Question

If $f(x+y+z)=f\left(x\right).f\left(y\right).f\left(z\right)$ for all $x,y,z$ and $f\left(2\right)=4,f^{\prime }\left(0\right)=3$, then $f^{\prime }\left(2\right)$ equals

• A. $12$
• B. $9$
• C. $16$
• D. $6$

Answer 1

Answer: (A)

$12$

$f(x+y+z)=f\left(x\right).f\left(y\right).f\left(z\right)$
putting $x=y=z=0$, we get
$f\left(0\right)={f\left(0\right)}^3$
$\Rightarrow f\left(0\right)[1-{f\left(0\right)}^2]=0$
$\Rightarrow f\left(0\right)=0or1or-1$
$f\left(0\right)=0$ is discarded because it would imply $f\left(x\right)=0$ for all 'x'.
Similarly $f\left(0\right)=-1$ is discarded because it would imply $f\left(2\right)$ is negative, which contradicts the value of $f\left(2\right)$ given in the question.
Hence, $f\left(0\right)=1$
Now in the equation, substitute y=2 and z=0
We get,
$f(x+2)=f\left(x\right)f\left(2\right)f\left(0\right)$
$f(x+2)=4f\left(x\right)$
$\Rightarrow f^{\prime }(x+2)=4f^{\prime }\left(x\right)$
Substitute $x=0$ to get
$f^{\prime }\left(2\right)=4f^{\prime }\left(0\right)=4\times 3=12$