Question

The function \[ f(x)=2x+\cot^{-1}x+\log\!\left(\sqrt{1+x^2}-x\right) \] is

• A. decreases on \((0,\infty)\)
• B. decreases on \((-\infty,0)\)
• C. neither increases nor decreases on \((-\infty,\infty)\)
• D. increases on \((-\infty,\infty)\) \bigskip

Hint:

A useful identity is \[ \sqrt{1+x^2}-x = \frac{1}{\sqrt{1+x^2}+x}. \] It converts complicated logarithmic derivatives into very simple forms.

Answer 1

The Correct Option is D

Concept: A function is increasing on an interval if \[ f'(x)>0 \] throughout that interval.

Step 1: Differentiate the function. Given \[ f(x)=2x+\cot^{-1}x+\log(\sqrt{1+x^2}-x). \] Differentiating, \[ f'(x) = 2-\frac{1}{1+x^2} + \frac{d}{dx} \log(\sqrt{1+x^2}-x). \]

Step 2: Simplify the logarithmic term. Use the identity \[ (\sqrt{1+x^2}-x)(\sqrt{1+x^2}+x)=1. \] Hence \[ \log(\sqrt{1+x^2}-x) = -\log(\sqrt{1+x^2}+x). \] Differentiating, \[ \frac{d}{dx} \log(\sqrt{1+x^2}-x) = -\frac{1}{\sqrt{1+x^2}}. \] Therefore, \[ f'(x) = 2-\frac{1}{1+x^2} -\frac{1}{\sqrt{1+x^2}}. \]

Step 3: Let \[ t=\sqrt{1+x^2}. \] Then \[ t\ge 1, \qquad \frac{1}{1+x^2}=\frac1{t^2}. \] Thus \[ f'(x) = 2-\frac1t-\frac1{t^2}. \] \[ = \frac{2t^2-t-1}{t^2}. \] Factorizing, \[ 2t^2-t-1 = (2t+1)(t-1). \] Hence \[ f'(x) = \frac{(2t+1)(t-1)}{t^2}. \]

Step 4: Determine the sign of \(f'(x)\). Since \[ t=\sqrt{1+x^2}\ge 1, \] we have \[ 2t+1>0, \qquad t-1\ge 0, \qquad t^2>0. \] Therefore \[ f'(x)\ge 0 \] for all \(x\), and \[ f'(x)>0 \] for all \(x\neq 0\). Hence the function is increasing on the entire real line.

Step 5: Write the final answer. \[ \boxed{\text{The function increases on }(-\infty,\infty).} \]