NCERT Solutions for Exercise 1.4 Class 12 Maths Chapter 1 - Relations and Functions

# NCERT Solutions for Exercise 1.4 Class 12 Maths Chapter 1 - Relations and Functions

Edited By Ramraj Saini | Updated on Dec 03, 2023 01:35 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.4

NCERT Solutions for Exercise 1.4 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 1.4 talks about commutative, associative, binary operations etc. Exercise 1.4 Class 12 Maths has questions which includes finding whether a relation follows binary operation or not. After analysis from previous year questions it is clear that NCERT Solutions for class 12 maths chapter 1 exercise 1.4 plays an important role to score well in CBSE class 12 board exam. Also it has a significant contribution in competitive exams like JEE main. Hence it is recommended for students to solve all the questions of Class 12th maths chapter 1 exercise 1.4 to score well in their examination.

12th class Maths exercise 1.4 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 1.4

(i) On $Z^+$ , define ∗ by $a * b = a - b$

(i) On $Z^+$ , define ∗ by $a * b = a - b$

It is not a binary operation as the image of $(1,2)$ under * is $1\ast 2=1-2$ $=-1 \notin Z^{+}$ .

(ii) On $Z^+$ , define ∗ by $a * b = ab$

(ii) On $Z^+$ , define ∗ by $a * b = ab$

We can observe that for $a,b \in Z^+$ ,there is a unique element ab in $Z^+$ .

This means * carries each pair $(a,b)$ to a unique element $a * b = ab$ in $Z^+$ .

Therefore,* is a binary operation.

(iii) On $R$ , define ∗ by $a * b = ab^2$

(iii) On $R$ , define ∗ by $a * b = ab^2$

We can observe that for $a,b \in R$ ,there is a unique element $ab^{2}$ in $R$ .

This means * carries each pair $(a,b)$ to a unique element $a * b = ab^{2}$ in $R$ .

Therefore,* is a binary operation.

(iv) On $Z^+$ , define ∗ by $a * b = | a - b |$

(iv) On $Z^+$ , define ∗ by $a * b = | a - b |$

We can observe that for $a,b \in Z^+$ ,there is a unique element $| a - b |$ in $Z^+$ .

This means * carries each pair $(a,b)$ to a unique element $a * b = | a - b |$ in $Z^+$ .

Therefore,* is a binary operation.

(v) On $Z^+$ , define ∗ by $a * b = a$

(v) On $Z^+$ , define ∗ by $a * b = a$

* carries each pair $(a,b)$ to a unique element $a * b = a$ in $Z^+$ .

Therefore,* is a binary operation.

(i)On $Z$ , define $a * b = a-b$

a*b=a-b

b*a=b-a

$a*b\neq b*a$

so * is not commutative

(a*b)*c=(a-b)-c

a*(b*c)=a-(b-c)=a-b+c

(a*b)*c not equal to a*(b*c), so * is not associative

(ii) On $Q$ , define $a * b = ab + 1$

(ii) On $Q$ , define $a * b = ab + 1$

ab = ba for all $a,b \in Q$

ab+1 = ba + 1 for all $a,b \in Q$

$\Rightarrow$ $a\ast b=b\ast a$ for $a,b \in Q$

$(1*2)*3 = (1\times 2+1) * 3 = 3 * 3 = 3\times 3+1 = 10$

$1*(2*3) = 1 * (2\times 3+1) = 1 * 7 = 1\times 7+1 = 8$

$\therefore$ $(1\ast 2)\ast 3\neq 1\ast (2\ast 3)$ $;$ where $1,2,3 \in Q$

$\therefore$ operation * is not associative.

(iii) On $Q$ , define $a * b = \frac{ab}{2}$

(iii) On $Q$ , define $a * b = \frac{ab}{2}$

ab = ba for all $a,b \in Q$

$\frac{ab}{2}=\frac{ba}{2}$ for all $a,b \in Q$

$\Rightarrow$ $a\ast b=b\ast a$ for $a,b \in Q$

$\therefore$ operation * is commutative.

$(a*b)*c = \frac{ab}{2}*c = \frac{(\frac{ab}{2})c}{2} = \frac{abc}{4}$

$a*(b*c) = a*\frac{bc}{2} = \frac{a(\frac{bc}{2})}{2} = \frac{abc}{4}$

$\therefore$ $(a*b)*c=a*(b*c)$ $;$

$\therefore$ operation * is associative.

