Question

A function $f:R\to R$ satisfies the equation $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in R,f\left(x\right)\ne 0$. Suppose that the function is differentiable at $x=0$ and $f^{\prime }\left(0\right)=2$. Then,

• A. $f^{\prime }\left(x\right)=f\left(x\right)$
• B. $f^{\prime }\left(x\right)=2f\left(x\right)$
• C. $2f^{\prime }\left(x\right)=f\left(x\right)$
• D. None of these

Answer 1

The correct option is B $f^{\prime }\left(x\right)=2f\left(x\right)$

Given,

$f(x+y)=f\left(x\right).f\left(y\right)\forall x,y\in R$.......(1).

Then $f(0+0)=f\left(0\right).f\left(0\right)$

or, $f\left(0\right)=1$ [ Since $f\left(x\right)\ne 0\forall x\in R$].......(2).

Also given,

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

or, $\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=2$

or, $\underset{h\to 0}{lim}\frac{f\left(0\right)f\left(h\right)-f\left(0\right)}{h}=2$ [ Using (1)]

or, $\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}=2$........(3) [Using (2)].

Now,

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

or, $f^{\prime }\left(x\right)=$$\underset{h\to 0}{lim}\frac{f\left(x\right).f\left(h\right)-f\left(x\right)}{h}$ [ Using (1)]

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

or, $f^{\prime }\left(x\right)=2f\left(x\right)$. [ Using (3)]