NCERT Solutions for Class 12 Maths Relations and Functions
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NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 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
NCERT Solutions:
Given that ∗ defined by a ∗ b = a – b in Z+.
Since the image of (1, 2) under * is 1 * 2 = 1 − 2 = −1 ∉ Z+, the operation * is not a binary operation.
(ii) On Z+, define ∗ by a ∗ b = ab
NCERT Solutions:
Given that ∗ defined by a ∗ b = ab in Z+.
Since for each a, b ∈ Z+, * carries each pair (a, b) to a unique element a * b = ab in Z+, then the operation * is a binary operation.
(iii) On Z+, define ∗ by a ∗ b = ab2
NCERT Solutions:
Given that ∗ defined by a ∗ b = ab2 in R.
Since for each a, b ∈ R, * carries each pair (a, b) to a unique element a * b = ab2 in R, then the operation * is a binary operation.
(iv) On Z+, define ∗ by a ∗ b = |a – b|
NCERT Solutions:
Given that ∗ defined by a ∗ b = | a – b | in Z+.
Since for each a, b ∈ Z+, * carries each pair (a, b) to a unique element a * b = | a – b | in Z+, then the operation * is a binary operation.
(v) On Z+, define ∗ by a ∗ b = a
NCERT Solutions:
Given that ∗ defined by a ∗ b = a in Z+.
Since for each a, b ∈ Z+, * carries each pair (a, b) to a unique element a * b = a in Z+, then the operation * is a binary operation.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 1.
For each binary operation ∗ defined below, determine whether ∗ is commutative or associative.
(i) On Z+, define ∗ by a ∗ b = a – b
NCERT Solutions:
Given that ∗ defined by a ∗ b = a – b in Z.
Let a = 2, b = 3 in Z. Then,
2 * 3 = 2 – 3 = – 1 and 3 * 2 = 3 – 2 = 1
We get that 2 * 3 ≠ 3 * 2
Thus, the given operation * is not commutative.
Let a = 2, b = 3, c = 4 in Z. Then,
(2 * 3) * 4 = (2 – 3) * 4 = – 1 * 4 = – 1 – 4 = –5
2 * (3 * 4) = 2 * (3 – 4) = 2 * –1 = 2 – (–1) = 2 + 1 = 3
We get that (2 * 3) * 4 ≠ 2 * (3 * 4)
Thus, the given operation * is not associative.
(ii) On Z+, define ∗ by a ∗ b = ab + 1
NCERT Solutions:
Given that ∗ defined by a ∗ b = ab + 1 in Q.
Let a = 2, b = 3 in Q. Then,
2 * 3 = 2×3 + 1 = 6 + 1 = 7 and 3 * 2 = 3×2 + 1 = 6 + 1 = 7
We get that 2 * 3 = 3 * 2
Thus, the given operation * is commutative.
Let a = 2, b = 3, c = 4 in Q. Then,
(2 * 3) * 4 = (2×3 + 1) * 4 = 7 * 4 = 7×4 + 1 = 29
2 * (3 * 4) = 2 * (3×4 + 1) = 2 * 13 = 2×13 + 1 = 27
We get that (2 * 3) * 4 ≠ 2 * (3 * 4)
Thus, the given operation * is not associative.
(iii) On Z+, define ∗ by a ∗ b = ab/2
NCERT Solutions:
Given that ∗ defined by a ∗ b = ab/2 in Q.
Let a = 2, b = 3 in Q. Then, 2 * 3 = (2 × 3)/2 = 3 and 3 * 2 = (3 × 2) / 2 = 3
We get that 2 * 3 = 3 * 2. Thus, the given operation * is commutative.
Let a = 2, b = 3, c = 4 in Q. Then,
We get that (2 * 3) * 4 = 2 * (3 * 4). Thus, the given operation * is associative.
(iv) On Z+, define ∗ by a ∗ b = 2ab
NCERT Solutions:
Given that ∗ defined by a ∗ b = 2ab in Z+.
