Functional equations, number of solutions
If
f(x+y) = f(x) + f(y) - xy - 1, x and y are real numbers, and f(1) = 1, then the number of solutions of f(n) = n, where n is a natural number is?
A. one
B. no solution
C. three
D. None of these12 Replies
@Apu
ans is option A
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.how 2 approach
Plug in y=1 and then its an AP with increasing decreasing difference. (Initial difference being -1, i.e. f(n+1)<f(n) for all natural n.
A?
Oh wait you posted answer
Ah, I just read the question. Since it is an A.P. with decreasing difference, the a.p. doesn't increase as fast as compared to linear increase.
Find f(2) by plugging in x=y=1 to find f(2)=0
Since this decreases quickly, after f(1)=1 there can be no solution to f(n)=n
Yup. That's the logical approach. I just brute forced a general term for the series
ahh wow
that's really nice
got it
thanks yall
+solved Opt SirLancelotDuLac
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