Consider the following statements regarding the function f(x) = |ln x| − |x − 1|

Statement 1: f(x) is differentiable for all x > 0

Statement 2: f(x) is increasing in (1, ∞)

Statement 3: f(x) is decreasing in (0, 1)

We are given the function:
$f\left(x\right)=|\ln x|-|x-1|$
defined for $x>0$.

Let's analyze the behaviors of the logarithmic term $|\ln x|$ and the linear term $|x-1|$ on different intervals.

Case 1: When $x\ge 1$
For $x\ge 1$, we have $\ln x\ge 0$ and $x-1\ge 0$.
Thus, the absolute values can be opened directly:
$f\left(x\right)=\ln x-(x-1)=\ln x-x+1$

Case 2: When $0<x<1$
For $0<x<1$, we have $\ln x<0$ and $x-1<0$.
Thus, the absolute values open with negative signs:
$f\left(x\right)=-\ln x-(-(x-1\left)\right)=-\ln x+x-1$

Now we evaluate each statement step-by-step.

Evaluating Statement 1: Differentiability at $x=1$
Since $f\left(x\right)$ is composed of differentiable functions in the intervals $(0,1)$ and $(1,\infty )$, we only need to check differentiability at the boundary point $x=1$.
Let us find the left-hand derivative (LHD) and right-hand derivative (RHD) at $x=1$.

For $0<x<1$:
$f^{\text{'}}\left(x\right)=\frac{d}{dx}(-\ln x+x-1)=-\frac{1}{x}+1$
Taking the limit as $x\to 1^{-}$:
$\text{LHD}=-\frac{1}{1}+1=0$

For $x>1$:
$f^{\text{'}}\left(x\right)=\frac{d}{dx}(\ln x-x+1)=\frac{1}{x}-1$
Taking the limit as $x\to 1^{+}$:
$\text{RHD}=\frac{1}{1}-1=0$

Since the Left-Hand Derivative equals the Right-Hand Derivative ($\text{LHD}=\text{RHD}=0$) at $x=1$, the function $f\left(x\right)$ is differentiable at $x=1$. Since it is differentiable everywhere else in $x>0$, Statement 1 is true.

Evaluating Statement 2: Monotonicity in $(1,\infty )$
For $x>1$, we have:
$f^{\text{'}}\left(x\right)=\frac{1}{x}-1$
Since $x>1$, we have $\frac{1}{x}<1$, which implies:
$f^{\text{'}}\left(x\right)<0$
Therefore, $f\left(x\right)$ is decreasing in $(1,\infty )$. Thus, Statement 2 is false.

Evaluating Statement 3: Monotonicity in $(0,1)$
For $0<x<1$, we have:
$f^{\text{'}}\left(x\right)=-\frac{1}{x}+1$
Since $0<x<1$, we have $\frac{1}{x}>1$, which implies:
$-\frac{1}{x}<-1\Rightarrow -\frac{1}{x}+1<0$
Thus, $f^{\text{'}}\left(x\right)<0$ for all $x\in (0,1)$.
Therefore, $f\left(x\right)$ is decreasing in $(0,1)$. Thus, Statement 3 is true.

Consequently, Statement 1 and Statement 3 are true, while Statement 2 is false.