Let a = 2, b = 3 in Z. Then, 2 * 3 = 2(2×3) = 26 = 64 and 3 * 2 = 2(3×2) = 26 = 64
We get that 2 * 3 = 3 * 2. Thus, the given operation * is commutative.
Let a = 2, b = 3, c = 4 in Z+. Then,
(2 * 3) * 4 = 2(2×3) * 4 = 64 * 4 = 2(64×4) = 2256
2 * (3 * 4) = 2 * 2(3×4) = 2 * 4096 = 2(2×4096) = 28192
We get that (2 * 3) * 4 ≠ 2 * (3 * 4)
Thus, the given operation * is not associative.
(v) On Z+, define ∗ by a ∗ b = ab
NCERT Solutions:
Given that ∗ defined by a ∗ b = ab in Z+.
Let a = 2, b = 3 in Z. Then, 2 * 3 = 23 = 8 and 3 * 2 = 32 = 9
We get that 2 * 3 ≠ 3 * 2. Thus, the given operation * is not commutative.
Let a = 2, b = 3, c = 4 in Z+. Then,
(2 * 3) * 4 = 23 * 4 = 8 * 4 = 84 = 4096 = 212
2 * (3 * 4) = 2 * 34 = 2 * 81 = 281
We get that (2 * 3) * 4 ≠ 2 * (3 * 4). Thus, the given operation * is not associative.
(vi) On Z+, define ∗ by a ∗ b = a/(b + 1)
NCERT Solutions:
Given that ∗ defined by a ∗ b = a/(b + 1) in R – {–1}.
Let a = 2, b = 3 in R – {–1}. Then, 2 * 3 = 2/(3 + 1) = ½ and 3 * 2 = 3/(2 + 1) = 1
We get that 2 * 3 ≠ 3 * 2. Thus, the given operation * is not commutative.
Let a = 2, b = 3, c = 4 in R – {–1}. Then,
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 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 ∧.
NCERT Solutions:
Given that the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}.
Thus, the operation table for the given operation is as below.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 4.
Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} is given by following multiplication table.
(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)
NCERT Solutions:
From the given table, we have 2 * 3 = 1, 1 * 4 = 1, 3 * 4 = 1 and 2 * 1 = 1. Thus, (2 * 3) = 1 * 4 = 1 and 2 * (3 * 4) = 2 * 1 = 1.
(ii) Is * commutative?
NCERT Solutions:
For every a, b ∈ {1, 2, 3, 4, 5}, we have a * b = b * a. Therefore, the operation * is commutative.
(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).
NCERT Solutions:
From the given table, we have 2 * 3 = 1, 4 * 5 = 1, and 1 * 1 = 1. Thus, (2 * 3) * (4 * 5) = 1 * 1 = 1.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 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.
NCERT Solutions:
Given that the binary operation ∗′ on the set {1, 2, 3, 4, 5} defined by a ∗′ b = HCF of a and b.
Thus, the operation table for the given operation ∗′ is as below.
In the question no 4, the operation table for the given operation ∗ is as below.
We observe that the operation tables for the operations * and *′ are the same. Thus, the operation *′ is same as the operation*.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 6.
Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b.
(i) Find 5 ∗ 7, 20 ∗ 16
NCERT Solutions:
Given that the binary operation on N given by a ∗ b = L.C.M. of a and b.
Thus, 5 * 7 = LCM of 5 and 7 = 35 and 20 * 16 = LCM of 20 and 16 = 80
(ii) Is * commutative?
NCERT Solutions:
Given that the binary operation on N given by a ∗ b = L.C.M. of a and b
Since LCM of (a, b) and LCM of (b, a) is same for all N, then a ∗ b = b * a
Thus, the given operation * is commutative.
(iii) Is * associative?
NCERT Solutions:
Given that the binary operation on N given by a ∗ b = L.C.M. of a and b
We have
(a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c
a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c
Thus, (a * b) * c = a * (b * c).
