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GATE CSE 2024 Session 2 Q16: Integral of f over [0, 2] when f(x) = 1 − f(2 − x)

Question

Let f(x)f(x) be a continuous function from R\mathbb{R} to R\mathbb{R} such that

f(x)=1−f(2−x)f(x) = 1 - f(2 - x)

Which one of the following options is the CORRECT value of ∫02f(x) dx\displaystyle\int_0^2 f(x)\,dx?

Try it: multiple choice: exactly one correct option

Select one option
Wrong answer costs 1/3 mark in GATE.
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Answer: B (official GATE 2024 final answer key)

Concept used: The substitution x→2−xx \to 2 - x leaves ∫02\int_0^2 unchanged.

First step: Let I=∫02f(x) dx=∫02f(2−x) dxI = \int_0^2 f(x)\,dx = \int_0^2 f(2-x)\,dx; adding gives 2I=∫02[f(x)+f(2−x)] dx=∫021 dx2I = \int_0^2 [f(x) + f(2-x)]\,dx = \int_0^2 1\,dx.

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Source: official GATE 2024 Session 2 CS question paper and final answer key, organised by IISc Bangalore. Question paper · Answer key