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

(C) $\mathrm{I}_{1}+\mathrm{nI}_{2}=4 \mathrm{n}^{2} \mathrm{k}$

(A) $\mathrm{I}_{1}+2 \mathrm{I}_{2}=4 \mathrm{nk}$

(B) $\mathrm{I}_{1}+2 \mathrm{I}_{2}=2 \mathrm{nk}$

(D) $\mathrm{I}_{1}+\mathrm{nI}_{2}=6 \mathrm{n}^{2} \mathrm{k}$

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Definite Integrals

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} = \int_{0}^{4 n k} f (x) dx

and

I_{2} = \int_{- k}^{3 k} f (x) d x

, then

Answer: (C)

From

f (x) + f (x + k) = n

, we get

f (x) = f (x + 2 k)

, so

f

is periodic with period

2 k

.

Integrating

f (x) + f (x + k) = n

from 0 to

k

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

\int_{0}^{k} f (x) d x + \int_{k}^{2 k} f (x) d x = nk

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

Therefore:

I_{1} = \int_{0}^{4 nk} f (x) d x = 2 n \int_{0}^{2 k} f (x) d x = 2 n \cdot nk = 2 n^{2} k

I_{2} = \int_{- k}^{3 k} f (x) d x = 2 \int_{0}^{2 k} f (x) d x = 2 nk

Checking the options:

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

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option