NCERT Solutions for Maths Class 12 Exercise 1.4

NCERT Solutions for Maths Class 12 Exercise 1.4

NCERT Solutions for Maths Class 12 Exercise 1.4

NCERT Solutions for Maths Class 12 Exercise 1.4

Maths Class 12 Ex 1.4 Question 1.

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
(ii) On Z+, define * by a * b = ab
(iii) On R, define * by a * b = ab²
(iv) On Z+, define * by a * b = |a – b|
(v) On Z+, define * by a * b = a

Solution:
(i) For all a, b
Z+ and a > b, a * b = a – b > 0, which belongs to Z+.
But if a < b, then a * b = a – b < 0, which does not belong to Z+.
Thus, the given operation * is not a binary operation.
(ii) For all a, b
Z+, a * b = ab > 0, which belongs to Z+.
Thus, the given operation * is a binary operation.
(iii) For all a, b
R, a * b = ab² also belongs to R.
Thus, the given operation * defined by a * b = ab² is a binary operation.
(iv) For all a, b
Z+, a * b = |a – b| also belongs to Z+.
Thus, the given operation * defined by a * b = |a – b| is a binary operation.
(v) For all a, b
Z+, a * b = a Z+.
Thus, the given operation * defined by a * b = a is a binary operation.

Maths Class 12 Ex 1.4 Question 2.

For each binary operation * defined below, determine whether * is binary, commutative or associative.
(i) On Z, define a * b = a – b
(ii) On Q, define a * b = ab + 1
(iii) On Q, define a * b = ab/2
(iv) On Z+, define a * b = 2ab
(v) On Z+, define a * b = ab
(vi) On R – {-1}, define a * b = a/(b + 1)

Solution:
(i) (a) On Z, operation * is defined as
a * b = a – b
b * a = b – a
Since a – b ≠ b – a
a * b ≠ b * a
Hence, the binary operation * is not commutative.

(b) Since a – (b – c) ≠ (a – b) – c a * (b * c) ≠ (a * b) * c
Hence, the binary operation * is not associative.

(ii) (a) On Q, operation * is defined as a * b = ab + 1
Then, ab + 1 = ba + 1
a * b = b * a
Hence, the binary operation * is commutative.

(b) We have (a * b) * c = (ab + 1) * c = (ab + 1)c + 1 = abc + c + 1
Again, a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a+ 1
Therefore, a * (b * c) ≠ (a * b) * c
Hence, the binary operation * is not associative.

(iii) (a) On Q, operation * is defined as a * b = ab/2

Again, b * a = ba/2. Since ab/2 = ba/2 a * b = b * a
Hence, the binary operation * is commutative.

(b) We have a * (b * c) = a * bc/2 = abc/4 and (a * b) * c = ab/2 * c = abc/4

Therefore, (a * b) * c = a * (b * c).
Hence, the binary operation * is associative.

(iv) (a) On Z+, operation * is defined as a * b = 2ab
Then, a * b = 2ab and b * a = 2ba = 2ab
Therefore, a * b = b * a.
Hence, the binary operation * is commutative.

(b) We have a * (b * c) = a * 2bc = 2a.2bc
Again, (a * b) * c = 2ab * c = 22ab.c
Thus, (a * b) * c ≠ a * (b * c).
Hence, the binary operation * is not associative.

(v) (a) On Z+, operation * is defined as a * b = ab
And b * a = ba.
Since, ab ≠ ba  a * b ≠ b * a.
Hence, the binary operation * is not commutative.

(b) We have (a * b) * c = ab * c = (ab)c

And a * (b * c) = a * bc = (a)bc
Thus, (a * b) * c ≠ a * (b * c)
Hence, the binary operation * is not associative.

(vi) (a) On R – {-1}, operation * is defined as a * b = a/(b + 1).

And b * a = b/(a + 1). Thus, a * b ≠ b * a.

Hence, the binary operation * is not commutative.

(b) Similarly, a * (b * c) ≠ (a * b) * c.

Hence, the binary operation * is not associative.

Maths Class 12 Ex 1.4 Question 3.

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

Solution:
The table for the binary operation ^ on the set {1, 2, 3, 4, 5} is as follows.

 ^ 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

Maths Class 12 Ex 1.4 Question 4.

Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.
(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) Is * commutative?
(iii) Compute (2 * 3) * (4 * 5).
(Hint: use the following table)

Table

 * 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

Solution:
(i) From the given table, we can see that 2 * 3 = 1 and 1 * 4 = 1
Therefore, (2 * 3) * 4 = 1 * 4 = 1
Again, we have 3 * 4 = 1. Therefore, 2 * (3 * 4) = 2 * 1 = 1
(ii) Let a, b
{1, 2, 3, 4, 5}

From the given table, we can see that a * a = a
Also, a * b = b * a = 1, when a or b or both are odd.
Again, 2 * 4 = 4 * 2 = 2, when a and b are even and a ≠ b.
Thus, a * b = b * a
Hence, the given binary operation * is commutative.
(iii) From the given table, we can see that 2 * 3 = 1 and 4 * 5 = 1.

