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The function f(x) = 10x from R to [0, ∞) is
- a)an identity function
- b)one-one and into
- c)a constant function
- d)one-one and onto
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The function f(x) =10x from R to [0, ∞) isa)an identity function...
Method to Solve :
A function's injectivity and surjectivity depends upon the domain and co-domain. A can be injective for a given domain but can be not-injective . As the domain is [0, inf) and the co-domain is R
The above function is one-one(injective) but not onto (surjective).
However if the co-domain too have been [0,inf) this function would have been surjective. And if the domain have too have been R ,|x| would not have been injective.
An injective function is a function that preserves distinctness , it never maps distinct elements of its domain to the same element of its co-domain.
as |x|= x in [0, inf)
f(x)=x in the given domain
Lets assume for two different x f(x) is equals
f(x1)=f(x2)
=> x1=x2
Hence f(x1) = f(x2) only if x1=x2
Moreover if you draw the graph it will be a straight line with slope 45 degrees. And if we draw a line parallel to x-axis it intersects the plotted line only at one place which means there are no two values in domain i.e. x for same value of y.
Hence the function is one-one in the given domain.
A function is said to be surjective if every element in Y (co-domain) has a corresponding element in X (domain)
Now as the co-domain is R but the f(x) can take only positive values therefore it is not onto. As for all negative values in the co-domain there is no x in the domain mapping these values.
As f(x) =x in [0,inf)
As x>=0
Therefore f(x) >=0
Hence in the given Domain and for the given Co-domain the function f(x)= |x| is injective but not surjective
The above function is one-one(injective) but not onto (surjective).
However if the co-domain too have been [0,inf) this function would have been surjective. And if the domain have too have been R ,|x| would not have been injective.
An injective function is a function that preserves distinctness , it never maps distinct elements of its domain to the same element of its co-domain.
as |x|= x in [0, inf)
f(x)=x in the given domain
Lets assume for two different x f(x) is equals
f(x1)=f(x2)
=> x1=x2
Hence f(x1) = f(x2) only if x1=x2
Moreover if you draw the graph it will be a straight line with slope 45 degrees. And if we draw a line parallel to x-axis it intersects the plotted line only at one place which means there are no two values in domain i.e. x for same value of y.
Hence the function is one-one in the given domain.
A function is said to be surjective if every element in Y (co-domain) has a corresponding element in X (domain)
Now as the co-domain is R but the f(x) can take only positive values therefore it is not onto. As for all negative values in the co-domain there is no x in the domain mapping these values.
As f(x) =x in [0,inf)
As x>=0
Therefore f(x) >=0
Hence in the given Domain and for the given Co-domain the function f(x)= |x| is injective but not surjective
Most Upvoted Answer
The function f(x) =10x from R to [0, ∞) isa)an identity function...
Explanation:
Function f(x) = 10x is defined from R (set of real numbers) to [0, ) (set of non-negative real numbers).
To determine whether the function is one-one, onto, or both, we can use the following definitions:
- One-one function: A function f(x) is one-one if each element of the function's codomain (output set) has at most one preimage (input value).
- Onto function: A function f(x) is onto if each element of the function's codomain has at least one preimage in the domain (input set).
Let's check whether f(x) satisfies these conditions:
One-one:
Suppose f(x1) = f(x2) for some x1, x2 in R. Then, we have:
10x1 = 10x2
x1 = x2
Therefore, f(x) is one-one.
Onto:
Let y be an arbitrary element of [0, ). We need to find an x in R such that f(x) = y. Since y is non-negative, we can choose x = y/10. Then:
f(x) = 10(x) = 10(y/10) = y
Therefore, f(x) is onto.
Conclusion:
Since f(x) is both one-one and onto, it is a bijective function.
Function f(x) = 10x is defined from R (set of real numbers) to [0, ) (set of non-negative real numbers).
To determine whether the function is one-one, onto, or both, we can use the following definitions:
- One-one function: A function f(x) is one-one if each element of the function's codomain (output set) has at most one preimage (input value).
- Onto function: A function f(x) is onto if each element of the function's codomain has at least one preimage in the domain (input set).
Let's check whether f(x) satisfies these conditions:
One-one:
Suppose f(x1) = f(x2) for some x1, x2 in R. Then, we have:
10x1 = 10x2
x1 = x2
Therefore, f(x) is one-one.
Onto:
Let y be an arbitrary element of [0, ). We need to find an x in R such that f(x) = y. Since y is non-negative, we can choose x = y/10. Then:
f(x) = 10(x) = 10(y/10) = y
Therefore, f(x) is onto.
Conclusion:
Since f(x) is both one-one and onto, it is a bijective function.
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The function f(x) =10x from R to [0, ∞) isa)an identity functionb)one-one and intoc)a constant functiond)one-one and ontoCorrect answer is option 'B'. Can you explain this answer?
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