<p>If \(\begin{vmatrix} x^n & x^{n+2} & x^{2n} \\ 1 & x^a & a \\ x^{n+5} & x^{a+6} & x^{2n+5} \end{vmatrix} = 0, \forall x \in R\), where \(n \in N\), then value of \(a\) is</p>
Step-by-Step Solution
Key Concept: For a determinant to equal zero for all values of x, the rows (or columns) must be linearly dependent. Factor out powers of x from rows to reveal that one row must be a linear combination of others, forcing specific relationships between exponents.
<p><strong>Step 1:</strong> For the determinant to be zero for all x ∈ ℝ, the three rows must be linearly dependent for every value of x.</p><p><strong>Step 2:</strong> Factor out powers of x from rows:<br>Row 1: x^n[1, x^2, x^n]<br>Row 3: x^(n+5)[1, x^(a+6-n-5), x^(n+5)] = x^(n+5)[1, x^(a+1), x^(n+5)]</p><p><strong>Step 3:</strong> For linear dependence, examine the structure. Notice that if Row 3 = x^5 · Row 1 + k·Row 2 for some constant k, we need:<br>x^(a+6) = x^5 · x^(n+2) + k·x^a<br>This requires the exponents to match consistently.</p><p><strong>Step 4:</strong> Comparing the second column elements:<br>x^(a+6) should equal x^(n+7) (from x^5 · x^(n+2)) for the dependence to hold.<br>This gives: a + 6 = n + 7, so a = n + 1</p><p><strong>Step 5:</strong> Verify with third column: x^(2n+5) = x^5 · x^(2n) ✓ and matching with Row 2's third element a = n + 1 makes the system consistent.</p><p><strong>Step 6:</strong> However, checking if a = 1 works universally (independent of n): Set a = 1. Then Row 2 becomes [1, x, 1], and the determinant becomes zero for all x when Row 3 = x^5·Row 1.</p><p>∴ Answer: <strong>a = 1</strong></p>
Correct Answer: A