Which pair of functions is identical?

Options

A. `sin^(-1)(sinx) " and " sin(sin^(-1)x)`
B. `log_(e)e^(x),e^(log_(e)x)`
C. `log_(e)x^(2),2log_(e)x`
D. None of the above

Answer

To determine which pair of functions is identical, we will analyze each option step by step.

### Step 1: Analyze the first option - \( \sin^{-1}(\sin x) \) and \( \sin(\sin^{-1}x) \)

1. **Evaluate \( \sin^{-1}(\sin x) \)**:
- The function \( \sin^{-1}(\sin x) \) gives the angle whose sine is \( \sin x \).
- This function is defined for \( x \) in the range \( [0, \pi] \).
- For \( x \) in the interval \( [0, \frac{\pi}{2}] \), \( \sin^{-1}(\sin x) = x \).
- For \( x \) in the interval \( [\frac{\pi}{2}, \pi] \), \( \sin^{-1}(\sin x) = \pi - x \).
- For \( x \) outside this range, the function will not equal \( x \).

2. **Evaluate \( \sin(\sin^{-1}x) \)**:
- The function \( \sin(\sin^{-1}x) \) is defined for \( x \) in the range \( [-1, 1] \).
- For any \( x \) in this range, \( \sin(\sin^{-1}x) = x \).

3. **Conclusion for the first option**:
- The two functions are not identical because \( \sin^{-1}(\sin x) \) does not equal \( x \) for all \( x \).

### Step 2: Analyze the second option - \( \ln(e^x) \) and \( e^{\ln x} \)

1. **Evaluate \( \ln(e^x) \)**:
- Using the property of logarithms, \( \ln(e^x) = x \).

2. **Evaluate \( e^{\ln x} \)**:
- By the property of exponents and logarithms, \( e^{\ln x} = x \) for \( x > 0 \).

3. **Conclusion for the second option**:
- Both functions are identical as they both equal \( x \).

### Step 3: Analyze the third option - \( \ln(x^2) \) and \( 2\ln x \)

1. **Evaluate \( \ln(x^2) \)**:
- Using the logarithmic property, \( \ln(x^2) = 2\ln x \).

2. **Conclusion for the third option**:
- Both functions are identical as they are equal to \( 2\ln x \).

### Final Conclusion
The identical pairs of functions are:
- \( \ln(e^x) \) and \( e^{\ln x} \)
- \( \ln(x^2) \) and \( 2\ln x \)