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If f(x) = 0, when x is rational
x, when x is irrational,
g(x) = x, when x is rational
0, when x is irrational
then f - g is
x, when x is irrational,
g(x) = x, when x is rational
0, when x is irrational
then f - g is
- a)One-One onto
- b)One-One not onto
- c)Not one-one but onto
- d)Not one-one not onto
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
If f(x) = 0, when x is rational x, when x is irrational, g(x) = x, w...
Explanation:
To determine whether the function f-g is one-one and onto, we need to first find the expression for f-g.
Finding f-g:
(f-g)(x) = f(x) - g(x)
When x is rational, f(x) = x and g(x) = x, so (f-g)(x) = x - x = 0
When x is irrational, f(x) = 0 and g(x) = 0, so (f-g)(x) = 0 - 0 = 0
Therefore, (f-g)(x) = 0 for all x, rational or irrational.
One-One:
For a function to be one-one, different inputs must produce different outputs.
But in this case, (f-g)(x) = 0 for all x, which means that different inputs produce the same output.
Therefore, the function f-g is not one-one.
Onto:
For a function to be onto, every element in the range must be mapped to by some element in the domain.
In this case, the range of f-g is {0}, which means that every element in the range is indeed mapped to by some element in the domain.
Therefore, the function f-g is onto.
Therefore, the correct answer is option 'A': One-One onto.
To determine whether the function f-g is one-one and onto, we need to first find the expression for f-g.
Finding f-g:
(f-g)(x) = f(x) - g(x)
When x is rational, f(x) = x and g(x) = x, so (f-g)(x) = x - x = 0
When x is irrational, f(x) = 0 and g(x) = 0, so (f-g)(x) = 0 - 0 = 0
Therefore, (f-g)(x) = 0 for all x, rational or irrational.
One-One:
For a function to be one-one, different inputs must produce different outputs.
But in this case, (f-g)(x) = 0 for all x, which means that different inputs produce the same output.
Therefore, the function f-g is not one-one.
Onto:
For a function to be onto, every element in the range must be mapped to by some element in the domain.
In this case, the range of f-g is {0}, which means that every element in the range is indeed mapped to by some element in the domain.
Therefore, the function f-g is onto.
Therefore, the correct answer is option 'A': One-One onto.
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If f(x) = 0, when x is rational x, when x is irrational, g(x) = x, w...
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