<p>If the system of equations \(2x - y + z = 0\), \(x - 2y + z = 0\) 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 coefficient matrix must be singular (determinant = 0). This determines 'a', and then use the functional equation f(5+x) + f(x) = 2 to establish symmetry about x = 2.5 for evaluating the definite integral.
<p><strong>Step 1: Find the value of a</strong></p><p>For infinitely many solutions, the coefficient matrix determinant must equal zero:</p><p>$$\begin{vmatrix} 2 & -1 & 1 \\ 1 & -2 & 1 \\ a & -1 & 2 \end{vmatrix} = 0$$</p><p>Expanding: $2(-4+1) + 1(2-a) + 1(-1+2a) = 0$</p><p>$-6 + 2 - a - 1 + 2a = 0$</p><p>$a - 5 = 0 \Rightarrow a = 5$</p><p><strong>Step 2: Use the functional equation</strong></p><p>Given: $f(5+x) + f(x) = 2$ for all $x \in \mathbb{R}$</p><p>This means $f(5+x) = 2 - f(x)$, showing that $f$ is symmetric about the point $(2.5, 1)$</p><p><strong>Step 3: Evaluate the integral</strong></p><p>We need: $\displaystyle\int_0^{-10} f(x)\,dx$ (since $-2a = -10$)</p><p>Substitute $u = 5 - x$ in $\displaystyle\int_0^{5} f(x)\,dx$:</p><p>$$\displaystyle\int_0^{5} f(x)\,dx = \int_5^{0} f(5-u)(-du) = \int_0^{5} f(5-u)\,du$$</p><p>Using $f(5-u) = 2 - f(u)$:</p><p>$$\displaystyle\int_0^{5} f(x)\,dx = \int_0^{5} [2-f(x)]\,dx = 10 - \int_0^{5} f(x)\,dx$$</p><p>$$2\int_0^{5} f(x)\,dx = 10 \Rightarrow \int_0^{5} f(x)\,dx = 5$$</p><p>By similar symmetry: $\displaystyle\int_5^{10} f(x)\,dx = 5$</p><p>Therefore: $\displaystyle\int_0^{-10} f(x)\,dx = -\int_{-10}^{0} f(x)\,dx = -10$ (using periodicity/symmetry properties)</p><p>∴ Answer: <strong>A</strong></p>
Correct Answer: A