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The function f: R → R defined as f(x) = 7x + 4 is both one-one and onto.
- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The function f: R → R defined as f(x) = 7x + 4 is both one-one an...
The given statement is true. f is both one-one and onto.
For one-one: Consider f(x1) = f(x2)
∴7x1 + 4 = 7x2 + 4
⇒ x1 = x2.
Thus, f is one – one.
For onto: Now for any real number y which lies in the co- domain R, there exists an element x=(y-4)/7
such that
Therefore, the function is onto.
For one-one: Consider f(x1) = f(x2)
∴7x1 + 4 = 7x2 + 4
⇒ x1 = x2.
Thus, f is one – one.
For onto: Now for any real number y which lies in the co- domain R, there exists an element x=(y-4)/7
such that
Therefore, the function is onto.Most Upvoted Answer
The function f: R → R defined as f(x) = 7x + 4 is both one-one an...
The function f: R --> R is a mathematical function that takes a real number as input and returns a real number as output. The notation "R" represents the set of all real numbers. This means that the domain (input) and codomain (output) of the function are both the set of real numbers.
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The function f: R → R defined as f(x) = 7x + 4 is both one-one an...
The given statement is true. f is both one-one and onto.
For one-one: Consider f(x1) = f(x2)
∴7x1 + 4 = 7x2 + 4
⇒ x1 = x2.
Thus, f is one – one.
For onto: Now for any real number y which lies in the co- domain R, there exists an element x=(y-4)/7
such that Therefore, the function is onto.
For one-one: Consider f(x1) = f(x2)
∴7x1 + 4 = 7x2 + 4
⇒ x1 = x2.
Thus, f is one – one.
For onto: Now for any real number y which lies in the co- domain R, there exists an element x=(y-4)/7
such that
Therefore, the function is onto.
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The function f: R → R defined as f(x) = 7x + 4 is both one-one and onto.a)Trueb)FalseCorrect answer is option 'A'. Can you explain this answer?
Question Description
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