Question

If \[ f(x) = 2x - \tan^{-1} x - \log(x + \sqrt{x^2 + 1}) \] is monotonically increasing, then

• A. \( x<0 \)
• B. \( x \in \mathbb{R} - \{0\} \)
• C. \( x \in \mathbb{R} \)
• D. \( x>0 \)

Hint:

To check where a function is monotonically increasing, compute its derivative and find where it is non-negative.

Answer 1

The Correct Option is C

To determine where the function is monotonically increasing, we need to compute the derivative of the function \( f(x) \). Given: \[ f(x) = 2x - \tan^{-1} x - \log(x + \sqrt{x^2 + 1}) \] First, calculate the derivative: 1. The derivative of \( 2x \) is 2. 2. The derivative of \( \tan^{-1} x \) is \( \frac{1}{1 + x^2} \). 3. The derivative of \( \log(x + \sqrt{x^2 + 1}) \) is found using the chain rule: \[ \frac{d}{dx} \log(x + \sqrt{x^2 + 1}) = \frac{1}{x + \sqrt{x^2 + 1}} \left( 1 + \frac{x}{\sqrt{x^2 + 1}} \right) \] Simplifying: \[ \frac{d}{dx} \log(x + \sqrt{x^2 + 1}) = \frac{1}{\sqrt{x^2 + 1}} \] Thus, the derivative of \( f(x) \) is: \[ f'(x) = 2 - \frac{1}{1 + x^2} - \frac{1}{\sqrt{x^2 + 1}} \] Now, for \( f(x) \) to be monotonically increasing, we need \( f'(x) \geq 0 \). The derivative simplifies to: \[ f'(x) = 2 - \frac{1}{1 + x^2} - \frac{1}{\sqrt{x^2 + 1}} \geq 0 \] After analyzing this expression, we find that the function is monotonically increasing for all real values of \( x \). Therefore, the correct answer is \( x \in \mathbb{R} \), or option (3).