Question

Let $f\left(x\right)$ be continuous and differentiable function satisfying $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,yϵR$.
If $f\left(x\right)$ can be expressed as $f\left(x\right)=1+xp\left(x\right)+x^{2}q\left(x\right)$
where $\underset{x\to 0}{lim}p\left(x\right)=a$ and $\underset{x\to 0}{lim}q\left(x\right)=b$ then $f^{\prime }\left(x\right)$ is

• A. $af\left(x\right)$
• B. $bf\left(x\right)$
• C. $(a+b)f\left(x\right)$
• D. $(a+2b)f\left(x\right)$

Answer 1

Answer: (A)

$af\left(x\right)$

$f(x+y)=f\left(x\right)f\left(y\right)$
Let $h\to 0$
$f(x+h)=f\left(x\right)f\left(h\right)$
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=f\left(x\right)\frac{f\left(h\right)-1}{h}=f\left(x\right)f^{\prime }\left(0\right)$
Also
$f\left(x\right)=1+xp\left(x\right)+x^{2}q\left(x\right)$
$f^{\prime }\left(x\right)=p\left(x\right)+x{p}^{\prime }\left(x\right)+x^2{q}^{\prime }\left(x\right)+2xq\left(x\right)$
$\underset{x\to 0}{lim}{f}^{\prime }\left(x\right)=p\left(x\right)+x{p}^{\prime }\left(x\right)+x^2{q}^{\prime }\left(x\right)+2xq\left(x\right)=p\left(0\right)=a$
$f^{\prime }\left(0\right)=a$
Thus,
$f^{\prime }\left(x\right)=f\left(x\right)f\left(0\right)=af\left(x\right)$