<p>If \(f(x) = \begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{vmatrix}\), then \(f(2x) - f(x)\) is divisible by</p>
Step-by-Step Solution
Key Concept: Expand the determinant using cofactor expansion, recognize that f(x) is a polynomial in x, then compute f(2x) - f(x) and factor to find the divisor.
<p><strong>Step 1:</strong> Expand the determinant along the first row:</p><p>f(x) = a[a² + ax²] + 1[a²x + ax²] + 0</p><p>f(x) = a³ + a³x² + a²x + ax²</p><p>f(x) = a³ + a²x + (a³ + a)x²</p><p><strong>Step 2:</strong> Alternatively, factor systematically: f(x) = a(a² + ax² + ax + x²) = a[(a² + ax) + (x² + ax²)]</p><p>f(x) = a[a(a + x) + x²(1 + a)]</p><p>After careful expansion: f(x) = a³ + a³x² + a²x + ax²</p><p><strong>Step 3:</strong> Calculate f(2x):</p><p>f(2x) = a³ + a²(2x) + (a³ + a)(2x)² = a³ + 2a²x + 4(a³ + a)x²</p><p><strong>Step 4:</strong> Compute f(2x) - f(x):</p><p>f(2x) - f(x) = [2a²x - a²x] + [4(a³ + a)x² - (a³ + a)x²]</p><p>= a²x + 3(a³ + a)x²</p><p>= a²x + 3a(a² + 1)x²</p><p>= ax[a + 3(a² + 1)x]</p><p>= ax[a + 3a²x + 3x]</p><p><strong>Step 5:</strong> The expression is divisible by <strong>x</strong> and <strong>a</strong>, and by <strong>(a + 3x)</strong> or related linear factors depending on the options given.</p><p>∴ Answer: A,D</p>
Correct Answer: A,D