For any real number $$x$$, let $$[x]$$ denote the largest integer less than or equal to $$x$$. Let $$f$$ be a real-valued function defined on the interval $$[-10, 10]$$ by
$$f(x) = \begin{cases} x - [x], & \text{if } [x] \text{ is odd} \\ 1 + [x] - x, & \text{if } [x] \text{ is even} \end{cases}$$
Then, the value of $$\dfrac{\pi^2}{10} \displaystyle\int_{-10}^{10} f(x) \cos \pi x \, dx$$ is

We need to find $$\dfrac{\pi^2}{10} \displaystyle\int_{-10}^{10} f(x) \cos \pi x \, dx$$ where $$f(x) = \begin{cases} x - [x], & \text{if } [x] \text{ is odd} \\ 1 + [x] - x, & \text{if } [x] \text{ is even} \end{cases}$$.

Note that $$\{x\} = x - [x]$$ is the fractional part of $$x$$. So:

When $$[x]$$ is odd: $$f(x) = \{x\}$$

When $$[x]$$ is even: $$f(x) = 1 - \{x\}$$

This is a triangular wave function with period 2. On each interval $$[n, n+1)$$, $$f$$ is a sawtooth going from 0 to 1 (when $$n$$ is odd) or from 1 to 0 (when $$n$$ is even). The function is periodic with period 2 and symmetric.

Since $$f(x)$$ has period 2, and $$\cos \pi x$$ also has period 2, the integrand $$f(x)\cos \pi x$$ has period 2. The interval $$[-10, 10]$$ has length 20, which is exactly 10 periods.

$$\int_{-10}^{10} f(x) \cos \pi x \, dx = 10 \int_0^2 f(x) \cos \pi x \, dx$$

On $$[0, 1)$$: $$[x] = 0$$ (even), so $$f(x) = 1 - x$$.

On $$[1, 2)$$: $$[x] = 1$$ (odd), so $$f(x) = x - 1$$.

$$\int_0^2 f(x)\cos \pi x\, dx = \int_0^1 (1 - x)\cos \pi x\, dx + \int_1^2 (x - 1)\cos \pi x\, dx$$

Using integration by parts with $$u = 1 - x$$, $$dv = \cos \pi x\, dx$$:

$$I_1 = \left[\dfrac{(1 - x)\sin \pi x}{\pi}\right]_0^1 + \int_0^1 \dfrac{\sin \pi x}{\pi}\, dx$$ $$= 0 + \dfrac{1}{\pi}\left[-\dfrac{\cos \pi x}{\pi}\right]_0^1 = \dfrac{1}{\pi^2}(-\cos \pi + \cos 0) = \dfrac{1}{\pi^2}(1 + 1) = \dfrac{2}{\pi^2}$$

Substitute $$t = x - 1$$, so $$x = t + 1$$, $$dx = dt$$, limits: $$t$$ from 0 to 1:

$$I_2 = \int_0^1 t \cos \pi(t + 1)\, dt = -\int_0^1 t \cos \pi t\, dt$$

since $$\cos \pi(t+1) = -\cos \pi t$$.

Using integration by parts with $$u = t$$, $$dv = \cos \pi t\, dt$$:

$$\int_0^1 t\cos \pi t\, dt = \left[\dfrac{t \sin \pi t}{\pi}\right]_0^1 - \int_0^1 \dfrac{\sin \pi t}{\pi}\, dt = 0 - \dfrac{1}{\pi}\left[-\dfrac{\cos \pi t}{\pi}\right]_0^1$$ $$= -\dfrac{1}{\pi^2}(-\cos \pi + \cos 0) = -\dfrac{1}{\pi^2}(2) = -\dfrac{2}{\pi^2}$$

So $$I_2 = -\left(-\dfrac{2}{\pi^2}\right) = \dfrac{2}{\pi^2}$$.

$$\int_0^2 f(x)\cos \pi x\, dx = \dfrac{2}{\pi^2} + \dfrac{2}{\pi^2} = \dfrac{4}{\pi^2}$$ $$\int_{-10}^{10} f(x)\cos \pi x\, dx = 10 \cdot \dfrac{4}{\pi^2} = \dfrac{40}{\pi^2}$$ $$\dfrac{\pi^2}{10} \cdot \dfrac{40}{\pi^2} = 4$$

The answer is Option A: $$4$$.