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

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 $a\ast b=a-b$

Answer:

(i) On $Z^{+}$ , define ∗ by $a\ast b=a-b$

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 $a\ast b=ab$

Answer:

(ii) On $Z^{+}$ , define ∗ by $a\ast b=ab$

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 $a\ast b=ab$ 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 $a\ast b=a{b}^2$

Answer:

(iii) On $R$ , define ∗ by $a\ast b=a{b}^2$

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

This means * carries each pair $(a,b)$ to a unique element $a\ast b=a{b}^2$ 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 $a\ast b=|a-b|$

Answer:

(iv) On $Z^{+}$ , define ∗ by $a\ast b=|a-b|$

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 $a\ast b=|a-b|$ 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 $a\ast b=a$

Answer:

(v) On $Z^{+}$ , define ∗ by $a\ast b=a$

* carries each pair $(a,b)$ to a unique element $a\ast b=a$ 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 $a\ast b=a-b$

Answer:

a*b=a-b

b*a=b-a

$a\ast b\ne b\ast a$

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 $a\ast b=ab+1$

Answer:

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

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 $a,b\in Q$

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

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

$\therefore$ $(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ $;$ 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 $a\ast b=\frac{ab}{2}$

Answer:

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

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.

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

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

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

$\therefore$ operation * is associative.

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

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

Answer:

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

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.

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

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

$\therefore$ $(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ $;$ 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 $a\ast b=a^b$

Answer:

(v) On $Z^{+}$ , define $a\ast b=a^b$

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

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

$\therefore$ the operation is not commutative.

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

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

$\therefore$ $(2\ast 3)\ast 4\ne 2\ast (3\ast 4)$ $;$ 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 $a\ast b=\frac{a}{b+1}$

Answer:

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

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

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

$\therefore$ the operation is not commutative.

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

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

$\therefore$ $(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ $;$ 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 $\wedge$ .

Answer:

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

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

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

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 $(2\ast 3)\ast 4$ and $2\ast (3\ast 4)$

(Hint: use the following table)

Answer:

(i)

$(2\ast 3)\ast 4=1\ast 4=1$

$2\ast (3\ast 4)=2\ast 1=1$

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)

Answer:

(ii)

For every $a,b\in \{1,2,3,4,5\}$ , we have $a\ast b=b\ast a$ . 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)

Answer:

(iii) (2 ∗ 3) ∗ (4 ∗ 5).

from the above table

$(2\ast 3)\ast (4\ast 5)=1\ast 1=1$

Question:5 Let ∗′ be the binary operation on the set $\{1,2,3,4,5\}$ defined by $a{\ast }^{\prime }b=H.C.F.ofaandb$ . Is the operation ∗′ same as the operation ∗ defined
in Exercise 4 above? Justify your answer.

Answer:

$a{\ast }^{\prime }b=H.C.F.ofaandb$ for $a,b\in \{1,2,3,4,5\}$

The operation table is as shown below:

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

Answer:

a*b=LCM of a and b

(i) 5 ∗ 7, 20 ∗ 16

$5\ast 7=L.C.Mof5and7=35$

$20\ast 16=L.C.Mof20and16=80$

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

(ii) Is ∗ commutative?

Answer:

$a\ast b=L.C.M.ofaandb$

(ii) $L.C.M.ofaandb=L.C.M.ofbanda$ for all $a,b\in N$

$\therefore a\ast b=b\ast a$

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?

Answer:

a $\ast$ b = L.C.M. of a and b

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

$(a\ast b)\ast c=(L.C.Mofaandb)\ast c=L.C.Mofa,bandc$

$a\ast (b\ast c)=a\ast (L.C.Mofbandc)=L.C.Mofa,bandc$

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

Hence, the operation is associative.

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

(iv) the identity of ∗ in N

Answer:

$a\ast b=L.C.M.ofaandb$

(iv) the identity of ∗ in N

We know that $L.C.M.ofaand1=a=L.C.M.of1anda$

$\therefore$ $a\ast 1=a=1\ast a$ 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 ∗?

Answer:

$a\ast b=L.C.M.ofaandb$

An element a is invertible in N

if $a\ast b=e=b\ast a$

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 $a\ast b=L.C.M.ofaandb$ a binary operation? Justify your answer.

Answer:

$a\ast b=L.C.M.ofaandb$

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

Operation table is as shown below:

From the table, we can observe that

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

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

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

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

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

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?

Answer:

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 $a,b\in A$

$\therefore a\ast b=b\ast a$

Hence, operation ∗ is commutative.

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

$(a\ast b)\ast c=(H.C.Fofaandb)\ast c=H.C.Fofa,b,c.$

$a\ast (b\ast c)=a\ast (H.C.Fofbandc)=H.C.Fofa,b,c.$

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

Hence, ∗ is associative.

