हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down the relation by listing all the pairs. Check whether it is symmetric

Advertisements
Advertisements

प्रश्न

On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down the relation by listing all the pairs. Check whether it is symmetric

योग
Advertisements

उत्तर

Given N = set of natural numbers

R is the relation defined by a R b if 2a + 3b = 30

3b = 30 – 2a ⇒ b = `(30 - 2a)/3` a, b ∈ N

a = 1, b = `(30 - 2)/3 = 28/3 ∉ "N"`

a = 2, b = `(30 - 4)/3 = 26/3 ∉ "N"`

a = 3, b = `(30 - 6)/3 = 24/3` = 8 ∈ N

∴ (3, 8) ∈ R

a = 4, b = `(30 - 8)/3 = 22/3 ∉ "N"`

a = 5, b = `(30 - 10)/3 = 20/3 ∉ "N"`

a = 6, b = `(30 - 12)/3 = 18/3` = 6 ∈ N

∴ (6, 6) ∈ R

a = 7, b = `(30 - 14)/3 = 16/3 ∉ "N"`

a = 8, b = `(30 - 16)/3 = 14/3 ∉ "N"`

a = 9, b = `(30 - 18)/3 = 12/3` = 4 ∈ N

∴ (9, 4) ∈ R

a = 10, b = `(30 - 20)/3 = 10/3 ∉ "N"`

a = 11, b = `(30 - 22)/3 = 8/3 ∉ "N"`

a = 12, b = `(30 - 24)/3 = 6/3` = 2 ∈ N

∴ (12, 2) ∈ R

a = 13, b = `(30 - 26)/3 = 4/3 ∉ "N"`

a = 14, b = `(30 - 28)/3 = 2/3 ∉ "N"`

a = 15, b = `(30 - 30)/3 = 0/3` = 0 ∈ N

When a > 15, b negative and does not belong to N.

∴ R = {(3, 8), (6, 6), (9, 4), (12, 2)}.

R is not symmetric since for (3, 8) ∈ R, (8, 3) ∉ R

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Sets, Relations and Functions - Exercise 1.2 [पृष्ठ १८]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 1 Sets, Relations and Functions
Exercise 1.2 | Q 5. (ii) | पृष्ठ १८

संबंधित प्रश्न

Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Is the following true?

f is a relation from A to B

Justify your answer in case.


Find the inverse relation R−1 in each of the cases:

(ii) R = {(xy), : xy ∈ N, x + 2y = 8}


Determine the domain and range of the relations:

(ii) \[S = \left\{ \left( a, b \right) : b = \left| a - 1 \right|, a \in Z \text{ and}  \left| a \right| \leq 3 \right\}\]

 


Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b2}. Is the statement true?

(a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R

Justify your answer in case.


Let R be a relation on N × N defined by
(ab) R (cd) ⇔ a + d = b + c for all (ab), (cd) ∈ N × N

(iii) (ab) R (cd) and (cd) R (ef) ⇒ (ab) R (ef) for all (ab), (cd), (ef) ∈ N × N

 

If A = [1, 3, 5] and B = [2, 4], list of elements of R, if
R = {(xy) : xy ∈ A × B and x > y}


If R = {(x, y) : x, y ∈ Z, x2 + y2 ≤ 4} is a relation on Z, then the domain of R is ______.


A relation R is defined from [2, 3, 4, 5] to [3, 6, 7, 10] by : x R y ⇔ x is relatively prime to y. Then, domain of R is


If `(x + 1/3, y/3 - 1) = (1/2, 3/2)`, find x and y


Write the relation in the Roster Form. State its domain and range

R5 = {(x, y)/x + y = 3, x, y∈ {0, 1, 2, 3}


Write the relation in the Roster Form. State its domain and range

R7 = {(a, b)/a, b ∈ N, a + b = 6}


Multiple Choice Question :

Let n(A) = m and n(B) = n then the total number of non-empty relation that can be defined from A to B is ________.


Find the domain of the function f(x) = `sqrt(1 + sqrt(1 - sqrt(1 - x^2)`


Discuss the following relation for reflexivity, symmetricity and transitivity:

On the set of natural numbers the relation R defined by “xRy if x + 2y = 1”


Let P be the set of all triangles in a plane and R be the relation defined on P as aRb if a is similar to b. Prove that R is an equivalence relation


Is the given relation a function? Give reasons for your answer.

f = {(x, x) | x is a real number}


Is the given relation a function? Give reasons for your answer.

g = `"n", 1/"n" |"n"` is a positive integer


Let n(A) = m, and n(B) = n. Then the total number of non-empty relations that can be defined from A to B is ______.


Let S = {x ∈ R : x ≥ 0 and `2|sqrt(x) - 3| + sqrt(x)(sqrt(x) - 6) + 6 = 0}`. Then S ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×