Question 1

Given $f(x) = \sqrt{x} \ln x - x + 1$.

Calculating the first derivative:

$f'(x) = \frac{1}{2\sqrt{x}} \ln x + \sqrt{x} \cdot \frac{1}{x} - 1 = \frac{\ln x + 2}{2\sqrt{x}} - 1$.

To find critical points, set $f'(x) = 0 \implies \ln x + 2 = 2\sqrt{x}$.

Let $g(x) = \ln x + 2 - 2\sqrt{x}$.

$g'(x) = \frac{1}{x} - \frac{1}{\sqrt{x}} = \frac{1-\sqrt{x}}{x}$.

$g'(x) = 0$ at $x=1$. For $x \in (0, 1)$, $g'(x) > 0$ and for $x > 1$, $g'(x) < 0$.

Thus, $g(x)$ has a maximum at $x=1$ where $g(1) = 0 + 2 - 2 = 0$.

Since the maximum of $g(x)$ is $0$, $g(x) \le 0$ for all $x \in (0, \infty)$.

This implies $f'(x) = \frac{g(x)}{2\sqrt{x}} \le 0$ for all $x$.

Since $f'(x)$ is zero only at $x=1$ and negative elsewhere, $f(x)$ is strictly decreasing and has no local extrema.

Checking (A): $f''(x) = \frac{d}{dx}(\frac{\ln x + 2}{2\sqrt{x}} - 1) = \frac{\frac{1}{x}(2\sqrt{x}) - (\ln x + 2)\frac{1}{\sqrt{x}}}{4x} = \frac{2 - \ln x - 2}{4x\sqrt{x}} = -\frac{\ln x}{4x\sqrt{x}}$.

In $(0, 1)$, $\ln x < 0$, so $f''(x) > 0$. Thus $f'(x)$ is increasing in $(0, 1)$. Hence (A) is false and (D) is true.