GATE CS 2024 Set 2 — Question 16

Given the functional equation:$$f(x) = 1 - f(2 - x) \implies f(x) + f(2 - x) = 1$$Let $I = \int_0^2 f(x) dx$.
Using the property of definite integrals $\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$, we substitute $x$ with $2 - x$:$$I = \int_0^2 f(2 - x) dx$$Adding the two expressions for $I$:$$2I = \int_0^2 f(x) dx + \int_0^2 f(2 - x) dx$$$$2I = \int_0^2 (f(x) + f(2 - x)) dx$$Substituting $f(x) + f(2 - x) = 1$:$$2I = \int_0^2 1 \, dx = [x]_0^2 = 2 - 0 = 2$$$$2I = 2 \implies I = 1$$Thus, the value of the integral is 1.