Question

Let $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x$ and $y,$ suppose $f\left(5\right)=2$ and $f^{\prime }\left(0\right)=3,$ then $f^{\prime }\left(5\right)=?$

• A. 6
• B. 7
• C. 4
• D. 8

Answer 1

The correct option is A 6

$f(x+y)=f\left(x\right)f\left(y\right)\forall x$ and $y$ ...(i)

Put $x=5$ and $y=0$ in (i) $\Rightarrow f\left(5\right)f\left(0\right)$

This gives $f\left(0\right)=1......(\because f(5)=2\ne 0)$
Now $f^{\prime }\left(5\right)=\underset{h\to 0}{lim}\frac{f(5+h)-f\left(5\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(5\right)f\left(h\right)-f\left(5\right)}{h}=f\left(5\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$
$f\left(5\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h-0}=f\left(5\right)f^{\prime }\left(0\right)$
$\therefore f^{\prime }\left(5\right)=2\times 3=6$