Question

A function f is such that f(x+y) = 2 f(x).f(y). If f(x) is differentiable at x=0, then f(0) can be

• A. 0
• B. 0.5
• C. 1
• D. 2

Answer 1

Answer: (A), (B)

The correct options are
A

0

B

0.5

$f(x+y)=2f\left(x\right).f\left(y\right)f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}{f}^{\prime }\left(x\right)={lim}_{h\to 0}\frac{2f\left(x\right).f\left(h\right)-f\left(x\right)}{h}(\because f(x+h)=2f(x).f(h\left)\right)\Rightarrow f^{\prime }\left(x\right)=f\left(x\right).{lim}_{h\to 0}\frac{\left(2f\right(h)-1)}{h}{f}^{\prime }\left(0\right)=f\left(0\right){lim}_{h\to 0}\frac{\left(2f\right(h)-1)}{h}$
Since f(x) is differentiable (and hence continuous) at x = 0,
$f^{\prime }\left(0\right)=\frac{f\left(0\right).\left(2f\right(0)-1)}{{lim}_{h\to 0}h}$
For the limit to exist finitely,
$f\left(0\right)=0or2f\left(0\right)-1=0\Rightarrow f\left(0\right)=0orf\left(0\right)=\frac{1}{2}$