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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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Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. Then relation R is
We have the relation R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} on the set A = {1, 2, 3, 4}. Definitions: 1. Function: A relation R is a function if each element of A occurs as the first component in at most one pair of R. 2. Transitive: A relation R is transitive if for all a, b, c ∈ A, wheneverRead more
We have the relation R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} on the set A = {1, 2, 3, 4}.
Definitions:
1. Function: A relation R is a function if each element of A occurs as the first component in at most one pair of R.
2. Transitive: A relation R is transitive if for all a, b, c ∈ A, whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
3. Symmetric: A relation R is symmetric if for all a, b ∈ A, whenever (a, b) ∈ R, then (b, a) ∈ R.
4. Reflexive: A relation R is reflexive if for all a ∈ A, (a, a) ∈ R.
Analysis:
1. Is R a function?
– The element 2 occurs as the first element in both (2, 4) and (2, 3), which goes against the definition of a function. Thus, R is not a function.
2. Is R transitive?
– Since (1, 3) ∈ R and (3, 1) ∈ R, we would need (1, 1) ∈ R for it to be transitive, but it isn’t there.
– For (2, 4) ∈ R and (4, 2) ∈ R, we would need (2, 2) ∈ R, which is absent too.
– Thus, R is not transitive.
3. Is R symmetric?
– Do we have (a, b) ∈ R implies (b, a) ∈ R?
– (1, 3) ∈ R, and (3, 1) ∈ R (symmetric pair).
– (4, 2) ∈ R, but (2, 4) ∈ R (symmetric pair).
– All needed symmetric pairs are there. So, R is symmetric.
4. Is R reflexive?
– See if (a, a) ∈ R for every a ∈ A:
– (1, 1), (2, 2), (3, 3), and (4, 4) are not in R.
– Hence, R is not reflexive.
Conclusion:
The relation R is:
– Not a function
– Not transitive
– Symmetric
– Not reflexive
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Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Based on the given information, f is best defined as:
Let A = {1, 2, 3} and B = {4, 5, 6, 7}. Let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. 1. Injective (One-to-one): Each element of A goes to a distinct element in B. There are no two elements of A going to the same element in B. Therefore, f is injective. 2. Surjective (Onto): Not all elRead more
Let A = {1, 2, 3} and B = {4, 5, 6, 7}.
Let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B.
1. Injective (One-to-one): Each element of A goes to a distinct element in B. There are no two elements of A going to the same element in B.
Therefore, f is injective.
2. Surjective (Onto):
Not all elements of B are covered by f. For instance, 7 is not mapped by any element of A.
Therefore, f is not surjective.
3. Bijective (One-to-one and Onto):
Since f is not surjective, it cannot be bijective.
4. Function:
f is a function since every element of A is related to exactly one element of B.
Therefore, the correct answer is: Injective function.
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