Question:

Step 1: Determine the domain of $f(x)$. For $\tan^{-1}\left(\frac{2x}{\sqrt{9-4x^2}}\right)$, we need $9 - 4x^2 > 0 \Rightarrow |x| < \frac{3}{2}$. For $\cos^{-1}\left(\frac{x}{3}\right)$, we need $|x| \leq 3$. Combined domain: $x \in \left(-\frac{3}{2}, \frac{3}{2}\right)$. Step 2: Simplify $f(x)$. Let $x = \frac{3}{2}\sin\theta$, then: $$\frac{2x}{\sqrt{9-4x^2}} = \frac{3\sin\theta}{3\cos\theta} = \tan\theta$$ So $\tan^{-1}\left(\frac{2x}{\sqrt{9-4x^2}}\right) = \theta = \sin^{-1}\left(\frac{2x}{3}\right)$. Step 3: Express $f(x)$ in simplified form: $$f(x) = \sin^{-1}\left(\frac{2x}{3}\right) - \cos^{-1}\left(\frac{x}{3}\right)$$ Step 4: Analyze the range of $f(x)$ on $\left(-\frac{3}{2}, \frac{3}{2}\right)$. At $x = 0$: $f(0) = 0 - \frac{\pi}{2} = -\frac{\pi}{2}$. At $x \to \frac{3}{2}^-$: $f \to \frac{\pi}{2} - \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$. At $x \to -\frac{3}{2}^+$: $f \to -\frac{\pi}{2} - \cos^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{2} - \frac{2\pi}{3} = -\frac{7\pi}{6}$. Step 5: The range of $f(x)$ is $\left(-\frac{7\pi}{6}, \frac{\pi}{6}\right)$ (open interval). The integral values of $k$ in this range are $k \in \{-3, -2, -1, 0\}$, giving 4 integral values. So option (3) is correct. Step 6: For $f(x) = \ln(-x)$, we need $x < 0$ (for $\ln(-x)$ to be defined), so $x \in \left(-\frac{3}{2}, 0\right)$. On this interval, $f(x)$ is strictly increasing (from $-\frac{7\pi}{6}$ to $-\frac{\pi}{2}$) and $\ln(-x)$ is strictly decreasing. Since one is increasing and the other decreasing, they intersect exactly once. So option (2) is correct.