Therefore, (2 * 3) * (4 * 5) = 1 * 1 = 1.

Maths Class 12 Ex 1.4 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 in question no. 4 above? Justify your answer.

Solution:
The binary operation *’ on the set {1, 2, 3, 4, 5} is defined by a *’ b = H.C.F. of a and b.
Let us prepare a table for the operation *’.

 *’ (HCF of a, b) 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

Since the table obtained above is the same as the table given in question no. 4, therefore, the operation *’ is the same as the operation * in question no. 4.

Maths Class 12 Ex 1.4 Question 6.

Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find
(i) 5 * 7, 20 * 16
(ii) Is * commutative?
(iii) Is * associative?
(iv) Find the identity of * in N
(v) Which elements of N are invertible for the operation *?

Solution:
The binary operation * on the set N is defined by a * b = L.C.M. of a and b.
(i) We have 5 * 7 = L.C.M. of 5 and 7 = 35
Again, 20 * 16 = L.C.M. of 20 and 16 = 80

(ii) We have a * b = L.C.M. of a and b and b * a = L.C.M. of b and a

Since L.C.M. of a and b = L.C.M. of b and a
Therefore, a * b = b * a.

Hence, the binary operation * is commutative.

(iii) We have a * (b * c) = L.C.M. of a, b, c and (a * b) * c = L.C.M. of a, b, c
Therefore, a * (b * c) = (a * b) * c
Hence, the binary operation * is associative.

(iv) Since 1 * a = a * 1 = a = L.C.M. of 1 and a.
Therefore, the identity of * in N is 1.

(v) Let * : N × N → N defined as a * b = L.C.M. of (a, b)
For a = 1, b = 1, a * b = L.C.M. of (a, b) = 1.
Otherwise, a * b ≠ 1
Therefore, the element 1 of N is invertible for the operation *.

Maths Class 12 Ex 1.4 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.

Solution:
The operation * on the set {1, 2, 3, 4, 5} is defined as a * b = L.C.M. of a and b.

Here, 4 * 5 = L.C.M. of 4 and 5 = 20, which does not belong to the given set {1, 2, 3, 4, 5}.
Therefore, the given operation * is not a binary operation.

Maths Class 12 Ex 1.4 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?

Solution:
The binary operation * on set N is defined as a * b = H.C.F. of a and b
(a) We have a * b = H.C.F. of a and b, and b * a = H.C.F. of b and a

Since H.C.F. of a and b = H.C.F. of b and a
Therefore, a * b = b * a
Hence, the binary operation * is commutative.

(b) We have a * (b * c) = a * (H.C.F. of b, c) = H.C.F. of a and (H.C.F. of b, c) = HCF of a, b and c.
Similarly, (a * b) * c = H.C.F. of a, b and c
Therefore, (a * b) * c = a * (b * c).
Hence, the binary operation * is associative.

(c) We have 1 * a = a * 1 = 1 ≠ a
Therefore, it does not exist any identity for this binary operation on N.

Maths Class 12 Ex 1.4 Question 9.

Let * be a binary operation on the set Q of rational numbers as follows:
(i) a * b = a – b
(ii) a * b = a² + b²
(iii) a * b = a + ab
(iv) a * b = (a – b)²
(v) a * b = ab/4
(vi) a * b = ab²
Find which of the binary operations are commutative and which are associative.

Solution:
(i) The binary operation * on the set Q is defined as
(a) a * b = a – b and b * a = b – a
But a – b ≠ b – a
Therefore, a * b ≠ b * a
Hence, the binary operation * is not commutative.
(b) a * (b * c) = a * (b – c) = a – (b – c) = a – b + c

And (a * b) * c = (a – b) * c = a – b – c

Since a – b + c ≠ a – b – c
Therefore, a * (b * c) ≠ (a * b) * c
Hence, the binary operation * is not associative.

(ii) The binary operation * on the set Q is defined as

(a) a * b = a² + b² and b * a = b² + a² = a² + b².
Therefore, a * b = b * a.
Hence, the binary operation * is commutative.
(b) a * (b * c) = a * (b² + c²) = a² + (b² + c²)²
And (a * b) * c = (a² + b²) * c = (a² + b²)² + c²
Thus, a * (b * c) ≠ (a * b) * c
Hence, the binary operation * is not associative.

