Advertisements
Advertisements
Question
Determine the domain and range of the relations:
(ii) \[S = \left\{ \left( a, b \right) : b = \left| a - 1 \right|, a \in Z \text{ and} \left| a \right| \leq 3 \right\}\]
Advertisements
Solution
(ii) \[S = \left\{ \left( a, b \right) : b = \left| a - 1 \right|, a \in Z \text{ and } \left| a \right| \leq 3 \right\}\]
Now,
a = -3 -2 ,-1,0,1,2,3
\[b = \left| - 3 - 1 \right| = 4\]
\[b = \left| - 2 - 1 \right| = 3\]
\[b = \left| - 1 - 1 \right| = 2\]
\[b = \left| 0 - 1 \right| = 1\]
\[b = \left| 1 - 1 \right| = 0\]
\[b = \left| 2 - 1 \right| = 1\]
\[b = \left| 3 - 1 \right| = 2\]
Thus, we have:
b = 4, 3, 2, 1, 0, 1, 2
Or,
S = {(-3,4)(-2,3) ,(-1,2) ,(0,1),(1,0),(2,1),(3,2)}
Domain (S) = {-3,-2,-1,0,1,2,3}
Range (S) = {0, 1, 2, 3, 4}
APPEARS IN
RELATED QUESTIONS
Find the inverse relation R−1 in each of the cases:
(ii) R = {(x, y), : x, y ∈ N, x + 2y = 8}
Let A = (3, 5) and B = (7, 11). Let R = {(a, b) : a ∈ A, b ∈ B, a − b is odd}. Show that R is an empty relation from A into B.
Determine the domain and range of the relations:
(i) R = {(a, b) : a ∈ N, a < 5, b = 4}
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 implies (b, a) ∈ R
Justify your answer in case.
For the relation R1 defined on R by the rule (a, b) ∈ R1 ⇔ 1 + ab > 0. Prove that: (a, b) ∈ R1 and (b , c) ∈ R1 ⇒ (a, c) ∈ R1 is not true for all a, b, c ∈ R.
If A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}, write (A − C) × (B − C).
Let A = [1, 2, 3], B = [1, 3, 5]. If relation R from A to B is given by = {(1, 3), (2, 5), (3, 3)}, Then R−1 is
Let A = {1, 2, 3, 4), B = {4, 5, 6}, C = {5, 6}. Verify, A × (B ∩ C) = (A × B) ∩ (A × C)
Let A = {1, 2, 3, 4), B = {4, 5, 6}, C = {5, 6}. Verify, A × (B ∪ C) = (A × B) ∪ (A × C)
Let A = {6, 8} and B = {1, 3, 5}
Show that R1 = {(a, b)/a ∈ A, b ∈ B, a − b is an even number} is a null relation. R2 = {(a, b)/a ∈ A, b ∈ B, a + b is odd number} is an universal relation
Write the relation in the Roster Form. State its domain and range
R1 = {(a, a2)/a is prime number less than 15}
Select the correct answer from given alternative.
Let R be a relation on the set N be defined by {(x, y)/x, y ∈ N, 2x + y = 41} Then R is ______.
Select the correct answer from given alternative.
If (x, y) ∈ R × R, then xy = x2 is a relation which is
Answer the following:
R = {1, 2, 3} → {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} Check if R is symmentric
Answer the following:
Show that the following is an equivalence relation
R in A = {x ∈ N/x ≤ 10} given by R = {(a, b)/a = b}
Let A = {1, 2, 3, 7} and B = {3, 0, –1, 7}, the following is relation from A to B?
R3 = {(2, –1), (7, 7), (1, 3)}
Let A = {1, 2, 3, 4, …, 45} and R be the relation defined as “is square of ” on A. Write R as a subset of A × A. Also, find the domain and range of R
A Relation R is given by the set `{(x, y)/y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}`. Determine its domain and range
Represent the given relation by
(a) an arrow diagram
(b) a graph and
(c) a set in roster form, wherever possible
{(x, y) | x = 2y, x ∈ {2, 3, 4, 5}, y ∈ {1, 2, 3, 4}
Multiple Choice Question :
If there are 1024 relation from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is
Let X = {a, b, c, d} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered pairs to be included to R to make it reflexive
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 equivalence
Choose the correct alternative:
The relation R defined on a set A = {0, −1, 1, 2} by xRy if |x2 + y2| ≤ 2, then which one of the following is true?
Choose the correct alternative:
Let f : R → R be defined by f(x) = 1 − |x|. Then the range of f is
Given R = {(x, y) : x, y ∈ W, x2 + y2 = 25}. Find the domain and Range of R.
If R1 = {(x, y) | y = 2x + 7, where x ∈ R and – 5 ≤ x ≤ 5} is a relation. Then find the domain and Range of R1.
Is the given relation a function? Give reasons for your answer.
f = {(x, x) | x is a real number}
Let S = {x ∈ R : x ≥ 0 and `2|sqrt(x) - 3| + sqrt(x)(sqrt(x) - 6) + 6 = 0}`. Then S ______.
A function from set \(A\) to set \(B\) is a relation in which:
Which statement violates the condition for a function?
Which statement is true in general for ordered pairs?
For a relation \(R\), the domain is the set of all:
The Cartesian product \(A\times A\) has \(9\) elements and contains \((-1,0)\) and \((0,1)\). What is \(A\)?
For \(A=\{-1,0,1\}\), which ordered pair is a remaining element of \(A\times A\) besides \((-1,0)\) and \((0,1)\)?
Why is \(R=\{(1,4),(2,3),(3,2),(4,1)\}\) not a function on \(A=\{1,2,3,4,5\}\)?
