Step 1: For (A)

\( y = \dfrac{1}{2 - \sin 3x} \)
Since \( \sin 3x \in [-1,1] \), we get \( 2 - \sin 3x \in [1,3] \).
Hence \( y \in \left[\tfrac{1}{3}, 1\right] \). → Matches with (III).

Step 2: For (B)

\( y = \dfrac{x^2 + x + 2}{x^2 + x + 1} = 1 + \dfrac{1}{x^2 + x + 1} \)
Since denominator is always positive, \( y > 1 \).
Minimum denominator = \(\tfrac{3}{4}\) at \(x = -\tfrac{1}{2}\).
So maximum \( y = 1 + \tfrac{1}{3/4} = \tfrac{7}{3} \).
Thus, Range = \((1, \tfrac{7}{3}] \). → Matches with (I).

Step 3: For (C)

\( y = \sin x - \cos x = \sqrt{2}\sin\!\left(x - \tfrac{\pi}{4}\right) \)
Hence, Range = \([-\sqrt{2}, \sqrt{2}] \). → Matches with (IV).

Step 4: For (D)

\( y = \cot^{-1}(-x) - \tan^{-1}(x) + \sec^{-1}(x) \)
Simplifying with inverse trig identities gives Range:
\(\left[\tfrac{\pi}{2}, \pi\right) \cup \left(\pi, \tfrac{3\pi}{2}\right]\). → Matches with (II).

The function is

\( f(x) = [x]^n , \quad n \geq 2 \)

The GIF \([x]\) is discontinuous at all integers. Raising it to the integer power \(n \geq 2\) does not remove this discontinuity, because the jump still exists at each integer value of \(x\).

For non-integer \(x\), the function is constant over intervals \((m, m+1)\) where \(m \in \mathbb{Z}\), so it is continuous within each open interval between integers.

Final Answer: The function is discontinuous at all integers.