A) 0
B) 1
C) 2
D) more than 2
Correct Answer: B
Solution :
| \[\operatorname{Given}\,f\left( x+y \right)=f\left( x \right)+f\left( y \right)-xy-1\]\[\forall x,y\in R\] |
| \[f\left( 1 \right)=1\] |
| \[f\left( 2 \right)=f\left( 1+1 \right)=f\left( 1 \right)+f\left( 1 \right)-1-1=0\] |
| \[f\left( 3 \right)=f\left( 2+1 \right)=f\left( 2 \right)+f\left( 1 \right)-2.1-1=-2\] |
| \[f\left( n+1 \right)=f\left( n \right)+f\left( 1 \right)-n-1=f\left( n \right)-n<f\left( n \right)\] |
| Thus\[f\left( 1 \right)>f\left( 2 \right)>f\left( 3 \right)>.....\] |
| and \[f\left( 1 \right)=1\therefore f\left( 1 \right)=1\] and \[f\left( n \right)<1,\] for \[n>1\] |
| Hence \[f\left( n \right)=n,n\in N\]has only one |
| Solution \[n=1\] |
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