Question

Let $f(x+y)=f\left(x\right)f\left(y\right)$ for all x and y. If $f\left(5\right)=2$ and $f^{\prime }\left(0\right)=3$

then $f^{\prime }\left(5\right)$is equal to

• A. $5$
• B. $8$
• C. $12$
• D. $6$

Answer 1

Answer: (D)

$6$

$f(x+h)=f\left(x\right)f\left(h\right)$
Putting $x=0,y=5$ in the given equation, we get

$f(0+5)=f\left(0\right)f\left(5\right)$

$⟹f\left(5\right)\left[f\right(0)-1]=0$

$⟹f\left(0\right)=1$
Consider, $f^{\prime }\left(5\right)=\underset{h\to 0}{lim}\frac{f(5+h)-f\left(5\right)}{h}$

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

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

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

$=f\left(5\right)f^{\prime }\left(0\right)=2\times 3=6$

$\therefore f^{\prime }\left(5\right)=6$