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

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. The NCERT chapter Relations and Functions also consists the following exercise for practice.

Relations and Functions Exercise 1.1

Relations and Functions Exercise 1.2

Relations and Functions Exercise 1.3

Relations and Functions Miscellaneous Exercise

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

Question:1(i) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(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 $1\ast 2=1-2$ $=-1 \notin Z^{+}$ .

Question:1(ii) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(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.

Question:1(iii) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(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.

Question:1(iv) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(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.

Question:1(v) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this

(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.

Question:2(i) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.

(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

Question:2(ii) For each operation ∗ defined below, determine whether ∗ is binary, commutative or 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.

Question:2(iii) For each operation ∗ defined below, determine whether ∗ is binary, commutative or 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.

Question:2(iv) For each operation ∗ defined below, determine whether ∗ is binary, commutative or 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.

Question:2(v) For each operation ∗ defined below, determine whether ∗ is binary, commutative or 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.

Question:2(vi) For each operation ∗ defined below, determine whether ∗ is binary, commutative or 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.

Question:3 Consider the binary operation $\wedge$ on the set $\{1, 2, 3, 4, 5\}$ defined by $a \wedge b = min \{a, b\}$ . Write the operation table of the operation $\wedge$ .

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

$a \wedge b = min \{a, b\}$ 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

Question:4(i) Consider a binary operation ∗ on the set $\{1, 2, 3, 4, 5\}$ given by the following multiplication table (Table 1.2).

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

(Hint: use the following table)

(i)

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

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

Question:4(ii) Consider a binary operation ∗ on the set $\{1, 2, 3, 4, 5\}$ given by the following multiplication table (Table 1.2).

(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.

Question:4(iii) Consider a binary operation ∗ on the set { $\{1, 2, 3, 4, 5\}$ given by the following multiplication table (Table 1.2).

(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).

(Hint: use the following table)

(iii) (2 ∗ 3) ∗ (4 ∗ 5).

from the above table

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

Question:5 Let ∗′ be the binary operation on the set $\{1, 2, 3, 4, 5\}$ defined by $a *' b = H.C.F. \;of\;a\;and\;b$ . Is the operation ∗′ same as the operation ∗ defined

$a *' b = H.C.F. \;of\;a\;and\;b$ 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

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

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

Question:6(ii) Let ∗ be the binary operation on N given by $a * b = L.C.M. \;of \;a\; and \;b$ . Find

(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.

Question:6(iii) Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(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.

Question:6(iv) Let ∗ be the binary operation on N given by $a* b = L.C.M.\; of \;a\; and \;b$ . Find

(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.

Question 6(v) Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(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

Question:7 Is ∗ defined on the set $\{1, 2, 3, 4, 5\}$ by $a * b = L.C.M. \;of \;a\; and \;b$ a binary operation? Justify your answer.

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

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.

Question:8 Let ∗ be the binary operation on N defined by a ∗ b = H.C.F. of a and b. Is ∗ commutative? Is ∗ associative? Does there exist identity for this binary
operation on N?

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.

Question:9(i) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(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.

Question:9(ii) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(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.

Question:9(iii) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(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.

Question:9(iv) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(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.

Question:9(v) Let ∗ be a binary operation on the set Q of rational numbers as follows:

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

Question:9(vi) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(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.

Question:10 Find which of the operations given above has identity.

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.

Question:11 Let $A = N \times N$ and ∗ be the binary operation on A defined by $(a, b) * (c, d) = (a + c, b + d)$ 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.

Question:12(i) State whether the following statements are true or false. Justify.

(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.

Question:12(ii) State whether the following statements are true or false. Justify.

(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.

Question:13 Consider a binary operation ∗ on N defined as $a * b = a^3 + b^3$ . Choose the correct answer.

(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.

Also Read| NCERT Notes For Class 12 Mathematics Chapter 1

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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