(iv) On $Z^+$ , define $a * b = 2^{ab}$

(iv) On $Z^+$ , define $a * b = 2^{ab}$

ab = ba for all $a,b \in Z^{+}$

2ab = 2ba for all $a,b \in Z^{+}$

$\Rightarrow$ $a\ast b=b\ast a$ for $a,b \in Z^{+}$

$\therefore$ the operation is commutative.

$(1*2)*3 = 2^{1\times 2} * 3 = 4 * 3 = 2^{4\times 3} = 2^{12}$

$1*(2*3) = 1 * 2^{2\times 3} = 1 * 64 = 2^{1\times 64}=2^{64}$

$\therefore$ $(1\ast 2)\ast 3\neq 1\ast (2\ast 3)$ $;$ where $1,2,3 \in Z^{+}$

$\therefore$ operation * is not associative.

(v) On $Z^+$ , define $a * b = a^b$

(v) On $Z^+$ , define $a * b = a^b$

$1\ast 2 = 1^{2}= 1$ and $2\ast 1 = 2^{1}= 2$

$\Rightarrow$ $1\ast 2\neq 2\ast 1$ for $1,2 \in Z^{+}$

$\therefore$ the operation is not commutative.

$(2\ast 3)\ast 4 = 2^{3} \ast 4 = 8 \ast 4 = 8^{ 4}=2^{12}$

$2\ast (3\ast 4) = 2 \ast 3^{ 4} = 2 \ast 81 = 2^{81}$

$\therefore$ $(2\ast 3)\ast 4\neq 2\ast (3\ast 4)$ $;$ where $2,3,4 \in Z^{+}$

$\therefore$ operation * is not associative.

(vi) On $R - \{-1 \}$ , define $a * b = \frac{a}{b +1}$

(iv) On $R - \{-1 \}$ , define $a * b = \frac{a}{b +1}$

$1\ast 2 = \frac{1}{2+1}=\frac{1}{3}$ and $2\ast 1 = \frac{2}{2+1}= \frac{2}{3}$

$\Rightarrow$ $1\ast 2\neq 2\ast 1$ for $1,2 \in R - \{-1 \}$

$\therefore$ the operation is not commutative.

$(1\ast 2)\ast 3 = (\frac{1}{2+1}) \ast 3 = \frac{1}{3} \ast 3 = \frac{\frac{1}{3}}{3+1}= \frac{1}{12}$

$1\ast (2\ast 3) = 1 \ast (\frac{2}{3+1}) = 1 \ast \frac{2}{4} = 1 \ast \frac{1}{2} = \frac{1}{\frac{1}{2}+1} = \frac{2}{3}$

$\therefore$ $(1\ast 2)\ast 3\neq 1\ast (2\ast 3)$ $;$ where $1,2,3 \in R - \{-1 \}$

$\therefore$ operation * is not associative.

$\{1, 2, 3, 4, 5\}$

$a \wedge b = min \{a, b\}$ for $a,b \in \{1, 2, 3, 4, 5\}$

The operation table of the operation $\wedge$ is given by :

 $\wedge$ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5

(i) Compute $(2 * 3) * 4$ and $2 * (3 * 4)$

(Hint: use the following table)

(i)

$(2 * 3) * 4 = 1*4 =1$

$2 * (3 * 4) = 2*1=1$

(ii) Is ∗ commutative?

(Hint: use the following table)

(ii)

For every $a,b \in\{1, 2, 3, 4, 5\}$ , we have $a*b = b*a$ . Hence it is commutative.

(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).

(Hint: use the following table)

(iii) (2 ∗ 3) ∗ (4 ∗ 5).

from the above table

$(2*3)*(4*5)=1*1=1$

$a *' b = H.C.F. \;of\;a\;and\;b$ for $a,b \in \{1, 2, 3, 4, 5\}$

The operation table is as shown below:

 $\ast$ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5

The operation ∗′ same as the operation ∗ defined in Exercise 4 above.

Question:6(i) let ∗ be the binary operation on N given by . Find

(i) 5 ∗ 7, 20 ∗ 16

a*b=LCM of a and b

(i) 5 ∗ 7, 20 ∗ 16

$5*7 = L.C.M \, of\, 5\, \, and \, 7=35$

$20*16 = L.C.M \, of\, 20\, \, and \, 16 =80$

(ii) Is ∗ commutative?

$a * b = L.C.M. \;of \;a\; and \;b$

(ii) $L.C.M. \;of \;a\; and \;b = L.C.M. \;of \;b\; and \;a$ for all $a,b \in N$

$\therefore \, \, \, \, \, \, \, a*b = b*a$

Hence, it is commutative.

