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).
Let \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^2 x}{1+5^x} \, dx \). Then:
\[ \int_0^{\frac{\pi}{4}} (\tan^3 x + \tan^5 x) \, dx \]
For \(1 \leq x<\infty\), let \(f(x) = \sin^{-1}\left(\frac{1}{x}\right) + \cos^{-1}\left(\frac{1}{x}\right)\). Then \(f'(x) =\)
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: