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Question: If f is a real valued function such that \[f(x + y) = f(x) + f(y)\] and \(f(1) = 5\), then the value...

If f is a real valued function such that f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y) and f(1)=5f(1) = 5, then the value of f(100)f(100)is
(1) 200200
(2) 300300
(3) 350350
(4) 400400
(5) 500500

Explanation

Solution

In the given question, a real valued function is given, along with a value for that function. To find the value of f(100)f(100), we need to substitute values for xxand yy. By doing this, we can figure out a pattern in which values have been given to the function. Hence, we can arrive at the final answer.

Complete step-by-step solution:
Consider the given function, f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y)
It is also given that f(1)=5f(1) = 5.
Now, let us substitute values for xxand yy.
If, x=2,y=1x = 2,y = 1,
The real valued function will be,
\eqalign{ & \Rightarrow f(1 + 1) = f(1) + f(1) \cr & \Rightarrow f(2) = 2f(1) \cr}
Substituting the value of f(1)f(1)in the above equation,
\eqalign{ & \Rightarrow f(2) = 2 \times 5 \cr & \Rightarrow f(2) = 10 \cr}
Now let us take another set of values for substituting in the function.
If, x=2,y=1x = 2,y = 1we get,
\eqalign{ & \Rightarrow f(2 + 1) = f(2) + f(1) \cr & \Rightarrow f(3) = f(2) + f(1) \cr}
We know the values of both f(1)andf(2)f(1)andf(2)
Therefore, by substituting, we get,
\eqalign{ & \Rightarrow f(3) = 10 + 5 \cr & \Rightarrow f(3) = 15 \cr}
From the above equations, we have figured out a pattern which is yielding us the values of the real valued function. That is,
For any positive integer, nn
f(n)=nf(1)f\left( n \right) = nf\left( 1 \right)
Now, we can find out the value of f(100)f\left( {100} \right)by substituting 100100in the place of nnin the above equation.
\eqalign{ & \Rightarrow f\left( {100} \right) = 100f\left( 1 \right) \cr & \Rightarrow f\left( {100} \right) = 100 \times 5 \cr & \Rightarrow f\left( {100} \right) = 500 \cr}
The final answer is f(100)=500f\left( {100} \right) = 500.
Hence, option (5) is the right answer.

Note: A real valued function will always have real values. This is done by a trial method, where values are substituted for the real valued function to get the pattern. Therefore, make sure that you start with small integers. After one or two such trials you will be able to figure out the pattern.