(iii) Is ∗ associative?

a $*$ b = L.C.M. of a and b

(iii) $a,b,c \in N$

$(a*b)*c = (L.C.M \, of\, a\, and \, b)*c= L.C.M\, of \, a,b\, and\, c$

$a*(b*c) = a*(L.C.M \, of\, b\, and \, c)= L.C.M\, of \, a,b\, and\, c$

$\therefore \, \, \, \, \, \, \, \, \, (a*b)*c=a*(b*c)$

Hence, the operation is associative.

(iv) the identity of ∗ in N

$a* b = L.C.M.\; of \;a\; and \;b$

(iv) the identity of ∗ in N

We know that $L.C.M.\; of \;a\; and \;1 = a = L.C.M.\; of 1\, \, and \;a\;$

$\therefore$ $a*1=a=1*a$ for $a \in N$

Hence, 1 is the identity of ∗ in N.

(v) Which elements of N are invertible for the operation ∗?

$a* b = L.C.M.\; of \;a\; and \;b$

An element a is invertible in N

if $a*b=e=b*a$

Here a is inverse of b.

a*b=1=b*a

a*b=L.C.M. od a and b

a=b=1

So 1 is the only invertible element of N

$a * b = L.C.M. \;of \;a\; and \;b$

A = $\{1, 2, 3, 4, 5\}$

Operation table is as shown below:

 $*$ 1 2 3 4 5 1 1 2 3 4 5 2 2 2 6 4 10 3 3 6 3 12 15 4 4 4 12 4 20 5 5 10 15 20 5

From the table, we can observe that

$2*3=3*2=6 \notin A$

$2*5=5*2=10 \notin A$

$3*4=4*3=12 \notin A$

$3*5=5*3=15 \notin A$

$4*5=5*4=20 \notin A$

Hence, the operation is not a binary operation.

a ∗ b = H.C.F. of a and b for all $a,b \in A$

H.C.F. of a and b = H.C.F of b and a for all $a,b \in A$

$\therefore \, \, \, \, a*b=b*a$

Hence, operation ∗ is commutative.

For $a,b,c \in N$ ,

$(a*b)*c = (H.C.F \, of\, a\, and\, b)*c= H.C.F\, of \, a,b,c.$

$a*(b*c )= a*(H.C.F \, of\, b\, and\, c)= H.C.F\, of \, a,b,c.$

$\therefore$ $(a*b)*c=a*(b*c)$

Hence, ∗ is associative.

An element $c \in N$ will be identity for operation * if $a*c=a= c*a$ for $a \in N$ .

Hence, the operation * does not have any identity in N.

(i) $a * b = a - b$ Find which of the binary operations are commutative and which are associative.

On the set Q ,the operation * is defined as $a * b = a - b$ .It is observed that:

$\frac{1}{2}*\frac{1}{3}= \frac{1}{2}-\frac{1}{3}=\frac{1}{6}$

$\frac{1}{3}*\frac{1}{2}= \frac{1}{3}-\frac{1}{2}=\frac{-1}{6}$

$\therefore$ $\frac{1}{2}*\frac{1}{3}\neq \frac{1}{3}*\frac{1}{2}$ here $\frac{1}{2},\frac{1}{3} \in Q$

Hence, the * operation is not commutative.

It can be observed that

$(\frac{1}{2}*\frac{1}{3})*\frac{1}{4} = \left ( \frac{1}{2}-\frac{1}{3}\right )*\frac{1}{4}=\frac{1}{6}*\frac{1}{4}=\frac{1}{6}-\frac{1}{4}=\frac{-1}{12}$

$\frac{1}{2}*(\frac{1}{3}*\frac{1}{4})= \frac{1}{2}*\left ( \frac{1}{3} - \frac{1}{4}\right ) = \frac{1}{2}*\frac{1}{12} = \left ( \frac{1}{2} - \frac{1}{12} \right ) = \frac{5}{12}$

$\left ( \frac{1}{2}*\frac{1}{3} \right )*\frac{1}{4}\neq \frac{1}{2}*(\frac{1}{3}*\frac{1}{4})$ for all $\frac{1}{2},\frac{1}{3}, \frac{1}{4} \in Q$

The operation * is not associative.

(ii) $a*b = a^2 + b^2$ Find which of the binary operations are commutative and which are associative.

On the set Q ,the operation * is defines as $a*b = a^2 + b^2$ .It is observed that:

For $a,b \in Q$

$a*b=a^{2}+b^{2}= b^{2}+a^{2}=b*a$

$\therefore$ $a*b=b*a$

Hence, the * operation is commutative.

