(x))= f(\frac{y-7}{10}) =10\frac{y-7}{10}+7=y-7+7=y$

$\therefore$ $gof(x)=I_R$ and $fog(x)=I_R$

Hence, $g : R \rightarrow R$ defined as $g(y)=\frac{y-7}{10}$

$f : W \rightarrow W$

$f (n) = n -1$ if n is odd

$f (n) = n +1$ if n is even.

For one-one:

Taking x as odd number and y as even number.

$f(x)=f(y)$

$x-1=y+1$

$x-y=2$

Now, Taking y as odd number and x as even number.

$f(x)=f(y)$

$x+1=y-1$

$x-y = - 2$

This is also impossible.

If both x and y are odd :

$f(x)=f(y)$

$x-1=y-1$

$x=y$

If both x and y are even :

$f(x)=f(y)$

$x+1=y+1$

$x=y$

$\therefore$ f is one-one.

Onto:

Any odd number 2r+1 in codomain of N is an image of 2r in domain N and any even number 2r in codomain N is the image of 2r+1 in domain N.

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Sice, f is invertible.

Let $g : W \rightarrow W$ as $m+1$ if m is even and $m-1$ if m is odd.

When x is odd.

$gof(x)=g(f(x))= g(n-1) =n-1+1=n$

When x is even

$gof(x)=g(f(x))= g(n+1) =n+1-1=n$

Similarly, m is odd

$fog(x)=f(g(x))= f(m-1) =m-1+1=m$

m is even ,

$fog(x)=f(g(x))= f(m+1) =m-1+1=m$

$\therefore$ $gof(x)=I_W$ and $fog(x)=I_W$

Hence, f is invertible and the inverse of f is g i.e. $f^{-1}=g$ , which is the same as f.

Hence, inverse of f is f itself.

This can be solved as following

f : R → R

$f(x) = x^{2}-3x+2$

$f(f(x)) = f(x^{2}-3x+2) = (x^{2}-3x+2)^{2} - 3(x^{2}-3x+2)+2$

$= (x^{4}+9x^{2}+4-6x^{3}-12x+4x^{2})-3x^{2}+9x-6+2$

$= x^{4} - 6x^{3}+10x^{2} - 3x$

The function $f : R \rightarrow \{x \in R : - 1 < x < 1\}$ defined by

$f(x) = \frac{x}{1 + |x|}$ , $x \in R$

One- one:

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

$\frac{x}{1+\left | x \right |}=\frac{y}{1+\left | y \right |}$

It is observed that if x is positive and y is negative.

$\frac{x}{1+x}= \frac{y}{1+y}$

Since x is positive and y is negative.

$x> y\Rightarrow x-y> 0$ but 2xy is negative.

$x-y\neq 2xy$

Thus, the case of x is positive and y is negative is removed.

Same happens in the case of y is positive and x is negative so this case is also removed.

When x and y both are positive:

$f(x)=f(y)$

$\frac{x}{1+x}= \frac{y}{1+y}$

$x(1+y)=y(1+x)$

$x+xy=y+xy$

$x=y$

When x and y both are negative : $f(x)=f(y)$

$\frac{x}{1-x}= \frac{y}{1-y}$

$x(1-y)=y(1-x)$

$x-xy=y-xy$

$x=y$

$\therefore$ f is one-one.

Onto:

Let $y \in R$ such that $-1< y< 1$

If y is negative, then $x= \frac{y}{y+1} \in R$

$f(x)=f(\frac{y}{y+1} )= \frac{\frac{y}{1+y}}{1+ | \frac{y}{1+y}|}$ $= \frac{\frac{y}{1+y}}{1+ \frac{-y}{1+y}}=\frac{y}{1+y-y}=y$

If y is positive, then $x= \frac{y}{1-y} \in R$

$f(x)=f(\frac{y}{1-y} )= \frac{\frac{y}{1-y}}{1+ | \frac{y}{1-y}|}$ $= \frac{\frac{y}{1-y}}{1+ \frac{-y}{1-y}}=\frac{y}{1-y+y}=y$

Thus, f is onto.

Hence, f is one-one and onto.

$f : R \rightarrow R$

$f (x) = x ^3$

One-one:

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

$x^{3}=y^{3}$

We need to prove $x=y$ .So,

• Let $x\neq y$ then there cubes will not be equal i.e. $x^{3}\neq y^{3}$ .

• It will contradict given condition of cubes being equal.

• Hence, $x=y$ and it is one -one which means it is injective

$f : N \rightarrow Z$

$g : Z \rightarrow Z$

$f (x) = x$

$g (x) = | x |$

One - one:

Since $g (x) = | x |$

$f(1)=\left | 1 \right | = 1$

$f(-1)=\left |- 1 \right | = 1$

As we can see $f(1)=f(-1)$ but $1\neq -1$ so $g(x)$ is not one-one.

