Question

A continuous function $f:R\to R$ satisfies the equation $f\left(x\right)=x+{\int }_0^{x}f\left(t\right)dt$. Which of the following options is true?

• A. $f(x+y)=f\left(x\right)+f\left(y\right)$
• B. $f(x+y)=f\left(x\right)f\left(y\right)$
• C. $f(x+y)=f\left(x\right)+f\left(y\right)+f\left(x\right)f\left(y\right)$
• D. $f(x+y)=f\left(xy\right)$

Answer 1

Answer: (C)

$f(x+y)=f\left(x\right)+f\left(y\right)+f\left(x\right)f\left(y\right)$

$f\left(x\right)=x+{\int }_0^{x}f\left(t\right)dt$

Differentiating both sides using Newtons Leibniz rule

$f^{\prime }\left(x\right)=1+f\left(x\right)$

Let $\frac{dt}{dx}=f^{\prime }\left(x\right)$

$\therefore \frac{dt}{dx}=1+t\Rightarrow \frac{dt}{1+t}=dx\Rightarrow \int \frac{dt}{1+t}=\int dx\Rightarrow ln(1+t)=x\Rightarrow ln(1+f\left(x\right))=x\Rightarrow 1+f\left(x\right)=e^x\Rightarrow f\left(x\right)=e^x-1$

Option $\left(C\right)$ satisfies our function as

$f(x+y)=e^{x+y}-1f\left(x\right)+f\left(y\right)+f\left(x\right)f\left(y\right)=e^x-1+e^y-1+(e^x-1)(e^y-1)f\left(x\right)+f\left(y\right)+f\left(x\right)f\left(y\right)=e^{x+y}-1$

Hence, option $C$ is correct.