A calculus problem by Sabhrant Sachan

$\begin{aligned} 1 & = \lim_{x \to 0} \frac{x(1+a\cos x) - b\sin x}{x^3} \\ & = \lim_{x \to 0} \left(\frac{1+a\color{#3D99F6}{\cos x}}{x^2} - \frac{b\color{#3D99F6}{\sin x}}{x^3} \right) \quad \quad \small \color{#3D99F6}{\text{Using Maclaurin series}} \\ & = \lim_{x \to 0} \left(\frac{1}{x^2}\left[1 + a \left(\color{#3D99F6}{1-\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...}\right) \right] - \frac{b}{x^3}\left[\color{#3D99F6}{x-\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...} \right] \right) \\ & = \lim_{x \to 0} \left(\frac{1}{x^2} + \frac{a}{x^2} - \frac{a}{2!} + \color{#D61F06}{\frac{ax^2}{4!} - \frac{ax^4}{6!} + ... } - \frac{b}{x^2} + \frac{b}{3!} - \color{#D61F06}{\frac{x^2}{5!} + \frac{x^4}{7!} - ... } \right) \quad \quad \small \color{#D61F06}{\text{Red terms}\to 0 \text{ as }x \to 0} \\ & = \frac{1}{x^2} + \frac{a}{x^2} - \frac{b}{x^2} - \frac{a}{2!} + \frac{b}{3!} \end{aligned}$

Equating coefficients, we have:

$\begin{cases} 1 + a - b = 0 & \implies b = 1 + a \\ - \dfrac{a}{2} + \dfrac{b}{6} = 1 & \implies - 3a + 1 + a = 6 \implies a = -\dfrac{5}{2} \implies b = - \dfrac{3}{2} \end{cases}$

$\implies \lfloor a^4b^4 \rfloor = \lfloor (-2.5)^4(-1.5)^4 \rfloor = \lfloor 197.753 \rfloor = 197 \implies \boxed{\text{None of these choices}}$