Let f : R rightarrow R and g : R rightarrow R | vedclass

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(B) Given $f(x+y) = f(x) + f(y) + f(x)f(y)$. Adding $1$ to both sides gives $1 + f(x+y) = (1 + f(x))(1 + f(y))$.Let $h(x) = 1 + f(x)$. Then $h(x+y) =

Vedclass · Accepted Answer

$A, B, D$

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(B) Given $f(x+y) = f(x) + f(y) + f(x)f(y)$. Adding $1$ to both sides gives $1 + f(x+y) = (1 + f(x))(1 + f(y))$.Let $h(x) = 1 + f(x)$. Then $h(x+y) = h(x)h(y)$.Since $f(x) = xg(x)$ and $\lim_{x 	o 0} g(x) = 1$,we have $\lim_{x 	o 0} f(x) = 0$,so $h(0) = 1 + f(0) = 1$.For $h(x+y) = h(x)h(y)$,the solution is $h(x) = e^{cx}$.Thus $1 + f(x) = e^{cx}$,or $f(x) = e^{cx} - 1$.Given $f(x) = xg(x)$,we have $g(x) = \frac{e^{cx}-1}{x}$.$\lim_{x 	o 0} g(x) = \lim_{x 	o 0} \frac{e^{cx}-1}{x} = c$.Since $\lim_{x 	o 0} g(x) = 1$,we have $c = 1$.Therefore,$f(x) = e^x - 1$.$f'(x) = e^x$,which is defined for all $x \in R$,so $(A)$ is $TRUE$.$f'(0) = e^0 = 1$,so $(D)$ is $TRUE$.$f'(1) = e^1 = e 
eq 1$,so $(C)$ is $FALSE$.For $g(x) = \frac{e^x-1}{x}$ for $x 
eq 0$ and $g(0)=1$,$g$ is differentiable at all $x 
eq 0$. At $x=0$,$g'(0) = \lim_{h 	o 0} \frac{\frac{e^h-1}{h}-1}{h} = \lim_{h 	o 0} \frac{e^h-1-h}{h^2} = \frac{1}{2}$. Thus $g$ is differentiable everywhere,so $(B)$ is $TRUE$.

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