Let $f$ be a non-zero real valued continuous function satisfying $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in…
Let $f$ be a non-zero real valued continuous function satisfying $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in \mathbb{R}$. If $f(2)=9$, then $f(6)$ is equal to
$3^2$
$3^6$
$3^4$
$3^3$
Solution
$
\because \quad f(x+y)=f(x) f(y), \forall x, y \in R
$
Put $x=y=1$, we get
$
\begin{aligned}
& f(2)=f(1) \cdot f(1)=9 \quad[\because f(2)=9] \\
& \Rightarrow \quad f(1)^2=9 \Rightarrow f(1)=3 \\
&
\end{aligned}
$
Now, put $x=2$ and $y=1$ in Eq. (i), we get
$
f(3)=f(2) \cdot f(1)=3^2 \cdot 3=3^3
$
Now, put $x=3$ and $y=1$ in Eq. (i), we get
$
f(4)=f(3) \cdot f(1)=3^3 \cdot 3=3^4
$
Again, put $x=4$ and $y=2$ in Eq. (i), we get
$
f(6)=f(4) \cdot f(2)=3^4 \cdot 3^2=3^6
$
Alternative Method
We have,
$
\begin{aligned}
f(x+y) & =f(x) f(y), \forall x, y \in R \\
f(2) & =9
\end{aligned}
$
and
Now,
$
\begin{aligned}
f(1+1) & =f(1) \cdot f(1) \\
f(2) & =\{f(1)\}^2 \\
\{f(1)\} & =\sqrt{\{f(2)\}}
\end{aligned}
$
$
\Rightarrow
$
$
\Rightarrow
$
Now,
$
\{f(1)\}=\sqrt{\{f(2)\}}
$
$
f(6)=f(1+1+1+1+1+1)
$
$
\begin{aligned}
& =f(1) \cdot f(1) \cdot f(1) \cdot f(1) \cdot f(1) \cdot f(1) \\
& =\{f(1)\}^6 \quad \text { [using Eq. (ii) }
\end{aligned}
$
[using Eq. (ii)]
$
=[\sqrt{\{f(2)\}}]^6=[f(2)]^3
$
$=(9)^3$
[using Eq. (i)]
$=(3)^6$