Hence, the given operation * is associative.
(iv) Find the identity of * in N.
NCERT Solutions:
Given that the binary operation on N given by a ∗ b = L.C.M. of a and b
Since LCM of (a, 1) and LCM of (1, a) is same for all N and that is 1,
then a ∗ 1 = 1 * a = LCM of a and 1 = a
Thus, 1 is the identity of the given operation *.
(v) Which elements of N are invertible for the operation ∗?
NCERT Solutions:
Given that the binary operation on N given by a ∗ b = L.C.M. of a and b
An element a in N is invertible with respect to the operation * if there exists an element b in N, such that a * b = c = b * a.
Thus, we have c = 1
This means that L.C.M of a and b = 1 = L.C.M of b and a. This case is possible only when a and b are equal to 1.
Thus, 1 is the only invertible element of N with respect to the operation *.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 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.
NCERT Solutions:
Given that ∗ defined on the set {1, 2, 3, 4, 5} by a ∗ b = L.C.M. of a and b
Thus, the operation table for the given operation ∗ is as below.
From the table, we observe that 6, 10, 12, 15, 20 ∉ A. Thus, the given operation * is not a binary operation.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 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?
NCERT Solutions:
Given that ∗ is the binary operation on N defined by a ∗ b = H.C.F. of a and b.
We know that H.C.F. of a and b = H.C.F. of b and a for all a, b ∈ N.
Thus, we get a * b = b * a. Hence, the given operation * is commutative.
For a, b, c ∈ N, we have
(a * b) * c = (H.C.F. of a and b) * c = H.C.F. of a, b and c
a *(b * c) = a *(H.C.F. of b and c) = H.C.F. of a, b, and c
Thus, we get (a * b) * c = a * (b * c). Hence, the given operation * is associative.
Now, an element e ∈ N will be the identity for the operation * if a * e = a = e* a for all a ∈ N. But this relation is not true for any a ∈ N.
Thus, the given operation * does not have any identity in N.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 9.
Let ∗ be a binary operation on the set Q of rational numbers as follows: Find which of the binary operations are commutative and which are associative.
(i) a * b = a – b
NCERT Solutions:
Given that ∗ be a binary operation as a ∗ b = a – b
Since a – b ≠ b – a, then a * b ≠ b * a. Thus, the given operation a ∗ b = a – b is not commutative.
For a, b, c, we have, (a * b) * c = (a – b) * c = (a – b) – c = a – b – c
a * (b * c) = a * (b – c) = a – (b – c) = a – b + c
Since a – b – c ≠ a – b + c, then (a * b) * c ≠ a * (b * c).
Thus, the given operation a ∗ b = a – b is not associative.
(ii) a * b = a2 + b2
NCERT Solutions:
Given that ∗ be a binary operation as a ∗ b = a2 + b2
Since a2 + b2 = b2 + a2, then a * b = b * a
Thus, the given operation a ∗ b = a2 + b2 is commutative.
For a, b, c, we have
(a * b) * c = (a2 + b2) * c = (a2 + b2)2 + c2
a * (b * c) = a * (b2 + c2) = a2 + (b2 + c2)2
Since (a2 + b2)2 + c2 ≠ a2 + (b2 + c2)2, then (a * b) * c ≠ a * (b * c).
Thus, the given operation a ∗ b = a2 + b2 is not associative.
(iii) a * b = a + ab
NCERT Solutions:
Given that ∗ be a binary operation as a ∗ b = a + ab
Since a + ab ≠ b + ba, then a * b ≠ b * a
Thus, the given operation a ∗ b = a + ab is not commutative.
For a, b, c, we have
(a * b) * c = (a + ab) * c = a + ab + (a + ab)c = a + ab + ac + abc
a * (b * c) = a * (b + bc) = a + a(b + bc) = a + ab + abc
Since a + ab + ac + abc ≠ a + ab + abc, then (a * b) * c ≠ a * (b * c).
