Question

Let $f:R\to R$ be continuous function satisfying $f(x)+f(x+k)=n$, for all $x\in R$ where $k>0$ and $n$ is a positive integer. If $I_1=\underset{0}{\overset{4nk}{\int }}f(x)dx$ and $I_2=\underset{-k}{\overset{3k}{\int }}f(x)dx$, then

• A. $I_1+2I_2=4nk$
• B. $I_1+2I_2=2nk$
• C. $I_1+n{I}_2=4n^{2}k$
• D. $I_1+n{I}_2=6n^{2}k$

Answer 1

Answer: (C)

$f(x)+f(x+k)=n$
$\Rightarrow f(x)=f(x+2k)$
$f(x)$ is periodic with period $2k$
$I_1=\underset{0}{\overset{4nk}{\int }}f(x)dx=2n\underset{0}{\overset{2k}{\int }}f(x)dx$
$I_2=\underset{-k}{\overset{3k}{\int }}f(x)dx=2{\int }_0^{2k}f(x)dx$
Now,
$f(x)+f(x+k)=n$
$\Rightarrow \underset{0}{\overset{k}{\int }}f(x)dx+\underset{0}{\overset{k}{\int }}f(x+k)dx=nk$
$\Rightarrow \underset{0}{\overset{k}{\int }}f(x)dx+\underset{k}{\overset{2k}{\int }}f(x)dx=nk$
$\Rightarrow \underset{0}{\overset{2k}{\int }}f(x)dx=nk$
$\Rightarrow I_1=2n^{2}k,I_2=2nk$
$\Rightarrow I_1+n{I}_2=4n^{2}k$