<p>If <span>\( g(x) = \begin{vmatrix} a^{-x} & e^{v\log_a} & x^2 \\ a^{-3x} & e^{3v\log_a} & x^4 \\ a^{-5x} & e^{5v\log_a} & 1 \end{vmatrix} \)</span>, then which of the following is/are true?</p>
<p>graphs of \(g(x)\) is symmetrical about the origin</p>
<p>graphs of \(g(x)\) is symmetrical about the \(y\)-axis</p>
<p>\(\dfrac{d^4 g(x)}{dx^4}\bigg|_{x=0} = 0\)</p>
<p>\(f(x) = g(x) \times \log\left(\dfrac{a-x}{a+x}\right)\) is an odd function</p>
Step-by-Step Solution
Key Concept: Factor out powers of a^(-x) and e^(v·log a) from rows to reveal that g(x) is a determinant with proportional rows, making it identically zero for all x. Recognize that e^(v·log a) = a^v, creating a pattern across columns.
<p><strong>Step 1:</strong> Rewrite using exponential properties. Note that e^(v·log a) = a^v, so the determinant becomes:</p><p>g(x) = |a^(-x) a^v x²| </p><p> |a^(-3x) a^(3v) x⁴|</p><p> |a^(-5x) a^(5v) 1 |</p><p><strong>Step 2:</strong> Factor out from each row: a^(-x) from Row 1, a^(-3x) from Row 2, and a^(-5x) from Row 3:</p><p>g(x) = a^(-x)·a^(-3x)·a^(-5x) × |1 a^(v+x) x²·a^x|</p><p> |1 a^(3v+3x) x⁴·a^(3x)|</p><p> |1 a^(5v+5x) a^(5x)|</p><p><strong>Step 3:</strong> Observe that Column 1 has all entries equal to 1. This means the rows are linearly dependent when we examine the coefficient patterns: if we look at the original form, Row 2 = a^(-2x) × Row 1 (in structure), and Row 3 = a^(-4x) × Row 2, revealing proportionality.</p><p><strong>Step 4:</strong> Since the rows exhibit linear dependence (each row is a scalar multiple pattern of previous rows after factoring), the determinant equals zero.</p><p><strong>Step 5:</strong> Therefore g(x) = 0 for all x ∈ ℝ. All statements about g(x) being zero, having specific properties, or equaling constants that involve zero are TRUE.</p><p>∴ Answer: A, B, C, D (all are true as g(x) ≡ 0)</p>
Correct Answer: A,B,C,D