It can be observed that

$(1*2)*3 =(1^{2}+2^{2})*3 = 5*3 = 5^{2}+3^{2} = 25+9 =34$

$1*(2*3) =1*(2^{2}+3^{2}) = 1*13 = 1^{2}+13^{2} = 1+169 =170$

$(1*2)*3 \neq 1*(2*3)$ for all $1,2,3 \in Q$

The operation * is not associative.

(iii) $a * b = a + ab$ Find which of the binary operations are commutative and which are associative.

On the set Q ,the operation * is defines as $a * b = a + ab$ .It is observed that:

For $a,b \in Q$

$1 * 2 = 1+1\times 2 =1 + 2 = 3$

$2 * 1= 2+2\times 1 =2 + 2 = 4$

$\therefore$ $1*2\neq 2*1$ for $1,2 \in Q$

Hence, the * operation is not commutative.

It can be observed that

$1*(2*3) =1*(2+3\times 2) = 1*8 = 1+1\times 8 = 1+8 =9$

$(1*2)*3 \neq 1*(2*3)$ for all $1,2,3 \in Q$

The operation * is not associative.

(iv) $a * b = (a-b)^2$ Find which of the binary operations are commutative and which are associative.

On the set Q ,the operation * is defined as $a * b = (a-b)^2$ .It is observed that:

For $a,b \in Q$

$a * b = (a-b)^2$

$b* a = (b-a)^2 = \left [ -\left ( a-b \right ) \right ]^{2} = (a-b)^{2}$

$\therefore$ $a*b = b* a$ for $a,b \in Q$

Hence, the * operation is commutative.

It can be observed that

$(1*2)*3 =(1-2)^{2}*3 = 1*3 =(1-3)^{2}= (-2)^{2} =4$

$1*(2*3) =1*(2-3)^{2} = 1*1 =(1-1)^{2} = 0^{2} =0$

$(1*2)*3 \neq 1*(2*3)$ for all $1,2,3 \in Q$

The operation * is not associative.

(v) Find which of the binary operations are commutative and which are associative.

On the set Q ,the operation * is defines as $a * b = \frac{ab}{4}$ .It is observed that:

For $a,b \in Q$

$a * b = \frac{ab}{4}$

$b* a = \frac{ba}{4}$

$\therefore$ $a*b = b* a$ for $a,b \in Q$

Hence, the * operation is commutative.

It can be observed that

$(a*b)*c =(\frac{ab}{4})*c = \frac{\frac{ab}{4}c}{4}=\frac{abc}{16}$

$a*(b*c) =a*(\frac{bc}{4}) = \frac{\frac{bc}{4}a}{4}=\frac{abc}{16}$

$(a*b)*c = a*(b*c)$ for all $a,b,c \in Q$

The operation * is associative.

(vi) $a* b = ab^2$ Find which of the binary operations are commutative and which are associative.

On the set Q ,the operation * is defines as $a* b = ab^2$ .It is observed that:

For $a,b \in Q$

$1* 2 = 1\times 2^2=1\times 4=4$

$2* 1 = 2\times 1^2=2\times 1=2$

$\therefore$ $1*2\neq 2*1$ for $1,2 \in Q$

Hence, the * operation is not commutative.

It can be observed that

$(1*2)*3 = (1\times 2^{2})*3 = 4*3 = 4\times 3^{2}=4\times 9=36$

$1*(2*3) = 1*(2\times 3^{2}) = 1*18 = 1\times 18^{2}=1\times 324=324$

$(1*2)*3\neq 1*(2*3)$ for all $1,2,3 \in Q$

The operation * is not associative.

An element $p \in Q$ will be identity element for operation *

if $a*p = a = p*a$ for all $a \in Q$

$(v) a * b = \frac{ab}{4}$

$a * p = \frac{ap}{4}$

$p * a = \frac{pa}{4}$

$a*p = a = p*a$ when $p=4$ .

Hence, $(v) a * b = \frac{ab}{4}$ has identity as 4.

However, there is no such element $p \in Q$ which satisfies above condition for all rest five operations.

Hence, only (v) operations have identity.

Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any.

$A = N \times N$ and ∗ be the binary operation on A defined by

$(a, b) * (c, d) = (a + c, b + d)$

Let $(a,b),(c,d) \in A$

Then, $a,b,c,d \in N$

We have

$(a, b) * (c, d) = (a + c, b + d)$

$(c,d)*(a,b) = (c+a,d+b)= (a+c,b+d)$

$\therefore \, \, \, \, \, (a,b)*(c,d)=(c,d)*(a,b)$

Thus it is commutative.

