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Question

If f(x+y)=f(x)+f(y)xy3 and f(1)=3, then the number of solutions for f(n)=3n, where nN is

A
0
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B
2
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C
1
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D
3
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Solution

The correct option is C 1
Putting x=n and y=1
f(n+1)=f(n)+f(1)n×13f(n+1)=f(n)n<f(n)f(n)>f(n+1)f(1)>f(n) for n>1
3>3n, which is not possible for n>1

So, there is no solution for n>1.
The only possible solution is n=1

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