Thus, the given operation a ∗ b = a + ab is not associative.
(iv) a * b = (a – b)2
NCERT Solutions:
Given that ∗ be a binary operation as a ∗ b = (a – b)2
Since (a – b)2 = (b – a)2, then a * b = b * a
Thus, the given operation a ∗ b = (a – b)2 is commutative.
For a, b, c, we have
(a * b) * c = (a – b)2 * c = ((a – b)2 – c)2
a * (b * c) = a * (b – c)2 = (a – (b – c)2)2
Since ((a – b)2 – c)2 ≠ (a – (b – c)2)2, then (a * b) * c ≠ a * (b * c).
Thus, the given operation a ∗ b = (a – b)2 is not associative.
(v) a * b = ab/4
NCERT Solutions:
Given that ∗ be a binary operation as a ∗ b = ab/4
Since ab/4 = ba/4, then a * b = b * a
Thus, the given operation a ∗ b = ab/4 is commutative.
For a, b, c, we have
Since abc/16 = abc/16, then (a * b) * c = a * (b * c). Thus, the given operation a ∗ b = ab/4 is associative.
(vi) a * b = ab2
NCERT Solutions:
Given that ∗ be a binary operation as a ∗ b = ab2
Since ab2 ≠ ba2, then a * b ≠ b * a
Thus, the given operation a ∗ b = ab2 is not commutative.
For a, b, c, we have
(a * b) * c = (ab2) * c = ab2c2
a * (b * c) = a * (bc2) = a(bc2)2 = ab2c4
Since ab2c2 ≠ ab2c4, then (a * b) * c ≠ a * (b * c).
Thus, the given operation a ∗ b = ab2 is not associative.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 10.
Show that none of the operations given above has identity.
NCERT Solutions:
We know that if a * e = a = e * a for all a ∈ Q, then the element e ∈ Q will be the identity element for the operation *.
However, there is no such element e ∈ Q with respect to each of the six operations given in question no 9.
Thus, none of the six operations in question 9 has identity.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 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.
NCERT Solutions:
Given that A = N × N and ∗ be the binary operation on A defined by
(a, b) ∗ (c, d) = (a + c, b + d).
For a, b, c, d ∈ N, we have
(a, b) ∗ (c, d) = (a + c, b + d) and (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
Since (a + c, b + d) = (a + c, b + d), then (a, b) ∗ (c, d) = (c, d) * (a, b)
Thus, the given operation (a, b) ∗ (c, d) = (a + c, b + d) is commutative.
For a, b, c, d, e, f ∈ N, we have
{(a, b) ∗ (c, d)} * (e, f) = {a + c, b + d} * (e, f) = (a + c + e, b + d + f)
And,
(a, b) ∗ {(c, d) * (e, f)} = (a, b) * {c + e, d + f}= (a + c + e, b + d + f)
Thus, {(a, b) ∗ (c, d)} * (e, f) = (a, b) ∗ {(c, d) * (e, f)}.
The given operation (a, b) ∗ (c, d) = (a + c, b + d) is associative.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 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
NCERT Solutions:
(i)
First we 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.
Therefore, statement (ii) is true.
NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4: Ques No 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?
NCERT Solutions:
Given that ∗ be a binary operation as a ∗ b = a3 + b3
Since a3 + b3 = b3 + a3, then a * b = b * a
Thus, the given operation a ∗ b = a3 + b3 is commutative.
For a, b, c, we have
(a * b) * c = (a3 + b3) * c = (a3 + b3)3 + c3
a * (b * c) = a * (b3 + c3) = a3 + (b3 + c3)3
Since (a3 + b3)3 + c3 ≠ a3 + (b3 + c3)3, then (a * b) * c ≠ a * (b * c).
Thus, the given operation a ∗ b = a3 + b3 is not associative.
Therefore, the correct option is B.
