Relations Question 3
Question: If $ f(x+y)=f(x)+f(y)-xy-1\forall x,y\in R $ and $ f(1)=1, $ then the number of solutions of $ f( n )=n, $
$ n\in N $ , is
Options:
A) 0
B) 1
C) 2
D) more then 2
Show Answer
Answer:
Correct Answer: B
Solution:
Given $ f(x+y)=f(x)+f(y)-xy-1\forall x,y\in R $
$ f(1)=1 $
$ f(2)=f(1+1)=f(1)+f(1)-1-1=0 $
$ f(3)=f(2+1)=f(2)+f(1)-2\times 1-1=-2 $
$ f(n+1)=f(n)+f(1)-n-1=f(n)-n<f(n) $
$ Thus,f(1)>f(2)>f(3)>….,andf(1)=1. $
$ Therefore,f(1)=1andf(n)<1,forn>1 $
Hence,$f(n)=n,n\in N,$has only one solution $n=1. $
BETA
