(A) \( y = \frac{1}{2-\sin(3x)} \) The range of \( \sin(3x) \) is \([-1, 1]\). Therefore, \( 2 - \sin(3x) \) ranges from \(1\) to \(3\). The reciprocal \( y \) will then range from \(\frac{1}{3}\) to \(1\). \[Range: \left[ \frac{1}{3}, 1 \right] (III) \] (B) \( y = \frac{x^2 + x + 2}{x^2 + x + 1}, x \in \mathbb{R} \) Let \( y = \frac{x^2 + x + 2}{x^2 + x + 1} \). To find the range, solve for \( x \): \[ y(x^2 + x + 1) = x^2 + x + 2 \] \[ yx^2 + yx + y = x^2 + x + 2 \] \[ (y - 1)x^2 + (y - 1)x + (y - 2) = 0 \] For real \( x \), the discriminant must be non-negative: \[ (y - 1)^2 - 4(y - 1)(y - 2) \geq 0 \] \[ (y - 1)[(y - 1) - 4(y - 2)] \geq 0 \] \[ (y - 1)(-3y + 7) \geq 0 \] Solving this inequality gives \( y \in \left(1, \frac{7}{3}\right) \). \[Range: \left(1, \frac{7}{3}\right) (I) \] (C) \( y = \sin(x) - \cos(x) \) Using trigonometric identities, \( y = \sqrt{2} \sin\left(x - \frac{\pi}{4}\right) \). The range of \( \sin \) is \([-1, 1]\), so the range of \( y \) is \([- \sqrt{2}, \sqrt{2}]\). \[Range: \left[-\sqrt{2}, \sqrt{2}\right] (IV) \] (D) \( y = \cot^{-1}(-x) - \tan^{-1}(x) + \sec^{-1}(x) \) The range of \( \cot^{-1}(-x) \) is \((0, \pi)\), the range of \( \tan^{-1}(x) \) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), and the range of \( \sec^{-1}(x) \) is \(\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]\). Combining these, the range of \( y \) is \(\left[\frac{\pi}{2}, \pi\right) \cup \left(\pi, \frac{3\pi}{2}\right]\). Range: \(\left[\frac{\pi}{2}, \pi\right) \cup \left(\pi, \frac{3\pi}{2}\right]\) (II) Matching: - (A) - III - (B) - I - (C) - IV - (D) - II Correct Answer: \(\boxed{A}\)