Let $f:(0, \pi) \to \mathbb{R}$ be a twice differentiable function such that $\lim_{t \to x} \frac{f(x)\sin t - f(t)\sin x}{t - x} = \sin^{2} x$ for all $x \in (0, \pi)$. If $f\left(\frac{\pi}{6}\right) = -\frac{\pi}{12}$, then which of the following statement(s) is (are) TRUE?

$\lim_{t \to x} \dfrac{f(x)\sin t - f(t)\sin x}{t - x} = \sin^{2} x, \qquad x \in (0,\pi)$ $f\!\left(\tfrac{\pi}{6}\right) = -\tfrac{\pi}{12}$ $\text{Find } f(x), \text{ then test (1)–(4)}$ $F(t) = f(x)\sin t - f(t)\sin x, \qquad F(x) = 0$ $\lim_{t \to x}\dfrac{F(t) - F(x)}{t - x} = F'(x)$ $F'(x) = f(x)\cos x - f'(x)\sin x = \sin^{2} x$ $\dfrac{f(x)\cos x - f'(x)\sin x}{\sin^{2} x} = 1$ $\dfrac{d}{dx}\!\left(\dfrac{f(x)}{\sin x}\right) = -\dfrac{f(x)\cos x - f'(x)\sin x}{\sin^{2} x} = -1$ $\dfrac{f(x)}{\sin x} = -x + C \Rightarrow f(x) = (C - x)\sin x$ $f\!\left(\tfrac{\pi}{6}\right) = \left(C - \tfrac{\pi}{6}\right)\cdot\tfrac{1}{2} = -\tfrac{\pi}{12} \Rightarrow C = 0$ $\boxed{f(x) = -x\sin x}$ $f\!\left(\tfrac{\pi}{4}\right) = -\dfrac{\pi}{4}\cdot\dfrac{1}{\sqrt{2}} = -\dfrac{\pi}{4\sqrt{2}}$ $\text{but statement says } +\dfrac{\pi}{4\sqrt{2}} \Rightarrow (1) \text{ FALSE}$ $f(x) < \dfrac{x^{4}}{6} - x^{2} \iff x\!\left(\sin x - x + \dfrac{x^{3}}{6}\right) > 0$ $\sin x > x - \dfrac{x^{3}}{6}\ (x>0) \Rightarrow (2) \text{ TRUE}$ $f'(x) = -\sin x - x\cos x$ $f'(\tfrac{\pi}{2}) = -1 < 0, \quad f'(\pi) = \pi > 0$ $\Rightarrow \exists\,\alpha \in (\tfrac{\pi}{2}, \pi) \text{ with } f'(\alpha) = 0 \Rightarrow (3) \text{ TRUE}$ $f''(x) = x\sin x - 2\cos x$ $f''(\tfrac{\pi}{2}) = \tfrac{\pi}{2}, \quad f(\tfrac{\pi}{2}) = -\tfrac{\pi}{2} \Rightarrow f''(\tfrac{\pi}{2}) + f(\tfrac{\pi}{2}) = 0$ $\Rightarrow (4) \text{ TRUE}$ $\boxed{\text{TRUE statements: } (2), (3), (4)}$ $f(x) = -x\sin x; \quad (1) \text{ false}, \quad (2),(3),(4) \text{ true}$