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If $f\left ( ab \right )=f\left ( a \right )f\left ( b \right )$ for all positive integers $a$ and $b$, then the largest possible value of $f\left (1\right )$ is 

  1. $1$
  2. $2$
  3. $0$
  4. $3$
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Given that,  $f(ab)=f(a)f(b);\forall$  $a,b \in \mathbb{Z}^{+}.$

Now, $f(1\cdot1)=f(1)f(1)$

$\Rightarrow f(1)=f^{2}(1)$

$\Rightarrow f^{2}(1)-f(1)=0$

$\Rightarrow f(1)(f(1)-1)=0$

$\Rightarrow f(1)=0;  f(1)=1$

$\therefore$ The largest possible value of $f(1)$ is $1.$

Correct Answer $:\text{A}$
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