Question

Which of the following functions are odd? 
\[ \begin{aligned} \text{I. } & f(x) = x \left( \frac{e^x -1}{e^x +1} \right) \\[8pt] \text{II. } & f(x) = k^x + k^{-x} + \cos x \\[8pt] \text{III. } & f(x) = \log \left( x + \sqrt{x^2 +1} \right) \end{aligned} \]

• A. II
• B. I, II
• C. III
• D. I

Hint:

A function is odd if its graph is symmetric about the origin. Check by substituting \( -x \) and verifying \( f(-x) = -f(x) \).

Answer 1

The Correct Option is C

Step 1: Checking odd function property A function \( f(x) \) is odd if \[ f(-x) = -f(x). \] Step 2: Analyzing each function - For \( f(x) = x \left( \frac{e^x -1}{e^x +1} \right) \), substituting \( -x \) does not yield \( -f(x) \). - For \( f(x) = k^x + k^{-x} + \cos x \), the presence of even and periodic terms does not satisfy odd function conditions. - For \( f(x) = \log \left( x + \sqrt{x^2 +1} \right) \), substituting \( -x \) gives \[ \log \left( -x + \sqrt{x^2 +1} \right) = -\log \left( x + \sqrt{x^2 +1} \right), \] satisfying the odd function condition.