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 I₁ = ∫₀⁴ⁿᵏ f(x)dx and I₂ = ∫₋ₖ³ᵏ f(x)dx, then :

Solution

Maths sir · 4 steps

Shift x by k once more and subtract: f repeats every 2k, so both integrals are counts of periods.

1. Find the period. Replace x by x + k to get f(x+k) + f(x+2k) = n. Subtract this from f(x) + f(x+k) = n to get f(x+2k) = f(x), so f is periodic with period 2k.
2. One period. Split ∫₀²ᵏ f(x) dx at k and put x = t + k in the second half. The two halves combine into one integral over [0, k].

   ∫₀²ᵏ f(x) dx = ∫₀ᵏ [ f(x) + f(x+k) ] dx = ∫₀ᵏ n dx = nk

3. Count periods. For a periodic function, every interval of length 2k gives the same integral nk, wherever it starts. [0, 4nk] holds 2n periods, so I₁ = 2n × nk = 2n²k. [-k, 3k] has length 4k, which is 2 periods, so I₂ = 2nk.
4. So. I₁ + nI₂ = 2n²k + 2n²k = 4n²k, which is option (c).

The idea to keep. When a relation links f(x) and f(x+k), shift once more and compare to find the period. Then a long integral is just the number of periods times the integral over one period.

Answer from the NTA final answer key. Solution by PrepWiser.