Plus Two Math's Solution Ex 1.4 Chapter1 Relations and Function

Plus Two Math's Solution Ex 1.4 Chapter1 Relations and Function

Relations and Functions of Mathematics are very important for doing well in Maths. In Plus Two we get questions on these relations and functions to do well in Maths. NCERT Solutions for Plus Two Maths Chapter 1 Relations and Functions Ex 1.4 is very important for students appearing for competitive exams.

Here we have given NCERT Solutions for Plus Two Maths Chapter 1 Relations and Functions Ex 1.4. We have provided step-by-step solutions along with detailed answers that help to understand the topic easily.

Board SCERT, Kerala
Text Book NCERT Based
Class Plus Two
Subject Math's Textbook Solution
Chapter Chapter 1
Exercise Ex 1.4
Chapter Name Relations and Function
Category Plus Two Kerala


Kerala Syllabus Plus Two Math's Textbook Solution Chapter  1 Relation and Function Exercises 1.4


Chapter  1  Relations and Function Solution



Kerala plus two maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

Chapter  1  Relations and Function Exercise   1.4

    Determine whether or not each of the definitions given below gives a binary operation. In the event that * is not a binary operation, give justification for this.

    (i) On Z+, define * by − b

    (ii) On Z+, define * by ab

    (iii) On R, define * by ab2

    (iv) On Z+, define * by = |− b|

    (v) On Z+, define * by a

    (i) On Z+, * is defined by * 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 ab ∈ Z+, there is a unique element ab in Z+.

    This means that * carries each pair (ab) to a unique element * b ab in Z+.

    Therefore, * is a binary operation.

    (iii) On R, * is defined by a * b = ab2.

    It is seen that for each ab ∈ R, there is a unique element ab2 in R.

    This means that * carries each pair (ab) to a unique element * b abin R.

    Therefore, * is a binary operation.

    (iv) On Z+, * is defined by * b = |a − b|.

    It is seen that for each ab ∈ Z+, there is a unique element |a − b| in Z+.

    This means that * carries each pair (ab) to a unique element * b =
    |a − b|in Z+.

    Therefore, * is a binary operation.

    (v) On Z+, * is defined by a * b = a.

    * carries each pair (ab) to a unique element * b ain Z+.

    Therefore, * is a binary operation.

    For each binary operation * defined below, determine whether * is commutative or associative.

    (i) On Z, define − b

    (ii) On Q, define ab + 1

    (iii) On Q, define 

    (iv) On Z+, define = 2ab

    (v) On Z+, define ab

    (vi) On − {−1}, define 

    (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 * b = ab + 1.

    It is known that:

    ab = ba &mnForE; a, b ∈ Q

    ⇒ ab + 1 = ba + 1 &mnForE; a, b ∈ Q

    ⇒ * b = * 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 * b 

    It is known that:

    ab = ba &mnForE; a, b ∈ Q

    ⇒ &mnForE; a, b ∈ Q

    ⇒ * 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 * b = 2ab.

    It is known that:

    ab = ba &mnForE; a, b ∈ Z+

    ⇒ 2ab = 2ba &mnForE; a, b ∈ Z+

    ⇒ * 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 * b = ab.

    It can be observed that:

     and 

    ∴ 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

    It can be observed that and 

    ∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ − {−1}

    Therefore, the operation * is not commutative.

    It can also be observed that:

    ∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ − {−1}

    Therefore, the operation * is not associative.

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

    The binary operation ∨ on the set {1, 2, 3, 4, 5} is defined as b = min {ab}

    &mnForE; ab ∈ {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


    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)

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

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

    (ii) For every a, b ∈{1, 2, 3, 4, 5}, we have * b = b * a. Therefore, the operation * is commutative.

    (iii) (2 * 3) = 1 and (4 * 5) = 1

    ∴(2 * 3) * (4 * 5) = 1 * 1 = 1

    Let*′ be the binary operation on the set {1, 2, 3, 4, 5} defined by *′ = H.C.F. of and b. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer.

    The binary operation *′ on the set {1, 2, 3 4, 5} is defined as *′ 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

    We observe that the operation tables for the operations * and *′ are the same.

    Thus, the operation *′ is same as the operation*.

    Let * be the binary operation on given by a * = L.C.M. of 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 are invertible for the operation *?

    The binary operation * on N is defined as * b = L.C.M. of a and b.

    (i) 5 * 7 = L.C.M. of 5 and 7 = 35

    20 * 16 = L.C.M of 20 and 16 = 80

    (ii) It is known that:

    L.C.M of a and b = L.C.M of b and a &mnForE; a, b ∈ N.

    a * b = * a

    Thus, the operation * is commutative.

    (iii) For a, b∈ N, we have:

    (* b) * c = (L.C.M of a and b) * c = LCM of ab, and c

    a * (b * c) = a * (LCM of b and c) = L.C.M of ab, and c

    ∴(* b) * c = a * (* c)

    Thus, the operation * is associative.

