COMEDK 2025 Evening Solved Paper

Concept:Find $x$ from domain restrictions of inverse trigonometric functions, then compute their sum to solve for $k$.Explanation:Step 1: Determine the common domain of all inverse functions.For $\sin^{-1}(x-1)$, domain: $-1 \le x-1 \le 1 \;\Rightarrow\; 0 \le x \le 2$.For $\cos^{-1}(x-3)$, domain: $-1 \le x-3 \le 1 \;\Rightarrow\; 2 \le x \le 4$.The only intersection is $x=2$.Step 2: Evaluate each term at $x=2$.$\sin^{-1}(2-1) = \sin^{-1}(1) = \frac{\pi}{2}$.$\cos^{-1}(2-3) = \cos^{-1}(-1) = \pi$.$\tan^{-1}\!\left(\frac{2}{2-2^2}\right) = \tan^{-1}\!\left(\frac{2}{-2}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}$.Step 3: Sum the values: $\frac{\pi}{2} + \pi - \frac{\pi}{4} = \frac{2\pi}{4} + \frac{4\pi}{4} - \frac{\pi}{4} = \frac{5\pi}{4}$.Step 4: Set equal to right-hand side: $\frac{5\pi}{4} = \cos^{-1}k + \pi$.Thus, $\cos^{-1}k = \frac{5\pi}{4} - \pi = \frac{\pi}{4}$.Hence, $k = \cos\!\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.Answer:$\boxed{\frac{1}{\sqrt{2}}}$ $\;$ (Option D).