retagged by
462 views

1 Answer

1 votes
1 votes
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 $:\text{A}$
edited by
Answer:

Related questions

1 votes
1 votes
1 answer
1
go_editor asked Mar 16, 2020
474 views
Let $f\left ( x \right )=x^{2}$ and $g\left ( x \right )=2^{x}$, for all real $x$. Then the value of $ f \left ( f\left ( g\left ( x \right ) \right )+g\left( f\left ( x...
1 votes
1 votes
1 answer
2
go_editor asked Mar 16, 2020
634 views
Let $f\left ( x \right )=2x-5$ and $g\left ( x \right )=7-2x.$ Then $ \mid f\left ( x \right ) + g\left ( x \right ) \mid=\mid f(x) \mid + \mid g\left ( x \right ) \mid$ ...
1 votes
1 votes
1 answer
4
1 votes
1 votes
1 answer
5