

NCERT Solutions for Class 12 Physics
NCERT Solutions Class 12 Maths Chapter 1 PDF Download Link
Solve The Following Questions NCERT Solutions for Class 12 Maths Chapter 1 Excercise 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 = ab 2 (iv) On Z + , define * by a * b = |a − b| (v) On Z + , define * by a * b = a Solution : (i) On Z + , * is defined by a * b = a − b. It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2 = −1 ∉ Z + . (ii) On Z + , * is defined by a * b = ab. It is seen that for each a, b ∈ Z + , there is a unique element ab in Z + . This means that * carries each pair (a, b) to a unique element a * b = ab in Z + . Therefore, * is a binary operation. (iii) On R, * is defined by a * b = ab 2 . It is seen that for each a, b ∈ R, there is a unique element ab 2 in R. This means that * carries each pair (a, b) to a unique element a * b = ab 2 in R. Therefore, * is a binary operation. (iv) On Z + , * is defined by a * b = |a − b|. It is seen that for each a, b ∈ Z + , there is a unique element |a − b| in Z + . This means that * carries each pair (a, b) to a unique element a * b = |a − b| in Z + . Therefore, * is a binary operation. (v) On Z + , * is defined by a * b = a. * carries each pair (a, b) to a unique element a * b = a in Z + . Therefore, * is a binary operation.NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.1
Question 2. For each binary operation * defined below, determine whether * is 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 = 2 ab
(v) On Z + , define a * b = a b
(vi) On R − {−1}, define a.b = a/b+1
Solution : (i) On Z , * is defined by a * b = a − b . It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1. ∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z Hence, the operation * is not commutative. Also we have: (1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4 1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2 ∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z Hence, the operation * is not associative. (ii) On Q , * is defined by a * b = ab + 1. It is known that: ab = ba &mnForE; a, b ∈ Q ⇒ ab + 1 = ba + 1 &mnForE; a, b ∈ Q ⇒ a * b = a * b &mnForE; a, b ∈ Q Therefore, the operation * is commutative. It can be observed that: (1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10 1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8 ∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Q Therefore, the operation * is not associative. (iii) On Q , * is defined by a * b = ab/2 It is known that: ab = ba &mnForE; a, b ∈ Q ⇒ ab/2 = ba/2 &mn ForE; a, b ∈ Q ⇒ a * b = b * a &mnForE; a, b ∈ Q Therefore, the operation * is commutative. For all a, b , c ∈ Q , we have:
Therefore, the operation * is associative.
(iv) On
Z
+
, * is defined by
a
*
b
= 2
ab
.
It is known that:
ab
=
ba
&mnForE;
a, b
∈
Z
+
⇒ 2
ab
= 2
ba
&mnForE;
a, b
∈
Z
+
⇒
a
*
b
=
b
*
a
&mnForE;
a, b
∈
Z
+
Therefore, the operation * is commutative.
It can be observed that:
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈
Z
+
Therefore, the operation * is not associative.
(v) On
Z
+
, * is defined by
a
*
b
=
a
b
.
It can be observed that:
1 x2 = 1
2
=1 and 2x 1 =2
1
= 2
∴ 1 * 2 ≠ 2 * 1 ; where 1, 2 ∈
Z
+
Therefore, the operation * is not commutative.
It can also be observed that:
∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈
Z
+
Therefore, the operation * is not associative.
(vi) On
R
, * − {−1} is defined by a.b = a/b+1
It can be observed that 1 x 2 = 1/2+1 = 1/2 and 2 x 1 = 2/1+1 = 2/2
∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈
R
− {−1}
Therefore, the operation * is not commutative.
It can also be observed that:
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈
R
− {−1}
Therefore, the operation * is not associative.
NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercise
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 binary operation ∨ on the set {1, 2, 3, 4, 5} is defined as a ∨ b = min {a, b} & mn For E; a, b ∈ {1, 2, 3, 4, 5}. Thus, the operation table for the given operation ∨ can be given as:| ∨ | 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 |
| * | 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 |
NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3
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 Exercise 4 above? Justify your answer. Solution : The binary operation *′ on the set {1, 2, 3 4, 5} is defined as a *′ b = H.C.F of a and b. The operation table for the operation *′ can be given as:| *′ | 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 |
| * | 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 |
(v) a * b = ab/4 (vi) a * b = ab 2
Find which of the binary operations are commutative and which are associative. Ans.
10. Find which of the operations given above has identity.
Solution : Let the identity be I.
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 :
A = N × N
* is a binary operation on A and is defined by:
(a, b) * (c, d) = (a + c, b + d)
Let (a, b), (c, d) ∈ A
Then, a, b, c, d ∈ N
We have:
(a, b) * (c, d) = (a + c, b + d)
(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
[Addition is commutative in the set of natural numbers]
∴(a, b) * (c, d) = (c, d) * (a, b)
Therefore, the operation * is commutative.
Now, let (a, b), (c, d), (e, f) ∈A
Then, a, b, c, d, e, f ∈ N
We have:
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) Define an operation * on N as:
a * b = a + b a, b ∈ N
Then, in particular, for b = a = 3, we have:
3 * 3 = 3 + 3 = 6 ≠ 3
Therefore, statement (i) is false.
(ii) R.H.S. = (c * b) * a
= (b * c) * a [* is commutative]
= a * (b * c) [Again, as * is commutative]
= L.H.S.
∴ a * (b * c) = (c * b) * a
Therefore, 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?
Solution :
On N, the operation * is defined as a * b = a
3
+ b
3
.
For, a, b, ∈ N, we have:
a * b = a
3
+ b
3
= b
3
+ a
3
= b * a [Addition is commutative in N]
Therefore, the operation * is commutative.
It can be observed that:
(1*2)*3 = (1
3
+2
3
)*3 = 9 * 3 = 93 + 33 = 729 + 27 = 756
Also, 1*(2*3) = 1*(2
3
+3
3
) = 1*(8 +27) = 1 × 35
= 1
3
+35
3
= 1 +(35)
3
= 1 + 42875 = 42876.
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ N
Therefore, the operation * is not associative.
Hence, the operation * is commutative, but not associative. Thus, the correct answer is B.
