Elitmus
Exam
Numerical Ability
Arithmetic
f(n) is a function...where f(f(n))+f(n)=2n+3. f(0)=1;find f(2012)
Read Solution (Total 4)
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- given, f(f(n)) + f(n) = 2n+3 -----(1)
also, f(0)=1
put n=0 in(1)
=> f(f(0))+ f(0)=2*0+3
=> f(1)+ 1 = 3
=> f(1)= 2
put n=1 in (1)
f(f(1)) + f(1) = 2*1 + 3
=> f(2) + 2 = 5
=> f(2) = 3
so on
f(3) = 4
f(4) = 5
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f(2012) = 2013 - 10 years agoHelpfull: Yes(54) No(0)
- f(f(n)+ f(n)=2n+3
put n=0 as we know f(0)=1;
f(1)+1=3
hence f(1)=2
similarly f(2012)= 2013 - 10 years agoHelpfull: Yes(0) No(0)
- ANSWER==> f(2012)=2013
given f(0)=1----
and
f(f(n))+f(n)=2n+3----
now put n= 0 in equ and we get
f(f(0))+ f(0)=2*0+3
f(1)+ 1 = 3
f(1)= 2
again for f(2)=3
f(3)=4, f(4)=5...... so f(2012)=2013
- 10 years agoHelpfull: Yes(0) No(0)
- ans:-2013
- 10 years agoHelpfull: Yes(0) No(0)
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