Question

If $f(x+y)=f\left(x\right)f\left(y\right)\forall x,y\in R$ and $f^{\prime }\left(0\right)=5,f\left(2\right)=6$, then $f^{\prime }\left(2\right)$

• A. $10$
• B. $20$
• C. $30$
• D. $40$

Answer 1

Answer: (C)

$30$

Put $y=h$
$f(x+h)=f\left(x\right)f\left(h\right)$
$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$f^{{}^{\prime }}\left(x\right)={lim}_{h\to 0}\frac{f\left(x\right)f\left(h\right)-f\left(x\right)}{h}$
$f^{{}^{\prime }}\left(x\right)={lim}_{h\to 0}f\left(x\right)\left(\frac{f\left(h\right)-1}{h}\right)\to${i}
Now in the given functional relation, putting $x=0,y=2$ we have
$f\left(2\right)=f\left(2\right)f\left(0\right)$
Given $f\left(2\right)=6$
$\therefore f\left(0\right)=1$
${lim}_{h\to 0}\frac{f\left(h\right)-f\left(0\right)}{h}=f^{{}^{\prime }}\left(0\right)\to${ii}
From {i} and {ii}, we get
$f^{{}^{\prime }}\left(x\right)=f\left(x\right)\left(f^{\prime }\right(0\left)\right)$
Substitute $x=2$,
$f^{\prime }\left(2\right)=f\left(2\right)f^{\prime }\left(0\right)$
$f^{\prime }\left(2\right)=6\times 5=30$