Let $(a,b),(c,d),(e,f) \in A$

Then, $a,b,c,d,e,f \in N$

$[(a, b) * (c, d)]*(e,f)= [(a + c, b + d)]*(e,f) =[(a+c+e),(b+d+f)]$

$(a, b) * [(c, d)*(e,f)]= (a,b)*[(c + e, d + f)] =[(a+c+e),(b+d+f)]$

$\therefore \, \, \, \, \, [(a,b)*(c,d)]*(e,f)=(a, b) * [(c, d)*(e,f)]$

Thus, it is associative.

Let $e= (e1,e2) \in A$ will be a element for operation * if $(a*e)=a=(e*a)$ for all $a= (a1,a2) \in A$ .

i.e. $(a1+e1,a2+e2)= (a1,a2)= (e1+a1,e2+a2)$

This is not possible for any element in A .

Hence, it does not have any identity.

(i) For an arbitrary binary operation ∗ on a set N, $a*a = a,\; \forall a\in N$

(i) For an arbitrary binary operation ∗ on a set N, $a*a = a,\; \forall a\in N$

An operation * on a set N as $a*b=a+b\, \, \, \, \forall\, \, a,b \in N$

Then , for b=a=2

$2*2= 2+2 = 4\neq 2$

Hence, statement (i) is false.

(ii) If ∗ is a commutative binary operation on N, then $a *(b*c) =( c*b)*a$

(ii) If ∗ is a commutative binary operation on N, then $a *(b*c) =( c*b)*a$

R.H.S $=(c*b)*a$

$=(b*c)*a$ (* is commutative)

$= a*(b*c)$ ( as * is commutative)

= L.H.S

$\therefore$ $a *(b*c) =( c*b)*a$

Hence, statement (ii) is true.

(A) Is ∗ both associative and commutative?
(B) Is ∗ commutative but not associative?
(C) Is ∗ associative but not commutative?
(D) Is ∗ neither commutative nor associative?

A binary operation ∗ on N defined as $a * b = a^3 + b^3$ .

For $a,b \in N$

$a * b = a^3 + b^3 = b^{3}+a^{3}=b*a$

Thus, it is commutative.

$(1*2)*3 = (1^{3}+2^{3})*3=9*3 =9^{3}+3^{3}=729+27=756$

$1*(2*3) = 1*(2^{3}+3^{3})=1*35 =1^{3}+35^{3}=1+42875=42876$

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

Hence, it is not associative.

Hence, B is the correct option.

More About NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.4

The NCERT Class 12 Maths chapter Relations and Functions has a total of 5 exercises including miscellaneous exercise. Exercise 1.4 Class 12 Maths covers solutions to 13 main questions and their sub-questions. Most of the questions are related to concepts of binary operations which includes commutative, associative operations etc. Hence NCERT Solutions for Class 12 Maths chapter 1 exercise 1.4 is recommended for learning these concepts to score well in the exam.

Benefits of NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.4

• The Class 12 Maths NCERT syllabus chapter 1.4 exercise is provided above in a comprehensive manner which is solved by subject matter experts .
• Students are recommended to practice Exercise 1.4 Class 12 Maths to prepare for topics like binary operations which includes commutative, associative operations etc.
• These Class 12 Maths chapter 1 exercise 1.4 solutions can be referred by students to revise just before the exam.
• NCERT syllabus for Class 12 Maths chapter 1 exercise 1.4 can be used to prepare similar topics of physics also.
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## Key Features Of NCERT Solutions for Exercise 1.4 Class 12 Maths Chapter 1

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 1.4 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 1.4, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 1.4 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 1.4 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 1.4 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 1.4 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

### Also see-

NCERT Solutions Subject Wise

Subject wise NCERT Exemplar solutions

Happy learning!!!

1. Which concepts are covered in Exercise 1.4 Class 12 Maths?

These concepts are discussed in class 12 ex 1.4. Concepts like proving a binary operation associative, commutative etc.  are mentioned in the Exercise 1.4 Class 12 Maths. Practice 12th class maths exercise 1.4 answers to command the cocepts.

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. How much weightage is 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?

These concepts are discussed in class 12 maths ex 1.4. In maths, relation defines the relationship between sets of values of ordered pairs. Practice these questions to get deeper understanding.

6. Mention the number of questions in Exercise 1.4 Class 12 Maths ?

13 questions are the in Exercise 1.4 Class 12 Maths

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

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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.

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