Question

Let f : R → R be a continuous function satisfying f(x) + f(x + k) = n, for all x ∈ R where k > 0 and n is a positive integer. If
$l_1 = \int_{0}^{4nk} f(x) \, dx$ and $l_2 = \int_{-k}^{3k} f(x) \, dx$, then

• A. $I_1+2I_2=4nk$
• B. $I_1+2I_2=2nk$
• C. $I_1+nI_2=4n^2k$
• D. $l_1+nl_2=6n^2k$

Answer 1

Answer: (C)

$I_1+nI_2=4n^2k$

Answer 2

The correct answer is (C) : $I_1​+nI_2​=4n^2k$
f : R → R and f(x) + f(x + k) = n ∀ x ∈ R
x → x + k
f(x + k) + f(x + 2k) = n
∴ f(x + 2k) = f(x)
So, period of f(x) is 2k
Now,
$I_1 = \int_{0}^{4nk} f(x) \, dx = 2n \int_{0}^{2k} f(x) \, dx$
$= 2n \left[ \int_{0}^{k} f(x) \, dx + \int_{k}^{2k} f(x) \, dx \right]$
x = t + k ⇒ dx = dt (in second integral)
$2n \left[ \int_{0}^{k} f(x) \, dx + \int_{0}^{k} f(t+k) \, dt \right] = 2n^2k$
Now,
$I_2 = \int_{-k}^{3k} f(x) \, dx = 2 \int_{0}^{2k} f(x) \, dx$
$I_2=2(nk)$
Therefore , $I_1+nI_2=4n^2k$