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Logical Reasoning
Number Series
F(n) is a function…where f(f(n))+f(n)=2n+3.F(n)=0;find f(2012) ?
A. 4123 B. 4027 C. 4131 D. 4321
Read Solution (Total 3)
-
- f(n)=0
putting n=0
f(0)=2n+3
f(f(2012))+f(2012)
=>f(0)=2n+3 where n=2012
B.4027 ans.... - 10 years agoHelpfull: Yes(7) No(0)
- f(n)=0
then putting n=0
f(f(n))+f(n)=2n+3
f(f(0))+f(0)=2n+3
then put n=2012
f(f(2012))+f(2012) so f(n)=0then also f(2012)=0
2n+3
2*2012+3=4027
- 10 years agoHelpfull: Yes(1) No(0)
- 4027 becouse it is given f(n)=0
so f(f(n))=f(f(0))=f(0)=0
hence f(f(n))+f(n)=2n+3
f(n)=0 this means f(2012)=0
0+0=2*2012+3=4027 - 10 years agoHelpfull: Yes(0) No(0)
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