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A function f: R → R defined as f(x) = 5x is
- a)Many-one and onto
- b)one-one but not onto
- c)neither one-one nor onto
- d)one-one and onto
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A function f: R → R defined as f(x) = 5x isa)Many-one and ...
Suppose x1 and x2 are real numbers such that f(x1) = f(x2). (We need to show x1 = x2 .)
5x1 = 5x2
Dividing by 5 on both sides gives
x1 = x2 (function is one - one)
Let y ∈R. (We need to show that
x in R such that f(x) = y.)
If such a real number x exists, then 5x -2 = y and x = (y + 2)/5. x is a real number since sums and quotients (except for division by 0) of real numbers are real numbers. It follows that
f(x) = 5((y + 2)/5) by the substitution and the definition of f
= y + 2
= y by basic algebra
Hence, f is onto.
5x1 = 5x2
Dividing by 5 on both sides gives
x1 = x2 (function is one - one)
Let y ∈R. (We need to show that
x in R such that f(x) = y.)If such a real number x exists, then 5x -2 = y and x = (y + 2)/5. x is a real number since sums and quotients (except for division by 0) of real numbers are real numbers. It follows that
f(x) = 5((y + 2)/5) by the substitution and the definition of f
= y + 2
= y by basic algebra
Hence, f is onto.
Most Upvoted Answer
A function f: R → R defined as f(x) = 5x isa)Many-one and ...
To R is a rule that assigns to each real number x a unique real number f(x).
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A function f: R → R defined as f(x) = 5x isa)Many-one and ontob)one-one but not ontoc)neither one-one nor ontod)one-one and ontoCorrect answer is option 'D'. Can you explain this answer?
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