Question

Let $f(x) = 1 + x\log\left(x + \sqrt{x^2+1}\right) - \sqrt{x^2+1}, \; x \geq 0$. Then:

• A. $f(x)$ is increasing on $(0,\infty)$
• B. $f(x)$ is increasing only on $(10,\infty)$
• C. $f(x)$ is increasing only on $(0,e)$
• D. $f(x)$ is decreasing on $(0,\infty)$
• E. $f(x)$ is decreasing only on $(100,\infty)$

Hint:

$\log(x+\sqrt{x^2+1})$ is always positive for $x>0$.

Answer 1

The Correct Option is A

Concept:
• Differentiate and check sign

Step 1: Differentiate
Using identity: \[ \frac{d}{dx}\log(x+\sqrt{x^2+1}) = \frac{1}{\sqrt{x^2+1}} \] \[ f'(x) = \log(x+\sqrt{x^2+1}) \]

Step 2: Check sign
For $x>0$: \[ x+\sqrt{x^2+1}>1 \Rightarrow \log(\cdot)>0 \]

Step 3: Conclusion
\[ f'(x)>0 \Rightarrow f(x) \text{ increasing} \] Final Conclusion:
Option (A)