Question

Let $f(x+y)=f\left(x\right)\cdot f\left(y\right),\forall x,y\in R$, suppose that $f\left(3\right)=3,$ $f^{\prime }\left(0\right)=11$, then $f^{\prime }\left(3\right)$ is given by

• A. $22$
• B. $44$
• C. $28$
• D. $33$

Answer 1

The correct option is D $33$

$f^{\prime }\left(3\right)=\underset{h\to 0}{lim}\frac{f(3+h)-f\left(3\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(3\right)\cdot f\left(h\right)-f\left(3\right)}{h}$ ......... As $f(x+y)=f\left(x\right).f\left(y\right)$
$\Rightarrow f^{\prime }\left(3\right)=3\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$ $[\because f(3)=3]$
Use L'Hospital rule,
$\Rightarrow f^{\prime }\left(3\right)=3\underset{h\to 0}{lim}\frac{f^{\prime }\left(h\right)}{1}$
$=3f^{\prime }\left(0\right)$ $[\because f^{\prime }(0)=11]$
$=3\times 11=33$