Question:

Step 1: Expand the determinant $f(x)$. The function $f(x)$ is given by the determinant: $$f(x) = \begin{vmatrix} \sin x & \cos x & \tan x \\ x^3 & x^2 & x \\ 2x & 1 & 0\end{vmatrix}$$ Expanding the determinant along the third row yields: $$f(x) = 2x(\cos x \cdot x - \tan x \cdot x^2) - 1(\sin x \cdot x - \tan x \cdot x^3) + 0$$ $$f(x) = 2x^2\cos x - 2x^3\tan x - x\sin x + x^3\tan x$$ $$f(x) = 2x^2\cos x - x^3\tan x - x\sin x$$ Step 2: Determine if $f(x)$ is an odd function. To determine the parity of $f(x)$, we evaluate $f(-x)$: $$f(-x) = 2(-x)^2\cos(-x) - (-x)^3\tan(-x) - (-x)\sin(-x)$$ Using the properties $\cos(-x) = \cos x$, $\sin(-x) = -\sin x$, and $\tan(-x) = -\tan x$: $$f(-x) = 2x^2\cos x - (-x^3)(-\tan x) - (-x)(-\sin x)$$ $$f(-x) = 2x^2\cos x - x^3\tan x - x\sin x$$ Since $f(-x) = f(x)$, the function $f(x)$ is an even function. Therefore, the statement "$f(x)$ is an odd function" is incorrect. Step 3: Evaluate $\lim_{x\to 0}\dfrac{f(x)}{x^2}$. We use the Maclaurin series expansions for $x \to 0$: $\cos x = 1 - \frac{x^2}{2!} + O(x^4)$ $\tan x = x + \frac{x^3}{3} + O(x^5)$ $\sin x = x - \frac{x^3}{3!} + O(x^5)$ Substitute these into the expression for $f(x)$: $$f(x) = 2x^2\left(1 - \frac{x^2}{2} + O(x^4)\right) - x^3\left(x + \frac{x^3}{3} + O(x^5)\right) - x\left(x - \frac{x^3}{6} + O(x^5)\right)$$ $$f(x) = (2x^2 - x^4 + O(x^6)) - (x^4 + \frac{x^6}{3} + O(x^8)) - (x^2 - \frac{x^4}{6} + O(x^6))$$ $$f(x) = 2x^2 - x^4 - x^4 - x^2 + \frac{x^4}{6} + O(x^6)$$ $$f(x) = (2x^2 - x^2) + (-x^4 - x^4 + \frac{x^4}{6}) + O(x^6)$$ $$f(x) = x^2 - \frac{11}{6}x^4 + O(x^6)$$ Now, we evaluate the limit: $$\lim_{x\to 0}\frac{f(x)}{x^2} = \lim_{x\to 0}\frac{x^2 - \frac{11}{6}x^4 + O(x^6)}{x^2}$$ $$\lim_{x\to 0}\frac{f(x)}{x^2} = \lim_{x\to 0}\left(1 - \frac{11}{6}x^2 + O(x^4)\right)$$ $$\lim_{x\to 0}\frac{f(x)}{x^2} = 1$$ Therefore, the statements "$\lim_{x\to 0}\dfrac{f(x)}{x^2} = 0$" and "$\lim_{x\to 0}\dfrac{f(x)}{x^2} = -1$" are incorrect. Step 4: Evaluate $f'(0)$. Since $f(x)$ is an even function and is differentiable, its derivative $f'(x)$ must be an odd function. For any differentiable odd function $g(x)$, it holds that $g(0) = 0$. Thus, $f'(0) = 0$. Alternatively, using the series expansion of $f(x)$ from Step 3: $f(x) = x^2 - \frac{11}{6}x^4 + O(x^6)$ Differentiating with respect to $x$: $f'(x) = \frac{d}{dx}\left(x^2 - \frac{11}{6}x^4 + O(x^6)\right)$ $f'(x) = 2x - \frac{11}{6}(4x^3) + O(x^5)$ $f'(x) = 2x - \frac{22}{3}x^3 + O(x^5)$ Substituting $x=0$: $f'(0) = 2(0) - \frac{22}{3}(0)^3 + O(0) = 0$ Therefore, the statement "$f'(0) = 0$" is correct.