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:
Since point \(P(\frac{1}{6}, \alpha)\) lies on the curve, its coordinates must satisfy the curve's equation.
We substitute \(x = \frac{1}{6}\) and \(y = \alpha\) into the equation.
Step 2: Key Formula or Approach:
Evaluate \(\sin^{-1}(3 \cdot \frac{1}{6})\).
Isolate the term \(\sin^{-1}(\alpha)\) and solve for \(\alpha\) using standard trigonometric values.
Step 3: Detailed Explanation:
Substitute the point into the equation:
\[ \sin^{-1}\left(3\left(\frac{1}{6}\right)\right) + 2\sin^{-1}(\alpha) = \frac{\pi}{2} \] Simplify the argument of the first term:
\[ \sin^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}(\alpha) = \frac{\pi}{2} \] We know that \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\):
\[ \frac{\pi}{6} + 2\sin^{-1}(\alpha) = \frac{\pi}{2} \] Subtract \(\frac{\pi}{6}\) from both sides:
\[ 2\sin^{-1}(\alpha) = \frac{\pi}{2} - \frac{\pi}{6} \] \[ 2\sin^{-1}(\alpha) = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \] Divide by 2:
\[ \sin^{-1}(\alpha) = \frac{\pi}{6} \] Take the sine of both sides:
\[ \alpha = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \] Step 4: Final Answer:
The value of \(\alpha\) is \(\frac{1}{2}\).