Question

The function \( f(x) = \log(x + \sqrt{x^2 + 1}) \) is

• A. an even function
• B. an odd function
• C. a periodic function
• D. neither an even nor an odd function

Hint:

To check if a function \( f(x) \) is even or odd, calculate \( f(-x) \). \[\begin{array}{rl} \bullet & \text{If \( f(-x) = f(x) \), the function is even.} \\ \bullet & \text{If \( f(-x) = -f(x) \), the function is odd.} \\ \end{array}\] For functions involving square roots like \( \log(\sqrt{A} \pm B) \), rationalizing the argument is a common and effective trick.

Answer 1

The Correct Option is B

To determine if a function is even or odd, we need to evaluate \( f(-x) \) and compare it to \( f(x) \). \[ f(-x) = \log(-x + \sqrt{(-x)^2 + 1}) = \log(-x + \sqrt{x^2 + 1}) \] This doesn't immediately look like \( f(x) \) or \( -f(x) \). Let's try rationalizing the argument of the logarithm. Multiply the argument by \( \frac{x + \sqrt{x^2+1}}{x + \sqrt{x^2+1}} \): \[ f(-x) = \log\left( (\sqrt{x^2 + 1} - x) \cdot \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1} + x} \right) \] \[ f(-x) = \log\left( \frac{(x^2+1) - x^2}{\sqrt{x^2 + 1} + x} \right) \] \[ f(-x) = \log\left( \frac{1}{x + \sqrt{x^2 + 1}} \right) \] Using the property of logarithms \( \log(1/a) = -\log(a) \): \[ f(-x) = -\log(x + \sqrt{x^2 + 1}) \] \[ f(-x) = -f(x) \] Since \( f(-x) = -f(x) \), the function is an odd function.