Let f be a continuous function on R satisfying f(x+y)=f(x)+f(y) for all x,y∈R with f(1) =2 and g be a function satisfying f(x)+g(x)= ex then the value of the integral ∫01 f(x)g(x)dx is

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1e−4

14(e−2)

2/3

12(e−3)

answer is C.

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Detailed Solution

First show that f(x)=2x,x∈RNow ∫01 f(x)g(x)dx=∫01 2xex−2xdx=2xex01−2∫01 exdx−43x301=2e−2(e−1)−43=23.

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