Let f:RR be a continuous function satisfying f(x)+f(x+k)=n for all xR, where k0 and n is …

Question

Let $f:\mathbb{R}	o\mathbb{R}$ be a continuous function satisfying $f(x)+f(x+k)=n$ for all $x\in\mathbb{R}$, where $k>0$ and $n$ is a positive integer. If $I_1=\displaystyle\int_0^{4nk}f(x)\,dx$ and $I_2=\displaystyle\int_{-k}^{3k}f(x)\,dx$, then:

Accepted Answer

$f: R \rightarrow R$ and $f(x)+f(x+k)=n \quad \forall x \in R$

$$
\begin{aligned}
&x \rightarrow x+k \\
&f(x+k)+f(x+2 k)=n \\
&\therefore \quad f(x+2 k)=f(x)
\end{aligned}
$$

So, period of $f(x)$ is $2 k$

$$
\begin{aligned}
&\text { Now, } I_{1}=\int_{0}^{4 n k} f(x) d x = 2 n \int_{0}^{2 k} f(x) d x \\
&=2 n\left[\int_{0}^{k} f(x) d x+\int_{k}^{2 k} f(x) d x\right] \\
&x=t+k \Rightarrow d x=d t \text { (in second integral) } \\
&=2 n\left[\int_{0}^{k} f(x) d x+\int_{0}^{k} f(t+k) d t\right] \\
&=2 n^{2} k
\end{aligned}
$$

Now, $I_2=\int_{-k}^{3 k} f(x) d x=2 \int_{0}^{2 k} f(x) d x$

$I_{2}=2(n k)$

$\therefore \quad l_{1}+n l_{2}=4 n^{2} k$

Solutions · 1