Question

Let $f(x+y)=f\left(x\right)f\left(y\right)\forall x,yϵR$ and $f\left(x\right)=1+xg\left(x\right)$ where $\underset{x\to 0}{lim}g\left(x\right)=1$. Then:

• A. f(x) is discontinuous at x=0
• B. f’(0) = 1
• C. $f^{\prime }\left(x\right)=f\left(x\right)\forall x$
• D. $f^{\prime }\left(x\right)=f"\left(x\right)\forall x$

Answer 1

The correct option is C $f^{\prime }\left(x\right)=f\left(x\right)\forall x$

$f(x+y)f\left(x\right)f\left(y\right)\Rightarrow f\left(0\right)=f(0{)}^2$
$f\left(x\right)=1+xg\left(x\right)\Rightarrow f\left(0\right)=1\Rightarrow f\left(0\right)=0or1\cdots \left(1\right)$
Let f(0)=1.At x=a(a is arbitary)
$f^{\prime }\left(a\right)=\underset{h\to 0}{lim}\frac{f(a+h)-f\left(a\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(a\right)f\left(h\right)-f\left(a\right)}{h}$
$=f\left(a\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$
$=f\left(a\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$
$=f\left(a\right)\underset{h\to 0}{lim}\frac{1+hg\left(h\right)-1}{h}$
$=f\left(a\right)\underset{h\to 0}{lim}=g\left(h\right)$
$=f\left(a\right)\times 1=f\left(a\right)$.
Hence $f^{\prime }\left(a\right)=f\left(a\right)\forall a\Rightarrow f^{\prime }\left(x\right)=f\left(x\right)\forall x$.