(iii) The binary operation * on the set Q is defined as

(a) a * b = a + ab and b * a = b + ba
Therefore, a * b ≠ b * a
Hence, the binary operation * is not commutative.
(b) a * (b * c) = a * (b + bc) = a + a(b + bc) = a + ab + abc

And (a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c = a + ab + ac + abc
Therefore, a * (b * c) ≠ (a * b) * c.
Hence, the binary operation * is not associative.

(iv) The binary operation * on the set Q is defined as

(a) a * b = (a – b)² and b * a = (b – a)² = (a – b)²

Therefore, a * b = b * a
Hence, the binary operation * is commutative.
(b) a * (b * c) = a * (b – c)² = [a – (b – c)²]²

And (a * b) * c = (a – b)² * c = [(a – b)² – c]²
Therefore, a * (b * c) ≠ (a * b) * c.
Hence, the binary operation * is not associative.

(v) The binary operation * on the set Q is defined as

(a) a * b = ab/4 and b * a = ba/4 = ab/4

Therefore, a * b = b * a
Hence, the binary operation * is commutative.
(b) a * (b * c) = a * bc/4 = abc/16

And (a * b) * c = ab/4 * c = abc/16
Therefore, a * (b * c) = (a * b) * c.
Hence, the binary operation * is associative.

(vi) The binary operation * on the set Q is defined as

(a) a * b = ab² and b * a = ba²

Therefore, a * b ≠ b * a
Hence, the binary operation * is not commutative.
(b) a * (b * c) = a * bc² = ab2c4

And (a * b) * c = ab² * c = ab2
Therefore, a * (b * c) ≠ (a * b) * c.
Hence, the binary operation * is not associative.

Maths Class 12 Ex 1.4 Question 10.

Find which of the operations given above has identity.

Solution:
(i) The binary operation * on the set Q is defined as a * b = a – b
For identity element e, a * e = e * a = a
Putting b = e, we get a * e = a – e ≠ a and e * a = e – a ≠ a
Thus the given operation * has no identity.

(ii) The binary operation * on the set Q is defined as a * b = a² + b² ≠ a

For identity element e, a * e = e * a = a
Putting b = e, we get a * e = a² + e² ≠ a
Thus, the given operation * has no identity.

(iii) The binary operation * on the set Q is defined as a * b = a + ab

For identity element e, a * e = e * a = a
Putting b = e, we get a * e = a + ae ≠ a
Thus, the given operation * has no identity.

(iv) The binary operation * on the set Q is defined as a * b = (a – b)²

For identity element e, a * e = e * a = a
Putting b = e, we get a * e = (a – e)² ≠ a for any value of e
Q.
Thus, the given operation * has no identity.

(v) The binary operation * on the set Q is defined as a * b = ab/4

For identity element e, a * e = e * a = a
Putting b = e, we get a * e = ae/4 and e * a = ea/4 ≠ a for any value of e
Q.
Thus, the given operation * has no identity.

(vi) The binary operation * on the set Q is defined as a * b = ab²

For identity element e, a * e = e * a = a

Putting b = e, we get a * e = ae² and e * a = ea² ≠ a for any value of e Q.
Thus, the given operation * has no identity.

Thus, the operations * given in question no. 9 have no identity.

Maths Class 12 Ex 1.4 Question 11.

Let A = N × 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.

Solution:
For A = N × N, the binary operation * is defined as (a, b) * (c, d) = (a + c, b + d)
(i) Now, (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
Therefore, (a, b) * (c, d) = (c, d) * (a, b)
Hence, the binary operation * is commutative.
(ii) Again, (a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f) = ((a + c + e), (b + d + f))
and [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = ((a + c + e, b + d + f))
Therefore, (a, b) * [(c, d) * (e, f)] = [(a, b) * (c, d)] * (e, f)
Hence, the binary operation * is associative.
(iii) For identity element e, a * e = e * a = a

Here, it is not true. Hence, the identity element does not exist.

Maths Class 12 Ex 1.4 Question 12.

State whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation * on a set N, a * a = a
a N.
(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a.

Solution:
(i) For an arbitrary binary operation * on a set N, it is given that
a * a = a
a N
Here, the binary operation * is not defined.
Hence, the given statement is false.
(ii) If * is a commutative binary operation on N, then c * b = b * c
(c * b) * a = (b * c) * a = a * (b * c)
Thus, a * (b * c) = (c * b) * a.
Hence, the given statement is true.

Maths Class 12 Ex 1.4 Question 13.

Consider a binary operation * on N defined as a * b = a³ + b³. 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?

Solution:
(B) Here, option (B) is correct because it is commutative but not associative.

Since a³ + b³ = b³ + a³ implies that a * b = b * a.

And a * (b * c) ≠ (a * b) * c.