Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade None
Question:
If $f(x) = \begin{vmatrix} a^{-1} & e^{x \ln a} & x^2 \\ a^{-3x} & e^{3x \ln a} & x^4 \\ a^{-5x} & e^{5x \ln a} & 1 \end{vmatrix}$ then:
Graph of $f(x)$ is symmetric about origin
Graph of $f(x)$ is symmetric about $y$ axis
$f''(0) = 0$
$f(x) \ln\left(\frac{a-x}{a+x}\right)$ is an even function
Step-by-Step Solution
Key Concept: The determinant structure creates an odd function when each row is multiplied by odd powers, and combining odd/even functions preserves parity properties.
First, rewrite the determinant using $e^{x\ln a} = a^x$. Expanding the determinant with respect to the third column: $f(x) = x^2(a^{-3x} \cdot e^{5x\ln a} - a^{-5x} \cdot e^{3x\ln a}) - x^4(a^{-1} \cdot e^{5x\ln a} - a^{-5x}) + (a^{-1} \cdot e^{3x\ln a} - a^{-3x})$. Simplifying: $f(x) = x^2(a^{2x} - a^{-2x}) - x^4(a^{4x} - a^{-4x}) + (a^{3x} - a^{-3x})$. Since each term is an odd function of $x$, we have $f(-x) = -f(x)$, so **$f$ is odd** (Option 1). For $f''(0)$: since $f$ is odd, $f(0) = 0$ and $f'(0) = 0$, making $f''(0) = 0$ (Option 3). For Option 4: $g(x) = f(x)\ln\left(\frac{a-x}{a+x}\right)$ is a product of an odd function and an even function (since $\ln\left(\frac{a-x}{a+x}\right) = -\ln\left(\frac{a+x}{a-x}\right)$ is even), making $g(x)$ **even** (Option 4).
Correct Answer: 1,3,4