Let f : N $\to$ N be a function such that f(m + n) = f(m) + f(n) for every m, n$\in$N. If f(6) = 18, then f(2) . f(3) is equal to :
Solution
f(m + n) = f(m) + f(n)<br><br>Put m = 1, n = 1<br><br>f(2) = 2f(1)<br><br>Put m = 2, n = 1<br><br>f(3) = f(2) + f(1) = 3f(1)<br><br>Put m = 3, n = 3<br><br>f(6) = 2f(3) $\Rightarrow$ f(3) = 9<br><br>$\Rightarrow$ f(1) = 3, f(2) = 6<br><br>f(2) . f(3) = 6 $\times$ 9 = 54
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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