Question

If $f(x+y)=f\left(x\right)\cdot f\left(y\right)$ for all $x$ and $y$ and $f\left(15\right)=2,f^{\prime }\left(0\right)=3$, then $f^{\prime }\left(5\right)$ will be

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is C $6$

Let $x=6,y=0\Rightarrow f(5+0)=f\left(5\right)\cdot f\left(0\right)$

$\Rightarrow f\left(5\right)=f\left(5\right)f\left(0\right)\Rightarrow f\left(0\right)=1$

$\therefore f^{\prime }\left(5\right)\underset{h\to 0}{lim}=\frac{f(5+h)-f\left(5\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(5\right)f\left(h\right)-f\left(5\right)}{h}$

$=\underset{h\to 0}{lim}2\left(\frac{f\left(h\right)-1}{h}\right)[\because f(5)=2]$

$=2\underset{h\to 0}{lim}\left(\frac{f\left(h\right)-f\left(0\right)}{h}\right)=2\times f^{\prime }\left(0\right)$

$=2\times 3=6$