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A relation R is defined on Z as: aRb if and only if a² – 7ab + 6b² = 0 Then, R is
We are given the relation R on ℤ, defined by: aRb if and only if a² - 7ab + 6b² = 0 We need to find out the properties of this relation: whether it is reflexive, symmetric, and/or transitive. Step 1: Check if the relation is reflexive. A relation R is reflexive if for every element a ∈ ℤ, we have aRRead more
We are given the relation R on ℤ, defined by:
aRb if and only if a² – 7ab + 6b² = 0
We need to find out the properties of this relation: whether it is reflexive, symmetric, and/or transitive.
Step 1: Check if the relation is reflexive.
A relation R is reflexive if for every element a ∈ ℤ, we have aRa. That is, we must check whether the equation a² – 7a ⋅ a + 6a² = 0 holds for every integer a.
Substitute a = b into the relation:
a² – 7a ⋅ a + 6a² = a² – 7a² + 6a² = 0
This reduces to:
0 = 0
Thus, the relation is reflexive because the equation holds for all a ∈ ℤ.
Step 2: Check if the relation is symmetric.
A relation R is symmetric if for every pair (a, b) ∈ ℤ, whenever aRb, we also have bRa. That is, if a² – 7ab + 6b² = 0 we need to see if b² – 7ba + 6a² = 0 also holds.
Since the relation aRb is defined by the equation a² – 7ab + 6b² = 0, we simply reverse a and b in the equation:
b² – 7ba + 6a² = 0
This is exactly the same as a² – 7ab + 6b² = 0 (just switching a and b does not affect the equation because it is symmetric in a and b).
Thus, the relation is symmetric.
Step 3: Determine if the relation is transitive.
A relation R is transitive if for every a, b, c ∈ ℤ, whenever aRb and bRc, we must have aRc. To check transitivity, we would need to check the equation a² – 7ab + 6b² = 0 and b² – 7bc + 6c² = 0, and then determine if a² – 7ac + 6c² = 0 holds. This is however complicated and does not ensure the transitivity simply by observing the structure of the equation (an algebraic check would be necessary).
For now, let us focus on the properties that have been given, namely, the reflexive and symmetric.
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Let f : R → R be defined by f(x) = 1/x, for all x ∈ R. Then, f is
Since dividing by zero is undefined, f: ℝ → ℝ, given by f(x) = 1/x, is not defined for x = 0. Since the function is not defined at x = 0, it is not well-defined on the entire set ℝ (since the domain of the function is not the whole ℝ). - (d) not defined. Click for more: https://www.tiwariacademy.comRead more
Since dividing by zero is undefined, f: ℝ → ℝ, given by f(x) = 1/x, is not defined for x = 0.
Since the function is not defined at x = 0, it is not well-defined on the entire set ℝ (since the domain of the function is not the whole ℝ).
– (d) not defined.
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The function f : R → R defined by f(x) = 4 + 3 cos x is
We have a function f: ℝ → ℝ defined as f(x) = 4 + 3 cos(x) Whether the given function is bijective, one-one, onto, or neither can be determined. Step 1 Check if the given function is one-one (injective). A function is said to be one-one (injective) if the different elements in the domain of a functiRead more
We have a function f: ℝ → ℝ defined as
f(x) = 4 + 3 cos(x)
Whether the given function is bijective, one-one, onto, or neither can be determined.
Step 1 Check if the given function is one-one (injective).
A function is said to be one-one (injective) if the different elements in the domain of a function go to different elements in the range i.e. f(x₁) = f(x₂) => x₁ = x₂.
Let’s see if the function is injective. Assume that f(x₁) = f(x₂), that is:
4 + 3 cos(x₁) = 4 + 3 cos(x₂)
We simplify this to obtain:
3 cos(x₁) = 3 cos(x₂)
cos(x₁) = cos(x₂)
The cosine function is periodic and not injective. That means that there may be different values of x₁ and x₂, like x₁ = 0 and x₂ = 2π, for which cos(x₁) = cos(x₂). In other words, the function is not injective.
Step 2: Determine whether the function is onto (surjective).
A function is onto (surjective) if for every element y ∈ ℝ, there exists an x ∈ ℝ such that f(x) = y.
For f(x) = 4 + 3 cos(x), the range of the cosine function is between -1 and 1, so the range of f(x) will be:
f(x) = 4 + 3 cos(x)
Since cos(x) lies between -1 and 1, the values of f(x) will lie between:
f(x) = 4 + 3(-1) = 1 and f(x) = 4 + 3(1) = 7
So, f(x) can only be taken in the interval [1, 7]. Since f(x) cannot take any value outside that interval, it is not onto because no values of x map to values of y that are outside that interval.
Conclusion:
The function f(x) = 4 + 3 cos(x) is neither one-one nor onto. So the correct option is
(d) neither one-one nor onto.
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The function f R → R defined as f(x) = x³ is:
The function f: ℝ → ℝ defined by f(x) = x³ is a function from the real numbers to the real numbers. Analyzing the function: 1. One-to-one (Injective): A function is one-to-one if different inputs map to different outputs. Suppose f(x₁) = f(x₂), i.e., x₁³ = x₂³. Then, x₁ = x₂. This demonstrates thatRead more
The function f: ℝ → ℝ defined by f(x) = x³ is a function from the real numbers to the real numbers.
Analyzing the function:
1. One-to-one (Injective):
A function is one-to-one if different inputs map to different outputs.
Suppose f(x₁) = f(x₂), i.e., x₁³ = x₂³.
Then, x₁ = x₂.
This demonstrates that f(x) = x³ is one-to-one (injective).
2. Onto (Surjective): A function is onto if every element in the target set, here ℝ, has a pre-image in the domain. For any real number y ∈ ℝ, we can find an x ∈ ℝ such that f(x) = x³ = y.
Specifically, x = ∛y. This demonstrates that the function f(x) = x³ is onto (surjective).
Conclusion: The function f(x) = x³ is one-to-one and onto, so the correct answer is: – One-one and onto.
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If a set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is
In order to find the number of one-one and onto mappings or bijective functions from set A to set B, consider the following: 1. One-one mapping: Each element of A should map to a unique element of B. 2. Onto mapping: Every element of B should be associated with at least one element of A. ObservationRead more
In order to find the number of one-one and onto mappings or bijective functions from set A to set B, consider the following:
1. One-one mapping: Each element of A should map to a unique element of B.
2. Onto mapping: Every element of B should be associated with at least one element of A.
Observations:
Set A has 5 elements (|A| = 5), and set B has 6 elements (|B| = 6).
– For a bijection (one-one and onto), the no. of elements in both the sets must be equal, that is, |A| = |B|.
Now, |A| ≠ |B| hence it is not possible to have a one-one and onto mapping.
Conclusion:
The number of one-one and onto mappings from A to B is:
0.
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