Thus , g(x) is not injective.

$gof : N \rightarrow Z$

$gof(x) = g(f(x)) = g(x) =$ $\left | x \right |$

Let $gof(x)=gof(y)\, \, \, \, \, x,y \in N$

$\left | x \right |=\left | y \right |$

Since, $x,y \in N$ so x and y are both positive.

$\therefore x=y$

Hence, gof is injective.

$f : N\rightarrow N$ and $g : N\rightarrow N$

$f(x) = x + 1$ and $g(x) = \left\{\begin{matrix} x -1 & if x > 1\\ 1 & if x = 1 \end{matrix}\right.$

Onto :

$f(x) = x + 1$

Consider element in codomain N . It is clear that this element is not an image of any of element in domain N .

$\therefore$ f is not onto.

$gof : N\rightarrow N$

$gof(x)= g(f(x))= g(x+1)= x+1-1 =x \, \, \, \, \, \, \, \, \, since\, x \in N\Rightarrow x+1> 1$

Now, it is clear that $y \in N$ , there exists $x=y \in N$ such that $gof(x)=y$ .

Hence, $gof$ is onto.

Given a non empty set X, consider P(X) which is the set of all subsets of X.

Since, every set is subset of itself , ARA for all $A \in P(x)$

$\therefore$ R is reflexive.

Let $ARB \Rightarrow A\subset B$

This is not same as $B\subset A$

If $A =\left \{ 0,1 \right \}$ and $B =\left \{ 0,1,2 \right \}$

then we cannot say that B is related to A.

$\therefore$ R is not symmetric.

If $ARB \, \, \, and \, \, \, BRC, \, \, then \, \, A\subset B \, \, \, and \, \, B\subset C$

this implies $A\subset C$ $= ARC$

$\therefore$ R is transitive.

Thus, R is not an equivalence relation because it is not symmetric.

Given $* : P(X) \times P(X) \rightarrow P(X)$ is defined as $A * B = A \cap B\;\; \forall A ,B\in P(X)$ .

As we know that $A \cap X=A=X\cap A\forall A \in P(X)$

$\Rightarrow$ $A*X =A=X*A \forall A in P(X)$

Hence, X is the identity element of binary operation *.

Now, an element $A \in P(X)$ is invertible if there exists a $B \in P(X)$ ,

such that $A*B=X=B*A$ (X is identity element)

i.e. $A\cap B=X=B\cap A$

This is possible only if $A=X=B$ .

Hence, X is only invertible element in $P(X)$ with respect to operation *

The number of all onto functions from the set $\{1, 2, 3, ... , n\}$ to itself is permutations on n symbols 1,2,3,4,5...............n.

Hence, permutations on n symbols 1,2,3,4,5...............n = n

Thus, total number of all onto maps from the set $\{1, 2, 3, ... , n\}$ to itself is same as permutations on n symbols 1,2,3,4,5...............n which is n.

(i) $F = \{(a, 3), (b, 2), (c, 1)\}$

$F:S\rightarrow T$

$F : \{a, b, c\}\rightarrow \left \{ 1,2,3 \right \}$ is defined as $F = \{(a, 3), (b, 2), (c, 1)\}$

$F(a)=3,F(b)=2,F(c)=1$

$\therefore \, \, F^{-1}:T\rightarrow S$ is given by

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

(ii) $F = \{(a, 2), (b, 1), (c, 1)\}$

$F:S\rightarrow T$

$F : \{a, b, c\}\rightarrow \left \{ 1,2,3 \right \}$ is defined as $F = \{(a, 2), (b, 1), (c, 1)\}$

$F(a)=2,F(b)=1,F(c)=1$ , F is not one-one.

So inverse of F does not exists.

Hence, F is not invertible i.e. $F^{-1}$ does not exists.

Given $* : R \times R \rightarrow R$ and $\circ : R \times R \rightarrow R$ is defined as
$a *b = |a - b|$ and $a \circ b = a \;\forall a,b \in R$

For $a,b \in R$ , we have

$a *b = |a - b|$

$b *a = |b - a| = \left | -(a-b) \right |=\left | a-b \right |$

$\therefore a*b = b *a$

$\therefore$ the operation is commutative.

