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Let \mathrm{f: \mathbf{R} \rightarrow \mathbf{R}} be a continuous function satisfying \mathrm{f(x)+f(x+\mathrm{k})=\mathrm{n}}, for all \mathrm{x \in \mathbf{R}} where \mathrm{k}>0$ and $\mathrm{n} is a positive integer. If \mathrm{I}_{1}=\int_{0}^{4 \mathrm{nk}} f(x) \mathrm{d} x$ and $\mathrm{I}_{2}=\int_{-\mathrm{k}}^{3 \mathrm{k}} f(x) \mathrm{d} x, then :

Option: 1

\mathrm{I}_{1}+2 \mathrm{I}_{2}=4 \mathrm{nk}


Option: 2

\begin{aligned} &\mathrm{I}_{1}+2 \mathrm{I}_{2}=2 \mathrm{nk} \\ \end{aligned}


Option: 3

\mathrm{I}_{1}+\mathrm{nI}_{2}=4 \mathrm{n}^{2} \mathrm{k} \\


Option: 4

\mathrm{I}_{1}+\mathrm{nI}_{2}=6 \mathrm{n}^{2} \mathrm{k}


Answers (1)

best_answer

\mathrm{f(x)+f(x+k)=n} \\

\mathrm{\text { Put } x \rightarrow x+k} \\

\mathrm{f(x+k)+f(x+2 k)=n}

Subtract these equations

\mathrm{f(x)-f(x+2 k)=0} \\

\mathrm{\Rightarrow f(x+2 k)=f(x) }

\mathrm{f(x) } is periodic with period \mathrm{2k }

\mathrm{\text { Now } I_{1}=\int_{0}^{4 n k} f(x) d x=\int_{0}^{2 n(2 k)} f(x) d x=2 n \int_{0}^{2 k} f(x) d x}

\mathrm{\text { and } I_{2}=\int_{-k}^{3 k} f(x) d x=\int_{-k}^{4 k-k} f(x) d x=\int_{0}^{4k} f(x) d x}

\mathrm{=\int_{0}^{2 \cdot(2 k)} f(x) d x=2 \int_{0}^{2 k} f(x) d x}

\mathrm{\text { Now } \int_{0}^{2 k} f(x) d x=\int_{0}^{k} f(x) d x+\int_{k}^{2 k} f(x) d x}

\mathrm{\text { Put } x=t+k}    in second integral

\mathrm{=\int_{0}^{k} f(x) d x+\int_{0}^{k} f(t+k) d t} \\

\mathrm{=\int_{0}^{k}(f(x)+f(x+k)) d x}

\mathrm{=\int_{0}^{k} n d x} \\

\mathrm{=n k} \\

\mathrm{\therefore I_{1} =2 n^{2} k, \quad I_{2}=2 n k}

Checking options,

\mathrm{I_{1}+n I_{2}=2 n^{2} k+2 n^{2} k=4 n^{2} k}

Hence correct answer is option 3

Posted by

Gaurav

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