Question:

<p><strong>Step 1: Recall key properties of inverse trigonometric functions.</strong></p><p>For $0 < x < 1$:</p><ul><li>$\sin^{-1}(x) \in (0, \pi/2)$ and is strictly increasing</li><li>$\cos^{-1}(x) \in (0, \pi/2)$ and is strictly decreasing</li><li>Fundamental identity: $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$</li></ul><p><strong>Step 2: Determine the range of $\sin^{-1}(x)$.</strong></p><p>When $0 < x < 1$: as $x \to 0^+$, $\sin^{-1}(x) \to 0^+$; as $x \to 1^-$, $\sin^{-1}(x) \to \frac{\pi}{2}^-$</p><p>Therefore: $\sin^{-1}(x) \in (0, \frac{\pi}{2})$</p><p><strong>Step 3: Determine the range of $\cos^{-1}(x)$.</strong></p><p>When $0 < x < 1$: as $x \to 0^+$, $\cos^{-1}(x) \to \frac{\pi}{2}^+$; as $x \to 1^-$, $\cos^{-1}(x) \to 0^+$</p><p>Therefore: $\cos^{-1}(x) \in (0, \frac{\pi}{2})$</p><p><strong>Step 4: Evaluate option (a): $[1, 2)$</strong></p><p>Since $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \approx 1.57$, and $\sin^{-1}(x) \in (0, \frac{\pi}{2})$, the sum satisfies: $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \in [1, 2)$ ✓</p><p>Option (a) is <strong>CORRECT</strong></p><p><strong>Step 5: Evaluate option (b): $[3, 4]$</strong></p><p>Since $\frac{\pi}{2} \approx 1.57$, this value does not lie in $[3,4]$. Option (b) is <strong>INCORRECT</strong></p><p><strong>Step 6: Evaluate option (c): $(-\infty, 2]$</strong></p><p>Since $\sin^{-1}(x) \in (0, \frac{\pi}{2})$ where $\frac{\pi}{2} \approx 1.57 < 2$, we have $\sin^{-1}(x) \in (-\infty, 2]$ ✓</p><p>Option (c) is <strong>CORRECT</strong></p><p><strong>Step 7: Evaluate option (d): $[-1, 1]$</strong></p><p>Since $\sin^{-1}(x) > 0$ for all $x \in (0,1)$, and $\sin^{-1}(x)$ never reaches values in $[-1, 0)$, option (d) is <strong>INCORRECT</strong></p><p><strong>∴ Answer: a,c</strong></p>