Kerala KEAM 5 June 2024 Solved Paper

We are given that: $$\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}.$$The range of the inverse sine function is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, so the maximum possible value of each of the individual terms $\sin^{-1} x$, $\sin^{-1} y$, and $\sin^{-1} z$ is $\frac{\pi}{2}$.Step 1: Since the sum of three inverse sines equals $\frac{3\pi}{2}$, we conclude that each of the terms must be equal to $\frac{\pi}{2}$. Thus: $$\sin^{-1} x = \frac{\pi}{2}, \quad \sin^{-1} y = \frac{\pi}{2}, \quad \sin^{-1} z = \frac{\pi}{2}.$$Step 2: Taking the sine of both sides of these equations, we find: $$x = \sin \frac{\pi}{2} = 1, \quad y = \sin \frac{\pi}{2} = 1, \quad z = \sin \frac{\pi}{2} = 1.$$Step 3: Therefore, the sum $x + y + z$ is: $$x + y + z = 1 + 1 + 1 = 3.$$Thus, the correct answer is option (E).Quick Tip: For problems involving inverse trigonometric functions, remember the range of each inverse function and the sum of angles to simplify the solution. Here, the sum of three inverse sines equaled the maximum possible value.