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Let f be a function satisfying f(x + y) = f(x) + f(y) for all x, y ∈ R, then f ‘ (x) =
- a)0 for all x ∈ R
- b)f (0) for all x ∈ R
- c)f ‘(0) for all x ∈ R
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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Let f be a function satisfying f(x + y) = f(x) + f(y) for all x, yW...
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Let f be a function satisfying f(x + y) = f(x) + f(y) for all x, yW...
Given:
- A function f(x y) = f(x) f(y) for all x, y R
To find:
- The value of f(x)
Explanation:
Let's analyze the given function and its properties to determine the value of f(x).
Property 1:
f(x y) = f(x) f(y) for all x, y R
This property states that the function f(x) satisfies the property of additivity. When two numbers are added, the function applied to the sum is equal to the sum of the function applied to each number individually.
Property 2:
f(0 y) = f(0) f(y) for all y R
Using the additivity property, we can rewrite the function as f(x y) = f(x) f(y) = f(0) f(x) f(y).
Applying this to the case where x = 0, we get f(0 y) = f(0) f(y) = f(0) f(0) f(y) = f(0)^2 f(y).
Property 3:
f(0) = 0 or f(0) = 1
Since f(0) is a constant term, it can only take the values 0 or 1.
Now let's consider two cases:
Case 1: f(0) = 0
If f(0) = 0, then using Property 2, we have f(0 y) = f(0)^2 f(y) = 0 f(y) = 0 for all y R.
Therefore, in this case, f(x) = 0 for all x R.
Case 2: f(0) = 1
If f(0) = 1, then using Property 2, we have f(0 y) = f(0)^2 f(y) = 1 f(y) = f(y) for all y R.
In this case, the value of f(x) can be any real number for all x R.
Conclusion:
From the above analysis, we can conclude that f(x) = 0 for all x R if f(0) = 0, and f(x) can be any real number for all x R if f(0) = 1.
Therefore, the correct answer is option 'C': f(0) for all x R.
- A function f(x y) = f(x) f(y) for all x, y R
To find:
- The value of f(x)
Explanation:
Let's analyze the given function and its properties to determine the value of f(x).
Property 1:
f(x y) = f(x) f(y) for all x, y R
This property states that the function f(x) satisfies the property of additivity. When two numbers are added, the function applied to the sum is equal to the sum of the function applied to each number individually.
Property 2:
f(0 y) = f(0) f(y) for all y R
Using the additivity property, we can rewrite the function as f(x y) = f(x) f(y) = f(0) f(x) f(y).
Applying this to the case where x = 0, we get f(0 y) = f(0) f(y) = f(0) f(0) f(y) = f(0)^2 f(y).
Property 3:
f(0) = 0 or f(0) = 1
Since f(0) is a constant term, it can only take the values 0 or 1.
Now let's consider two cases:
Case 1: f(0) = 0
If f(0) = 0, then using Property 2, we have f(0 y) = f(0)^2 f(y) = 0 f(y) = 0 for all y R.
Therefore, in this case, f(x) = 0 for all x R.
Case 2: f(0) = 1
If f(0) = 1, then using Property 2, we have f(0 y) = f(0)^2 f(y) = 1 f(y) = f(y) for all y R.
In this case, the value of f(x) can be any real number for all x R.
Conclusion:
From the above analysis, we can conclude that f(x) = 0 for all x R if f(0) = 0, and f(x) can be any real number for all x R if f(0) = 1.
Therefore, the correct answer is option 'C': f(0) for all x R.
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