Question:

<p><strong>Step 1: Substitute to form a quadratic inequality</strong></p><p>Let $t = \cot^{-1}x$. The inequality becomes:</p><p>$$t^2 - 5t + 6 > 0$$</p><p><strong>Step 2: Factor the quadratic</strong></p><p>$$t^2 - 5t + 6 = (t-2)(t-3)$$</p><p>So the inequality is: $(t-2)(t-3) > 0$</p><p><strong>Step 3: Solve the quadratic inequality</strong></p><p>$(t-2)(t-3) > 0$ when $t < 2$ or $t > 3$</p><p><strong>Step 4: Recall the range of inverse cotangent</strong></p><p>Since $t = \cot^{-1}x$, we have $t \in (0, \pi)$. Both $t=2$ and $t=3$ lie in $(0, \pi)$ since $0 < 2 < 3 < \pi \approx 3.14$.</p><p><strong>Step 5: Convert back to $x$ using monotonicity</strong></p><p>Since $\cot^{-1}x$ is strictly decreasing on $\mathbb{R}$:</p><p>• $\cot^{-1}x < 2$ means $x > \cot 2$ (inequality flips)</p><p>• $\cot^{-1}x > 3$ means $x < \cot 3$ (inequality flips)</p><p><strong>Step 6: Combine the solution</strong></p><p>From Step 3: $t < 2$ or $t > 3$</p><p>From Step 5: This translates to $x > \cot 2$ or $x < \cot 3$</p><p>Therefore: $x \in (-\infty, \cot 3) \cup (\cot 2, \infty)$</p><p><strong>∴ Answer: B</strong></p>