Question

\(\text{ If } f(x) = 2\left( \tan^{-1}(e^x) - \frac{\pi}{4} \right), \text{ then } f(x) \text{ is:}\)

• A. even and is strictly increasing in \( (0, \infty) \)
• B. even and is strictly decreasing in \( (0, \infty) \)
• C. odd and is strictly increasing in \( (-\infty, \infty) \)
• D. odd and is strictly decreasing in \( (-\infty, \infty) \)

Hint:

When analyzing the parity of a function, check if \( f(-x) = f(x) \) (even) or \( f(-x) = -f(x) \) (odd). For monotonicity, compute the derivative of the function: if \( f'(x) > 0 \), the function is strictly increasing, and if \( f'(x) < 0 \), the function is strictly decreasing. A positive derivative over the entire domain typically indicates an increasing function.

Answer 1

The Correct Option is C

To determine the nature of the function \( f(x) = 2\left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) \), we will investigate its properties in terms of evenness or oddness and monotonicity.

1. Check for Oddness:

A function \( f(x) \) is odd if \( f(-x) = -f(x) \) for all \( x \). Calculate \( f(-x) \):

\( f(-x) = 2\left( \tan^{-1}(e^{-x}) - \frac{\pi}{4} \right) \)

Note that \( \tan^{-1}(e^{-x}) = \frac{\pi}{2} - \tan^{-1}(e^x) \). Therefore,

\( f(-x) = 2\left( \left( \frac{\pi}{2} - \tan^{-1}(e^x) \right) - \frac{\pi}{4} \right) = 2\left( \frac{\pi}{2} - \frac{\pi}{4} - \tan^{-1}(e^x) \right)\)

\( f(-x) = \left( \pi - 2\tan^{-1}(e^x) \right) - \frac{\pi}{2} = -2\left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) = -f(x) \)

Since \( f(-x) = -f(x) \), the function \( f(x) \) is odd.

2. Check for Monotonicity:

To check whether the function is strictly increasing or decreasing, we calculate its derivative \( f'(x) \).

\( f(x) = 2\left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) \)

The derivative \( \frac{d}{dx}[\tan^{-1}(e^x)] = \frac{e^x}{1 + e^{2x}} \).

Therefore,

\( f'(x) = 2 \cdot \frac{e^x}{1 + e^{2x}} = \frac{2e^x}{1 + e^{2x}} \)

Observe that \( \frac{2e^x}{1 + e^{2x}} \) is positive for all \( x \). Thus, \( f(x) \) is strictly increasing for all \( (-\infty, \infty) \).

Conclusively, the function \( f(x) \) is odd and is strictly increasing over \( (-\infty, \infty) \).

Answer 2

Analyze the properties of \( f(x) = 2 \left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) \):

Step 1: Check for Parity (Even or Odd):

To determine if \( f(x) \) is even or odd, we check if \( f(-x) = f(x) \) for evenness or \( f(-x) = -f(x) \) for oddness.

We start by substituting \( -x \) into the function \( f(x) \):

\[ f(-x) = 2 \left( \tan^{-1}(e^{-x}) - \frac{\pi}{4} \right) \]

Using the property of the inverse tangent function, \( \tan^{-1}(e^{-x}) = \frac{\pi}{4} - \tan^{-1}(e^x) \), we get:

\[ f(-x) = 2 \left( \left( \frac{\pi}{4} - \tan^{-1}(e^x) \right) - \frac{\pi}{4} \right) \]

Simplifying:

\[ f(-x) = -2 \tan^{-1}(e^x) \]

Since \( f(x) = 2 \left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) \), we have:

\[ f(-x) = -f(x) \]

This confirms that \( f(x) \) is an odd function.

Step 2: Check for Monotonicity (Increasing or Decreasing):

To check whether \( f(x) \) is increasing or decreasing, we compute its derivative \( f'(x) \).

We differentiate \( f(x) = 2 \left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) \) with respect to \( x \):

\[ f'(x) = 2 \cdot \frac{d}{dx} \left( \tan^{-1}(e^x) \right) \]

The derivative of \( \tan^{-1}(e^x) \) is:

\[ \frac{d}{dx} \left( \tan^{-1}(e^x) \right) = \frac{e^x}{1 + e^{2x}} \]

Thus,

\[ f'(x) = \frac{2e^x}{1 + e^{2x}} \]

Since \( e^x > 0 \) for all \( x \), and the denominator \( 1 + e^{2x} > 0 \), we conclude that \( f'(x) > 0 \) for all \( x \in (-\infty, \infty) \). Therefore, \( f(x) \) is strictly increasing.

Conclusion:

\( f(x) \) is an odd function and strictly increasing over \( (-\infty, \infty) \), verifying that option (3) is correct.