$(1*2)*3 = (\left | 1-2 \right |)*3=1*3=\left | 1-3 \right |=2$

$1*(2*3) = 1*(\left | 2-3 \right |)=1*1=\left | 1-1 \right |=0$

$\therefore (1*2)*3\neq 1*(2*3)$ where $1,2,3 \in R$

$\therefore$ the operation is not associative

Let $a,b,c \in R$ . Then we have :

$a*(b \circ c) = a *b =\left | a-b \right |$

$(a*b )\circ(a* c) = \left | a-b \right | \circ \left | a-c \right | = \left | a-b \right |$

Hence, $a*(b \circ c)=(a*b )\circ(a* c)$

Now,

$1\circ (2*3) = 1\circ(\left | 2-3 \right |)=1\circ1=1$

$(1\circ 2)*(1 \circ 3) =1*1=\left | 1-1 \right |=0$

$\therefore 1 \circ(2*3)\neq (1\circ 2)*(1 \circ 3)$ for $1,2,3 \in R$

Hence, operation o does not distribute over operation *.

It is given that

$*: P(X) \times P(X) \rightarrow P(X)$ be defined as

$A * B = (A - B) \cup (B - A), A, B \ \epsilon \ P(X).$

Now, let $A \ \epsilon \ P(X)$ .
Then,

$A * \phi = (A - \phi) \cup (\phi - A) = A \cup \phi = A$
And

$\phi *A = (\phi - A) \cup (A - \phi) = \phi \cup A = A$

Therefore,

$A * \phi = \phi *A = A , A \ \epsilon \ P(X)$

Therefore, we can say that $\phi$ is the identity element for the given operation *.

Now, an element A $\epsilon$ P(X) will be invertible if there exists B $\epsilon$ P(X) such that

$A*B = \phi = B*A \ \ \ \ \ \ \ \ \ \ \ \ (\because \phi \ is \ an \ identity \ element)$

Now, We can see that

$%u200B%u200B%u200B%u200B%u200B%u200B%u200BA * A = (A -A) \cup (A - A) = \phi \cup \phi = \phi , A \ \epsilon \ P(X).$ $A * A = \left ( A - A \right ) \cup \left ( A-A \right ) = \phi \cup \phi = \phi ,$ such that $A \ \epsilon \ P(X)$

Therefore, by this we can say that all the element A of P(X) are invertible with $A^{-1}= A$

X = $\{0, 1, 2, 3, 4, 5\}$ as

$a * b = \left\{\begin{matrix} a + b &if\;a+b < 6 \\ a+ b -6 & if\;a+b\geq6 \end{matrix}\right.$

An element $c \in X$ is identity element for operation *, if $a*c=a=c*a \, \, \forall \, \, a \in X$

For $a \in X$ ,

$a *0 = a+0=a\, \, \, \, \, \, \, \, \, \, \, \, \left [ a \in X \Rightarrow a+0< 6 \right ]$

$0 *a = 0+a=a\, \, \, \, \, \, \, \, \, \, \, \, \left [ a \in X \Rightarrow a+0< 6 \right ]$

$\therefore \, \, \, \, \, \, \, \, \, a*0=a=0 *a$ $\forall a \in X$

Hence, 0 is identity element of operation *.

An element $a \in X$ is invertible if there exists $b \in X$ ,

such that $a*b=0=b*a$ i.e. $\left\{\begin{matrix} a + b =0=b+a &if\;a+b < 6 \\ a+ b -6=0=b+a-6 & if\;a+b\geq6\end{matrix}\right.$

means $a=-b$ or $b=6-a$

But since we have X = $\{0, 1, 2, 3, 4, 5\}$ and $a,b \in X$ . Then $a\neq -b$ .

$\therefore b=a-x$ is inverse of a for $a \in X$ .

Hence, inverse of element $a \in X$ , $a\neq 0$ is 6-a i.e. , $a^{-1} = 6-a$

Given :

$A = \{- 1, 0, 1, 2\}$ , $B = \{- 4, - 2, 0, 2\}$

$f, g : A \rightarrow B$ are defined by $f(x) = x^2 -x, x\in A$ and $g(x) = 2\left |x - \frac{1}{2} \right | - 1, x\in A$ .

It can be observed that

$f(-1) = (-1)^2 -(-1)=1+1=2$

$g(-1) = 2\left |-1 - \frac{1}{2} \right | - 1= 2\left | \frac{-3}{2} \right |-1=3-1=2$

$f(-1)=g(-1)$

$f(0) = (0)^2 -(0)=0+0=0$

$g(0) = 2\left |0 - \frac{1}{2} \right | - 1= 2\left | \frac{-1}{2} \right |-1=1-1=0$

$f(0)=g(0)$

$f(1) = (1)^2 -(1)=1-1=0$

$g(1) = 2\left |1 - \frac{1}{2} \right | - 1= 2\left | \frac{1}{2} \right |-1=1-1=0$

$f(1)=g(1)$

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

$g(2) = 2\left |2 - \frac{1}{2} \right | - 1= 2\left | \frac{3}{2} \right |-1=3-1=2$

$f(2)=g(2)$

$\therefore \, \, \, f(a)=g(a) \forall a\in A$

Hence, f and g are equal functions.

