If f(x)=\ln(x+\sqrt{1+x^{2}}), then which one of the following is correct?
- A. f(x)+f(-x)=0 ✓
- B. f(x)-f(-x)=0
- C. 2f(x)=f(-x)
- D. f(x)=2f(-x)
Correct Answer: A. f(x)+f(-x)=0
Explanation
Evaluate f(-x) = \ln(-x+\sqrt{1+x^2}). Adding f(x) and f(-x) yields \ln(x+\sqrt{1+x^2}) + \ln(-x+\sqrt{1+x^2}). Using the logarithm property \ln a + \ln b = \ln(ab), we get \ln((x+\sqrt{1+x^2})(-x+\sqrt{1+x^2})) = \ln(1+x^2-x^2) = \ln 1 = 0. Therefore, f(x)+f(-x)=0.
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