Let f: N arrow N be a function such that f ( m + n )= f ( m )+ f ( n ) for every m , n ∈ N. If f (6)=18 then f(2) ⋅ f(3) is equal to :

Question

Let f: N arrow N be a function such that f ( m + n )= f ( m )+ f ( n ) for every m , n ∈ N. If f (6)=18 then f(2) ⋅ f(3) is equal to : (A) 6 (B) 54 (C) 18 (D) 36

Tardigrade Admin · Accepted Answer

f( m + n )=f( m )+f( n ) Put m =1, n =1 f(2)=2 f(1) Put m =2, n =1 f(3)=f(2)+f(1)=3 f(1) Put m =3, n =3 f(6)=2 f(3) ⇒ f(3)=9 ⇒ f(1)=3, f(2)=6 f(2) ⋅ f(3)=6 × 9=54

Q. 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 $f (2) \cdot f (3)$ is equal to :

6

54

18

36

$f (m + n) = f (m) + f (n)$
Put $m = 1, n = 1$
$f (2) = 2 f (1)$
Put $m = 2, n = 1$
$f (3) = f (2) + f (1) = 3 f (1)$
Put $m = 3, n = 3$
$f (6) = 2 f (3) \Rightarrow f (3) = 9$
$\Rightarrow f (1) = 3, f (2) = 6$
$f (2) \cdot f (3) = 6 \times 9 = 54$

Q. Let f:N→N be a function such that f(m+n)=f(m)+f(n) for every m,n∈N. If f(6)=18 then f(2)⋅f(3) is equal to :

6

54

18

36

f(m+n)=f(m)+f(n) Put m=1,n=1 f(2)=2f(1) Put m=2,n=1 f(3)=f(2)+f(1)=3f(1) Put m=3,n=3 f(6)=2f(3)⇒f(3)=9 ⇒f(1)=3,f(2)=6 f(2)⋅f(3)=6×9=54

Q. 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 $f (2) \cdot f (3)$ is equal to :

$f (m + n) = f (m) + f (n)$
Put $m = 1, n = 1$
$f (2) = 2 f (1)$
Put $m = 2, n = 1$
$f (3) = f (2) + f (1) = 3 f (1)$
Put $m = 3, n = 3$
$f (6) = 2 f (3) \Rightarrow f (3) = 9$
$\Rightarrow f (1) = 3, f (2) = 6$
$f (2) \cdot f (3) = 6 \times 9 = 54$