<p><strong>549.</strong> If the system of equation \(2x - y + z = 0\), \(x - 2y + z = 0\) and \(ax - y + 2z = 0\) has infinitely many solutions and \(f(x)\) be a continuous function such that \(f(5+x) + f(x) = 2 \ \forall x \in R\), then \(\displaystyle\int_0^{-2a} f(x)\,dx\) is equal to:</p>
Step-by-Step Solution
Key Concept: For a homogeneous system to have infinitely many solutions, the determinant of coefficients must be zero. This constraint determines 'a', and the functional property f(5+x) + f(x) = 2 reveals that f is symmetric about x = 2.5, making the integral evaluable through symmetry arguments.
<p><strong>Step 1: Find 'a' from the determinant condition.</strong><br>For infinitely many solutions: <br>$$\begin{vmatrix} 2 & -1 & 1 \\ 1 & -2 & 1 \\ a & -1 & 2 \end{vmatrix} = 0$$<br>Expanding: 2(-4+1) + 1(2-a) + 1(-1+2a) = 0<br>-6 + 2 - a + 2a - 1 = 0<br>a = 5</p><p><strong>Step 2: Analyze the functional equation.</strong><br>Given: f(5+x) + f(x) = 2 for all x ∈ ℝ<br>This means f is symmetric about x = 2.5, and the average value is 1.</p><p><strong>Step 3: Evaluate the integral.</strong><br>We need: $$\int_0^{-10} f(x)\,dx$$<br>Substitute u = -x, so dx = -du:<br>$$\int_0^{-10} f(x)\,dx = -\int_0^{10} f(-u)\,du = \int_{10}^0 f(-u)\,du$$<br>From the functional equation with x → -x-5:<br>f(-x-5) + f(-x) = 2, so f(-u) + f(-u-5) = 2<br>Therefore: $$\int_0^{10} f(x)\,dx = \int_0^{10} [2 - f(5+x)]\,dx = 20 - \int_5^{15} f(x)\,dx = 20 - \int_0^{10} f(x)\,dx$$<br>This gives: $$\int_0^{10} f(x)\,dx = 10$$<br>Thus: $$\int_0^{-10} f(x)\,dx = -10$$</p><p>∴ Answer: A</p>
Correct Answer: A