Question

Find the domain of \[ f(x)=\sin^{-1}(2x-1) \]

• A. \([0,1]\)
• B. \([-1,1]\)
• C. \([0,\infty)\)
• D. \([-0.5,0.5]\)

Hint:

For inverse trigonometric functions, always restrict the inner expression to the valid range.

Answer 1

The Correct Option is A

Concept: For inverse sine function: \[ \sin^{-1}(u) \] the input value must satisfy: \[ -1\le u\le1 \]

Step 1: Apply inverse sine condition Given: \[ f(x)=\sin^{-1}(2x-1) \] Therefore: \[ -1\le2x-1\le1 \]

Step 2: Solve the inequality Add \(1\) throughout: \[ 0\le2x\le2 \] Divide throughout by \(2\): \[ 0\le x\le1 \]

Step 3: Write domain Hence the domain is: \[ [0,1] \] Final Answer: \[ \boxed{[0,1]} \]