Question

The range of f(x) = \(4\sin^{-1}(\frac{x^2}{x^2+1})\) is 

• A. \([0,\pi]\)
• B. \([0,\pi)\)
• C. \([0,2\pi]\)
• D. \([0,2\pi)\)

Answer 1

The Correct Option is D

Step 1: Write the given function. \[ f(x) = 4 \sin^{-1} \left( \frac{x^2}{x^2 + 1} \right) \] 
Step 2: Analyze the expression inside the inverse sine function. \[ 0 \leq \frac{x^2}{x^2 + 1} < 1 \] This means that the expression inside \( \sin^{-1} \) is valid as the sine function only takes values between 0 and 1. 
Step 3: Apply the inverse sine function. \[ 0 \leq \sin^{-1} \left( \frac{x^2}{x^2 + 1} \right) < \frac{\pi}{2} \] 
Step 4: Multiply by 4. \[ 0 \leq 4 \sin^{-1} \left( \frac{x^2}{x^2 + 1} \right) < 2\pi \] 
Step 5: Conclude the range. Thus, the range of \( f(x) \) is \( [0, 2\pi] \).