Question

If f be decreasing continuous function satisfying $f(x+y)=f\left(x\right)+f\left(y\right)-f\left(x\right)f\left(y\right)$ $\forall$ x ,y $\in$ $R$; $f\left(0\right)\ne 0$, then ${\int }_0^{1}f\left(x\right)dx$ is

• A. $1$
• B. $1-e$
• C. $2-e$
• D. None of these

Answer 1

Answer: (A)

$1$

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

$f\left(0\right)=f\left(0\right)+f\left(0\right)-f^2\left(0\right)$

$f\left(0\right)=2f\left(0\right)-f^2\left(0\right)$

$f^2\left(0\right)=f\left(0\right)$

$f\left(0\right)=1$

$y=0,x=$

$f\left(x\right)=f\left(x\right)+f\left(0\right)-f\left(x\right)$

$f\left(x\right)=1$

${\int }_0^{1}1dx$

$=1$