(A) 1
(B) 2
(C) 3
(D) 4

$A = \{1, 2, 3\}$

The smallest relations containing $(1, 2)$ and $(1, 3)$ which are
reflexive and symmetric but not transitive is given by

$R = \left \{ (1,1),(2,2),(3,3),(1,2),(1,3),(2,1),(3,1) \right \}$

$(1,1),(2,2),(3,3) \in R$ , so relation R is reflexive.

$(1,2),(2,1) \in R$ and $(1,3),(3,1) \in R$ , so relation R is symmetric.

$(2,1),(1,3) \in R$ but $(2,3) \notin R$ , so realation R is not transitive.

Now, if we add any two pairs $(2,3)$ and $(3,2)$ to relation R, then relation R will become transitive.

Hence, the total number of the desired relation is one.

Thus, option A is correct.

$A = \{1, 2, 3\}$

The number of equivalence relations containing $(1, 2)$ is given by

$R = \left \{ (1,1),(2,2),(3,3),(1,2),(2,1) \right \}$

We are left with four pairs $(2,3)$ , $(3,2)$ , $(1,3),(3,1)$ .

$(1,1),(2,2),(3,3) \in R$ , so relation R is reflexive.

$(1,2),(2,1) \in R$ and $(2,3),(3,2) \notin R$ , so relation R is not symmetric.

$(1,3),(3,1) \notin R$ , so realation R is not transitive.

Hence , equivalence relation is bigger than R is universal relation.

Thus the total number of equivalence relations cotaining $(1,2)$ is two.

Thus, option B is correct.

$f : R \rightarrow R$ is defined as $f(x) = \left\{\begin{matrix} 1 & x> 0 \\ 0 &x = 0 \\ -1& x < 0 \end{matrix}\right.$

$g : R \rightarrow R$ is defined as $g (x) = [x]$

Let $x \in (0, 1]$

Then we have , $[1]=1$ if x=1 and $[x]=0\, \, \, \, \, if\, 0< x< 1$

$\therefore \, \, gof(x)=g(f(x))=g(1)=\left [ 1 \right ]=1\, \, \, \,(since \, \, x> 0)$

Hence,for $x \in (0, 1]$ , $fog(x)=0$ and $gof(x)=1$ .

Hence , gof and fog do not coincide with $(0, 1]$ .

Binary operations on the set $\{a, b\}$ are is a function from $\{a, b\}\times \{a, b\}\rightarrow \{a, b\}$

i.e. * is a function from $\left \{ (a,b),(b,a),(a,a),(b,b) \right \} \rightarrow \{a, b\}$

Hence, the total number of binary operations on set $\{a, b\}$ is $2^{4}=16.$

Hence, option B is correct

More About NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercise

The NCERT Class 12 Maths chapter Relations and Functions has a total of 5 exercises including miscellaneous exercise. NCERT solutions for Class 12 Maths chapter 1 miscellaneous exercise covers solutions to 19 main questions and their sub-questions. This exercise covers more NCERT syllabus topics than preceding exercises and holds more weightage for exams also. Hence NCERT solutions for Class 12 Maths chapter 1 miscellaneous exercise is recommended for students to strengthen their conceptual understanding.

Benefits of NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercises

• By using NCERT book miscellaneous exercise chapter 1 Class 12 students will be able to get an holistic understanding of the topics mentioned in this chapter.
• Many questions mentioned in Class 12 Maths chapter 1 miscellaneous solutions are useful for the CBSE class 12 board exam and also for exams like JEE main, VITEEE, etc.
• Revision can also be facilitated by using NCERT solutions for Class 12 Maths chapter 10 miscellaneous exercises
• This exercise can also be used in similar physics topics also.
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## Key Features Of NCERT Solutions For Class 12 Chapter 1 Miscellaneous Exercise

• Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 1, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 chapter 1 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 1 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this class 12 maths ch 1 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for class 12 chapter 1 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 1 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!!!

1. Which concepts are covered in class 12 maths chapter 1 miscellaneous exercises

This exercise discusses concepts related to inverse, injective, onto functions.

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 held by the chapter relations and functions for CBSE board exam ?

The weightage of chapter relation and function is more than 5% in the CBSE board examination.

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

As students can assess from previous year questions that many questions are asked directly from NCERT exercise. Hence to score well in the examination, it is required to practice NCERT exercise for class 12 maths before the exam.

5. What are relations in Class 12 Maths?

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

6. Mention the number of questions in class 12 maths chapter 1 miscellaneous exercise?

Total 19 questions are the in miscellaneous exercise Class 12 Maths chapter 1

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

There are a total of 5 exercises including a miscellaneous exercise in the NCERT class 12 maths chapter 1.

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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