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

The Correct Option is C

To solve the problem, we are given a continuous function \(f: \mathbb{R} \rightarrow \mathbb{R}\) that satisfies the condition:

\(f(x) + f(x+k) = n\) for all \(x \in \mathbb{R}\), where \(k > 0\) and \(n\) is a positive integer.

We need to consider two integrals:
\(l_1 = \int_{0}^{4nk} f(x) \, dx\) and \(l_2 = \int_{-k}^{3k} f(x) \, dx\).

Let's explore what these integrals represent:

1. Given \(f(x) + f(x+k) = n,\) this relation suggests that the function \(f\) has a periodic or repetitive nature over intervals of length \(k\).
2. Consider evaluating the behavior of \(f\) over one such period, i.e., over the interval \([0, k]\). Over this interval, we can hypothesize a function structure that complies with \(f(x) + f(x+k) = n\).

Let's break down the computations for \(l_1\) and \(l_2\):

1. For \(l_1 = \int_{0}^{4nk} f(x) \, dx\):
   • Because the function satisfies \(f(x) + f(x+k) = n\), the function repeats over intervals of length \(k\).
   • The integral over such a complete set of repetitions (\(4n\) complete periods of \(k\)), should equal \(2nk\), multiplied by the periodic average of \(f\), which is \(\frac{nk}{2}\).
   • Hence, \(l_1 = 2n \times \frac{nk}{2} = 2n^2k\).
2. For \(l_2 = \int_{-k}^{3k} f(x) \, dx\):
   • Similarly, this interval also covers multiple periods, but starting at \(-k\).
   • A similar calculation as above yields the period and average structure yielding \(l_2 = n^2k\), but since it has fewer periods, we end this calculation with the recognition that integral balance over this minor length depends on the period alignment which resolves as 2 periodic contributions, thus, \(l_2 = 2nk\).

Combining these, the correct transformation appears as:

\(I_1 + nI_2 = 4n^2k\)

The integral \(l_1\) collects over a pure periodic progression ensuring a basic sum-distributive aggregation, where integral disruptions at the interval-specific periods map back periodic consistency, giving us the correct matchup to the answer \(I_1 + nI_2 = 4n^2k\).