GATE CS 2024 Set 2 Q6 — Maxima and Minima, Mean Value Theorem, Integration

We are given the functional equation:

$$f(x) = 1 - f(2-x)$$

We need to evaluate the definite integral:

$$I = \int_{0}^{2} f(x) \, dx$$

Using the standard property of definite integrals, namely:

$$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$$

For our limits $a = 0$ and $b = 2$, we have $a+b-x = 2-x$. Applying this property gives:

$$I = \int_{0}^{2} f(2-x) \, dx$$

From the given functional equation, we can express $f(2-x)$ as:

$$f(2-x) = 1 - f(x)$$

Substituting this back into the integral equation:

$$I = \int_{0}^{2} (1 - f(x)) \, dx$$

$$I = \int_{0}^{2} 1 \, dx - \int_{0}^{2} f(x) \, dx$$

$$I = [x]_{0}^{2} - I$$

$$I = 2 - I$$

$$2I = 2 \implies I = 1$$

Thus, the value of the integral is $1$, which corresponds to option B.