An element $c\in N$ will be identity for operation * if $a\ast c=a=c\ast a$ 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) $a\ast b=a-b$ Find which of the binary operations are commutative and which are associative.

Answer:

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

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

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

$\therefore$ $\frac{1}{2}\ast \frac{1}{3}\ne \frac{1}{3}\ast \frac{1}{2}$ here $\frac{1}{2},\frac{1}{3}\in Q$

Hence, the * operation is not commutative.

It can be observed that

$(\frac{1}{2}\ast \frac{1}{3})\ast \frac{1}{4}=(\frac{1}{2}-\frac{1}{3})\ast \frac{1}{4}=\frac{1}{6}\ast \frac{1}{4}=\frac{1}{6}-\frac{1}{4}=\frac{-1}{12}$

$\frac{1}{2}\ast (\frac{1}{3}\ast \frac{1}{4})=\frac{1}{2}\ast (\frac{1}{3}-\frac{1}{4})=\frac{1}{2}\ast \frac{1}{12}=(\frac{1}{2}-\frac{1}{12})=\frac{5}{12}$

$(\frac{1}{2}\ast \frac{1}{3})\ast \frac{1}{4}\ne \frac{1}{2}\ast (\frac{1}{3}\ast \frac{1}{4})$ 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) $a\ast b=a^2+b^2$ Find which of the binary operations are commutative and which are associative.

Answer:

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

For $a,b\in Q$

$a\ast b=a^2+b^2=b^2+a^2=b\ast a$

$\therefore$ $a\ast b=b\ast a$

Hence, the * operation is commutative.

It can be observed that

$(1\ast 2)\ast 3=(1^2+2^2)\ast 3=5\ast 3=5^2+3^2=25+9=34$

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

$(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ 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) $a\ast b=a+ab$ Find which of the binary operations are commutative and which are associative.

Answer:

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

For $a,b\in Q$

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

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

$\therefore$ $1\ast 2\ne 2\ast 1$ for $1,2\in Q$

Hence, the * operation is not commutative.

It can be observed that

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

$(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ 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) $a\ast b=(a-b{)}^2$ Find which of the binary operations are commutative and which are associative.

Answer:

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

For $a,b\in Q$

$a\ast b=(a-b{)}^2$

$b\ast a=(b-a{)}^2={[-(a-b)]}^2=(a-b{)}^2$

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

Hence, the * operation is commutative.

It can be observed that

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

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

$(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ 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) $a\ast b=\frac{ab}{4}$ Find which of the binary operations are commutative and which are associative.

Answer:

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

For $a,b\in Q$

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

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

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

Hence, the * operation is commutative.

It can be observed that

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

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

$(a\ast b)\ast c=a\ast (b\ast c)$ 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) $a\ast b=a{b}^2$ Find which of the binary operations are commutative and which are associative.

Answer:

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

For $a,b\in Q$

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

$2\ast 1=2\times 1^2=2\times 1=2$

$\therefore$ $1\ast 2\ne 2\ast 1$ for $1,2\in Q$

Hence, the * operation is not commutative.

It can be observed that

$(1\ast 2)\ast 3=(1\times 2^2)\ast 3=4\ast 3=4\times 3^2=4\times 9=36$

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

$(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ for all $1,2,3\in Q$

The operation * is not associative.

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

Answer:

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

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

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

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

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

$a\ast p=a=p\ast a$ when $p=4$ .

Hence, $(v)a\ast b=\frac{ab}{4}$ 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 $(a,b)\ast (c,d)=(a+c,b+d)$ Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any.

Answer:

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

$(a,b)\ast (c,d)=(a+c,b+d)$

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

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

We have

$(a,b)\ast (c,d)=(a+c,b+d)$

$(c,d)\ast (a,b)=(c+a,d+b)=(a+c,b+d)$

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

Thus it is commutative.

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

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

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

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

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

Thus, it is associative.

Let $e=(e1,e2)\in A$ will be a element for operation * if $(a\ast e)=a=(e\ast a)$ 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, $a\ast a=a,\mathrm{\forall }a\in N$

Answer:

(i) For an arbitrary binary operation ∗ on a set N, $a\ast a=a,\mathrm{\forall }a\in N$

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

Then , for b=a=2

$2\ast 2=2+2=4\ne 2$

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 $a\ast (b\ast c)=(c\ast b)\ast a$

Answer:

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

R.H.S $=(c\ast b)\ast a$

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

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

= L.H.S

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

Hence, statement (ii) is true.

Question:13 Consider a binary operation ∗ on N defined as $a\ast b=a^3+b^3$ . 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?

Answer:

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

For $a,b\in N$

$a\ast b=a^3+b^3=b^3+a^3=b\ast a$

Thus, it is commutative.

$(1\ast 2)\ast 3=(1^3+2^3)\ast 3=9\ast 3=9^3+3^3=729+27=756$

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

$\therefore (1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ where $1,2,3\in N$

Hence, it is not associative.