Question

The number of solutions of \[ \sin^{-1} x + \sin^{-1}(1-x) = \cos^{-1} x \] is:

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( 4 \)

Hint:

For inverse trig equations: \begin{itemize} \item Convert using identities. \item Check domain carefully. \end{itemize}

Answer 1

The Correct Option is B

Concept: Use identity: \[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \] Step 1: {\color{red}Rewrite RHS.} Equation: \[ \sin^{-1}x + \sin^{-1}(1-x) = \frac{\pi}{2} - \sin^{-1}x \] \[ 2\sin^{-1}x + \sin^{-1}(1-x) = \frac{\pi}{2} \] Step 2: {\color{red}Check domain.} Both inverse sine arguments in \( [-1,1] \): \[ x \in [0,1] \] Step 3: {\color{red}Test values.} Try symmetry \( x = \frac{1}{2} \): \[ \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \] LHS: \[ \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} \] RHS: \[ \cos^{-1}\frac{1}{2} = \frac{\pi}{3} \] Works. Step 4: {\color{red}Uniqueness.} Monotonic behavior ensures single solution.