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:

Always check the validity of solutions when dealing with inverse trigonometric functions due to their restricted domains and ranges.

Answer 1

The Correct Option is C

Step 1: Apply the identity \( \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \). The equation transforms to:\n\[\n\sin^{-1} x + \sin^{-1} (1 - x) = \frac{\pi}{2} - \sin^{-1} x\n\]\n\n
Step 2: Reorganize the equation.\n\[\n2 \sin^{-1} x + \sin^{-1} (1 - x) = \frac{\pi}{2}\n\]\n\[\n\sin^{-1} (1 - x) = \frac{\pi}{2} - 2 \sin^{-1} x\n\]\n\n
Step 3: Take the sine of both sides.\n\[\n1 - x = \sin\left(\frac{\pi}{2} - 2 \sin^{-1} x\right) = \cos(2 \sin^{-1} x)\n\]\n\n
Step 4: Use the identity \( \cos(2\alpha) = 1 - 2 \sin^2 \alpha \). Let \( \sin^{-1} x = \alpha \), which means \( \sin \alpha = x \).\n\[\n1 - x = 1 - 2 \sin^2 (\sin^{-1} x) = 1 - 2 x^2\n\]\n\n
Step 5: Solve the resulting quadratic equation.\n\[\n2x^2 - x = 0 \implies x(2x - 1) = 0\n\]\nThe potential solutions are \( x = 0 \) and \( x = \frac{1}{2} \).\n\n
Step 6: Check the solutions. For \( x = 0 \): \( \sin^{-1} 0 + \sin^{-1} 1 = 0 + \frac{\pi}{2} = \frac{\pi}{2} \), and \( \cos^{-1} 0 = \frac{\pi}{2} \). This is valid. For \( x = \frac{1}{2} \): \( \sin^{-1} \frac{1}{2} + \sin^{-1} \frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} \), and \( \cos^{-1} \frac{1}{2} = \frac{\pi}{3} \). This is also valid.
\n\nBoth solutions are confirmed.