Question

Let $f : R \rightarrow R$ 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 $I _{1}=\int\limits_{0}^{4 nk } f ( x ) dx$ and $I_{2}=\int\limits_{-k}^{3 k} f(x) d x$, then

• A. $I _{1}+2 I _{2}=4 nk$
• B. $I _{1}+2 I _{2}=2 nk$
• C. $I _{1}+ nI _{2}=4 n ^{2} k$
• D. $I _{1}+ nI _{2}=6 n ^{2} k$

Answer 1

Answer: (C)

$ f ( x )+ f ( x + k )= n $
$\Rightarrow f ( x )= f ( x +2 k )$
$f ( x )$ is periodic with period $2 k$
$I_{1}=\int\limits_{0}^{4 n k} f(x) d x=2 n \int\limits_{0}^{2 k} f(x) d x$
$I_{2}=\int\limits_{-k}^{3 k} f(x) d x=2 \int_{0}^{2 k} f(x) d x$
Now,
$f ( x )+ f ( x + k )= n$
$\Rightarrow \int\limits_{0}^{ k } f ( x ) dx +\int\limits_{0}^{ k } f ( x + k ) dx = nk$
$\Rightarrow \int\limits_{0}^{ k } f ( x ) dx +\int\limits_{ k }^{2 k } f ( x ) dx = nk$
$\Rightarrow \int\limits_{0}^{2 k } f ( x ) dx = nk$
$\Rightarrow I _{1}=2 n ^{2} k , I _{2}=2 nk$
$\Rightarrow I _{1}+ nI _{2}=4 n ^{2} k$