Let $f(x) = 1 + x\ln(x + \sqrt{x^2+1})$ and $g(x) = \sqrt{1+x^2}$. Then:
$f(x) > g(x)$ $\forall$ $x \in R^+$
$f(x) < g(x)$ $\forall$ $x \in R^-$
there exist $x = a > 0$ for which $f(x) < g(x)$
there exist $x = a < 0$ for which $f(x) > g(x)$
Step-by-Step Solution
Key Concept: Comparing two functions $f(x)$ and $g(x)$ is often best done by analyzing their difference, $h(x) = f(x) - g(x)$. By evaluating $h(x)$ at a convenient point and then analyzing the sign of its derivative $h'(x)$ over different intervals, one can determine the monotonicity of $h(x)$ and thus the relative values of $f(x)$ and $g(x)$. Recognizing that $\ln(x+\sqrt{x^2+1})$ is the inverse hyperbolic sine function ($\text{arsinh}(x)$) and its derivative is $\frac{1}{\sqrt{x^2+1}}$ is crucial for simplifying $h'(x)$.
Step 1: Define $h(x) = f(x) - g(x) = 1 + x\ln(x + \sqrt{x^2+1}) - \sqrt{1+x^2}$.
Step 2: Compute $h(0) = 1 + 0 - 1 = 0$.
Step 3: Differentiate $h(x)$:
$$h'(x) = \ln(x + \sqrt{x^2+1}) + x \cdot \frac{1}{\sqrt{x^2+1}} - \frac{x}{\sqrt{1+x^2}}$$
$$= \ln(x + \sqrt{x^2+1})$$
Step 4: For $x > 0$: $x + \sqrt{x^2+1} > 1$, so $\ln(x+\sqrt{x^2+1}) > 0$, meaning $h'(x) > 0$. Thus $h(x) > h(0) = 0$, i.e., $f(x) > g(x)$ for all $x \in R^+$. Option (a) is correct.
Step 5: For $x < 0$: $x + \sqrt{x^2+1} < 1$ (since $\sqrt{x^2+1} < 1 - x$ is not always true; more carefully, for $x < 0$, $\sinh^{-1}(x) < 0$, so $\ln(x+\sqrt{x^2+1}) < 0$), meaning $h'(x) < 0$. Thus $h(x) < h(0) = 0$, i.e., $f(x) < g(x)$ for all $x \in R^-$. Option (b) is correct.
Step 6: Since $f(x) > g(x)$ for all $x > 0$ and $f(x) < g(x)$ for all $x < 0$, options (c) and (d) are incorrect.
Correct Answer: 1, 2