Question:

Step 1: Determine the value of $a$ for which the system of equations has infinitely many solutions. A system of linear equations has infinitely many solutions if and only if the determinant of its coefficient matrix is zero. The coefficient matrix for the given system is: $$M = \begin{pmatrix} 2 & -1 & 1 \\ 1 & -2 & 1 \\ a & -1 & 2 \end{pmatrix}$$ Setting the determinant of $M$ to zero: $$\det(M) = 2((-2)(2) - (1)(-1)) - (-1)((1)(2) - (1)(a)) + 1((1)(-1) - (-2)(a)) = 0$$ $$2(-4 + 1) + (2 - a) + (-1 + 2a) = 0$$ $$2(-3) + 2 - a - 1 + 2a = 0$$ $$-6 + 2 - 1 + a = 0$$ $$-5 + a = 0$$ $$a = 5$$ Step 2: Analyze the functional equation $f(5+x) + f(x) = 2$. The given functional equation is $f(5+x) + f(x) = 2$. This can be rewritten as $f(x+5) = 2 - f(x)$. Now, consider $f(x+10)$: $$f(x+10) = f((x+5)+5) = 2 - f(x+5)$$ Substitute $f(x+5) = 2 - f(x)$ into the expression for $f(x+10)$: $$f(x+10) = 2 - (2 - f(x)) = f(x)$$ Thus, the function $f(x)$ is periodic with a period of $10$. For a continuous function $f(x)$ satisfying $f(c+x) + f(x) = 2k$, the integral over one period $2c$ is given by $\int_0^{2c} f(x)\,dx = 2ck$. In this case, $c=5$ and $k=1$. Therefore, $$\int_0^{10} f(x)\,dx = 2 \cdot 5 \cdot 1 = 10$$ Step 3: Compute the definite integral $\displaystyle\int_0^{-2a} f(x)\,dx$. Substitute the value of $a=5$ into the integral limit: $$\int_0^{-2a} f(x)\,dx = \int_0^{-2(5)} f(x)\,dx = \int_0^{-10} f(x)\,dx$$ Using the property of definite integrals $\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$: $$\int_0^{-10} f(x)\,dx = -\int_{-10}^0 f(x)\,dx$$ Since $f(x)$ is periodic with period $10$, the integral over any interval of length $10$ is the same. Specifically, $\int_{-10}^0 f(x)\,dx = \int_0^{10} f(x)\,dx$. Therefore, $$\int_0^{-10} f(x)\,dx = -\int_0^{10} f(x)\,dx$$ From Step 2, we know $\int_0^{10} f(x)\,dx = 10$. So, $$\int_0^{-10} f(x)\,dx = -10$$