Question

If $f(x+y)=f\left(x\right).f\left(y\right)$ and $f\left(5\right)=2$, $f^{\prime }\left(0\right)=3$ then $f^{\prime }\left(5\right)$ equals

• A. 0
• B. 1
• C. 6
• D. 2

Answer 1

The correct option is C 6

Given $f(x+y)=f\left(x\right)f\left(y\right)$

$\therefore$ $f(0+0)={\left(f\left(0\right)\right)}^2$

$\Rightarrow f\left(0\right)=0$ or $f\left(0\right)=1$ but $f\left(0\right)\ne 0$

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

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

$f^{\prime }\left(x\right)=f\left(x\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$

$\therefore$ $f^{\prime }\left(0\right)=f\left(0\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$

$3=\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$ $(\because f\left(0\right)=1)$

Now $f^{\prime }\left(x\right)=f\left(x\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$

$\therefore$ $f^{\prime }\left(5\right)=f\left(5\right)\times 3$

$=2\times 3=6$