Let 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 ${I_1} = \int\limits_0^{4nk} {f(x)dx}$ and ${I_2} = \int\limits_{ - k}^{3k} {f(x)dx}$, then :

$f: R \rightarrow R$ and $f(x)+f(x+k)=n \quad \forall x \in R$ <br/><br/> $$ \begin{aligned} &x \rightarrow x+k \\\\ &f(x+k)+f(x+2 k)=n \\\\ &\therefore \quad f(x+2 k)=f(x) \end{aligned} $$ <br/><br/> So, period of $f(x)$ is $2 k$ <br/><br/> $$ \begin{aligned} &\text { Now, } I_{1}=\int_{0}^{4 n k} f(x) d x = 2 n \int_{0}^{2 k} f(x) d x \\\\ &=2 n\left[\int_{0}^{k} f(x) d x+\int_{k}^{2 k} f(x) d x\right] \\\\ &x=t+k \Rightarrow d x=d t \text { (in second integral) } \\\\ &=2 n\left[\int_{0}^{k} f(x) d x+\int_{0}^{k} f(t+k) d t\right] \\\\ &=2 n^{2} k \end{aligned} $$ <br/><br/> Now, $I_2=\int_{-k}^{3 k} f(x) d x=2 \int_{0}^{2 k} f(x) d x$ <br/><br/> $I_{2}=2(n k)$ <br/><br/> $\therefore \quad l_{1}+n l_{2}=4 n^{2} k$