    (iv) It is known that:

    L.C.M. of a and 1 = a = L.C.M. 1 and a &mnForE; a ∈ N

    ⇒ a * 1 = a = 1 * a &mnForE; a ∈ N

    Thus, 1 is the identity of * in N.

    (v) An element a in N is invertible with respect to the operation * if there exists an element b in N, such that * b = e = b * a.

    Here, e = 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 *.

    Is * defined on the set {1, 2, 3, 4, 5} by = L.C.M. of and a binary operation? Justify your answer.

    The operation * on the set A = {1, 2, 3, 4, 5} is defined as

    a * b = L.C.M. of a and b.

    Then, the operation table for the given operation * can be given as:

    *

    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

    It can be observed from the obtained table that:

    3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A

    3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A

    Hence, the given operation * is not a binary operation.

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

    Let * be the binary operation on N defined by a * b = H.C.F of a and b b * a = H.C.F of b and a
    We know that , H.C.F of a and b = H.C.F of b and a e.g., a * b = b * a therefore, * is commutative.
    (1 * 2) * 3 = {H.C.F of 1 and 2 }* 3 = 1 * 3 = H.C.F of 1 and 3 = 1 1 * (2 * 3) = 1 * {H.C.F of 2 and 3 } = 1 * 1 = H.C.F of 1 and 1 = 1 e.g., (1 * 2) * 3 = 1 * (2 * 3) = 1 , where 1, 2 , 3 ∈ ?.therefore, * is associative .Now, an element  ∈ N  will be the identity for the operation.Now, if a * e = a = e * a, ∀ a ∈ N .
    But, this is not true for any a ∈ N .Therefore, the operation * does not have any identity in N.

    Let * be a binary operation on the set of rational numbers as follows:

    (i) − (ii) a2 + b2

    (iii) ab (iv) = (− b)2

    (v) (vi) ab2

    Find which of the binary operations are commutative and which are associative.

    (i) On Q, the operation * is defined as * b = a − b.

    It can be observed that:

    and 

     ; where

    Thus, the operation * is not commutative.

    It can also be observed that:

    Thus, the operation * is not associative.

    (ii) On Q, the operation * is defined as * b = a2 + b2.

    For a, b ∈ Q, we have:

    a * b = b * a

    Thus, the operation * is commutative.

    It can be observed that:
    (1*2)*3 =(12 + 22)*3 = (1+4)*3 = 5*3 = 52+32= 25 + 9 = 34(1*2)*3 =(12 + 22)*3 = (1+4)*3 = 5*3 = 52+32= 25 + 9 = 34
    1*(2*3)=1*(22+32) = 1*(4+9) = 1*13 = 12+ 132 = 1 + 169 = 1701*(2*3)=1*(22+32) = 1*(4+9) = 1*13 = 12+ 132 = 1 + 169 = 170
    ∴ (1*2)*3 ≠ 1*(2*3) , where 1, 2, 3 ∈ Q

    Thus, the operation * is not associative.

    (iii) On Q, the operation * is defined as * b = a + ab.

    It can be observed that:

    Thus, the operation * is not commutative.

    It can also be observed that:

    Thus, the operation * is not associative.

    (iv) On Q, the operation * is defined by a * b = (a − b)2.

    For ab ∈ Q, we have:

    * b = (a − b)2

    * a = (b − a)2 = [− (a − b)]2 = (a − b)2

    ∴ * b = b * a

    Thus, the operation * is commutative.

    It can be observed that:

    Thus, the operation * is not associative.

    (v) On Q, the operation * is defined as 

    For ab ∈ Q, we have:

    ∴ * b = * a

    Thus, the operation * is commutative.

    For a, b, c ∈ Q, we have:

    ∴(* b) * c = a * (* c)

    Thus, the operation * is associative.

    (vi) On Q, the operation * is defined as * b = ab2

    It can be observed that:

    Thus, the operation * is not commutative.

    It can also be observed that:

    Thus, the operation * is not associative.

    Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

    Find which of the operations given above has identity.

    An element ∈ Q will be the identity element for the operation * if

    * e = a = e * aa ∈ Q.

    We are given

    * b = ab4ab4

     a*e = aae4=a e=4Similarly, it can be checked for e*a=a, we get e=4Thus, e = 4 is the identity.

    State whether the following statements are true or false. Justify.

    (i) For an arbitrary binary operation * on a set N N.

    (ii) If * is a commutative binary operation on N, then * (c) = (b) * a

    (i) Define an operation * on N as:

    * 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. = (* b) * a

    = (* c) * [* is commutative]

    a * (* c) [Again, as * is commutative]

    = L.H.S.

    ∴ a * (* c) = (* b) * a

    Therefore, statement (ii) is true.

    Consider a binary operation * on defined as 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?

    On N, the operation * is defined as * b = a3 + b3.

    For, ab, ∈ N, we have:

    * b = a3 + b3 = b3 + a3 = * a [Addition is commutative in N]

    Therefore, the operation * is commutative.

    It can be observed that:

    ∴(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.


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