MathematicsMedium210×since 2002Q3429Let f : R →\to→ R be a continuous function satisfying f(x) + f(x + k) = n, for all x ∈\in∈ R where k > 0 and n is a positive integer. If I1=∫04nkf(x)dx{I_1} = \int\limits_0^{4nk} {f(x)dx}I1=0∫4nkf(x)dx and I2=∫−k3kf(x)dx{I_2} = \int\limits_{ - k}^{3k} {f(x)dx}I2=−k∫3kf(x)dx, then :AI1+2I2=4nk{I_1} + 2{I_2} = 4nkI1+2I2=4nkBI1+2I2=2nk{I_1} + 2{I_2} = 2nkI1+2I2=2nkCI1+nI2=4n2k{I_1} + n{I_2} = 4{n^2}kI1+nI2=4n2kDI1+nI2=6n2k{I_1} + n{I_2} = 6{n^2}kI1+nI2=6n2kCheck answerSkip