Question

If \( f(x) = \begin{vmatrix} 1 & 6 + x & 36 + x^2 \\ 0 & x - 3 & 3x^2 - 27 \\ 0 & 2x - 4 & 8x^2 - 32 \end{vmatrix} \), then \( \lim\limits_{x \to 1} \frac{f(x)}{f(-x)} \) is:

• A. 2
• B. -1
• C. 0
• D. 1

Hint:

Use determinant properties and row operations to simplify the matrix. If two rows become equal, the determinant is zero.

Answer 1

The Correct Option is C

Given the function: \[ f(x) = \begin{vmatrix} 1 & 6 + x & 36 + x^2 \\ 0 & x - 3 & 3x^2 - 27 \\ 0 & 2x - 4 & 8x^2 - 32 \end{vmatrix} \] We are asked to evaluate: \[ \lim_{x \to 1} \frac{f(x)}{f(-x)} \] Substituting \(x = 1\), we get: For the second row: \(x - 3 = -2\), \(3x^2 - 27 = 3 - 27 = -24\) For the third row: \(2x - 4 = -2\), \(8x^2 - 32 = 8 - 32 = -24\) The second and third rows become linearly dependent: \[ (0, -2, -24) \text{ and } (0, -2, -24) \] Thus, the determinant becomes zero at \(x = 1\). Similarly, at \(x = -1\), you also get linear dependence in the rows. Hence, \[ \lim_{x \to 1} \frac{f(x)}{f(-x)} = \frac{0}{0} \] This is an indeterminate form, requiring further expansion or simplification. Upon evaluating, we find: \[ \lim_{x \to 1} \frac{f(x)}{f(-x)} = 0 \]