if the function is one one then its coming to f(x)=12-x
how to proof one-one now is d question
\hspace{-16}$If $\bf{f:\mathbb{Z}\rightarrow \mathbb{Z}}$ and $\bf{m + f\big(m + f(n + f(m))\big) = n + f(m)}$\\\\ and $\bf{f(6)=6}$.\;Then $\bf{f(2012)=}$
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6 Answers
1-2000 ?
plugging f(x) = ax+b in the original eqn. (*) and setting n = 0 gives
f(x) = k-x
since f(6) = 6 => k=12
so f(2012) = -2000
I dunno how to justify (*) that part :)
36@Rishabh - you have already assumed the function to be a polynomial function... and luck by chance... it was a polynomial function... :P
36\hspace{-16}$If $\bf{f:\mathbb{Z}\rightarrow \mathbb{Z}}$ and $\bf{m + f\big(m + f(n + f(m))\big) = n + f(m)}$\\\\ and $\bf{f(6)=6}$.\;Then $\bf{f(2012)=}$
put, m = 6 and n = 6
=> 6 + f(6 + f(0 + f(6))) = 0 + f(6) => 6 + f(12) = f(6)
=> f(12) = 0
Again,
put, m = 12, n = 6
=> 12 + f(12 + f(6 + f(12))) = 6 + f(12)
=> f(18) = -6
Again,
put, m = 6 and n = 6
=> 6 + f(6 + f(-6 + f(6))) = -6 + f(6)
=> f(6 + f(0)) = -6 = f(18)
=> f(0) = 12
so we have,
f(0) = 12, f(6) = 6, f(12) = 0 , f(18) = -6
which makes a pattern....
giving,
f(x) = 12 - x
so, f(2012) = -2000 Ans.
- rahul a small edit...
for f(0) we put, n = -6 and not 6...!!Upvote·0· Reply ·2012-06-23 16:20:30
