Question

The point $P(\frac{1}{6}, \alpha)$, where $\alpha$ is a constant, lies on the curve with equation $\sin^{-1}(3x) + 2\sin^{-1}(y) = \frac{\pi}{2}, |x| \leq \frac{1}{3}, |y| \leq 1$, then the value of $\alpha$ is equal to

• A. $\frac{1}{2}$
• B. $2$
• C. $\frac{1}{4}$
• D. $4$
• E. $0$

Hint:

Always check if the substituted values result in standard angles like $\pi/6, \pi/4,$ or $\pi/3$, which makes solving for the unknown constant straightforward.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Substitute the coordinates of the point $(1/6, \alpha)$ into the equation of the curve to find the value of $\alpha$.

Step 2: Detailed Explanation:
1. Given equation: $\sin^{-1}(3x) + 2\sin^{-1}(y) = \pi/2$.
2. Substitute $x = 1/6$ and $y = \alpha$:
\[ \sin^{-1}\left(3 \cdot \frac{1}{6}\right) + 2\sin^{-1}(\alpha) = \frac{\pi}{2} \]
\[ \sin^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}(\alpha) = \frac{\pi}{2} \]
3. We know that $\sin^{-1}(1/2) = \pi/6$:
\[ \frac{\pi}{6} + 2\sin^{-1}(\alpha) = \frac{\pi}{2} \]
4. Isolate $2\sin^{-1}(\alpha)$:
\[ 2\sin^{-1}(\alpha) = \frac{\pi}{2} - \frac{\pi}{6} \]
\[ 2\sin^{-1}(\alpha) = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]
5. Solve for $\sin^{-1}(\alpha)$:
\[ \sin^{-1}(\alpha) = \frac{\pi}{6} \]
6. Find $\alpha$:
\[ \alpha = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

Step 3: Final Answer:
The value of $\alpha$ is $\frac{1}{2}$.