JEE Main 2022 · 28 Jun · Shift 2 · Mathematics Q8

← JEE Main 2022 · 28 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 8 of 90

Let $f:\mathbf R\to\mathbf R$ be a continuous function satisfying $f(x)+f(x+k)=n$ , for all $x\in\mathbf R$ 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 :

I_1+2I_2=4nk

I_1+2I_2=2nk

I_1+nI_2=4n^2k

I_1+nI_2=6n^2k

Answer and worked solution

Official answer: C

Period. $f(x+2k)=n-f(x+k)=f(x)$ , and over one period $\int_0^{2k}f=\int_0^k\left(f(x)+f(x+k)\right)dx=nk$ .

So $I_1=2n\cdot nk=2n^2k$ and $I_2=2nk$ : $I_1+nI_2=4n^2k$ .

(C)

Topics:Properties of definite integrals

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JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 8