Consider the following three statements for the function $$f: (0, \infty ) \rightarrow \mathbb R$$ defined by
$$f(x)= |\log_{e}{x}|-|x-1|:$$
(I)f is differentiable at all x > 0.
(II)f is increasing in (0, 1).
(III)f is decreasing in (1, $$\infty$$).
Then.

We split the function based on the critical point $$x=1$$:

• For $$0 < x < 1$$: $$\ln x$$ is negative and $$(x-1)$$ is negative.

$$f(x) = (-\ln x) - (-(x - 1)) = -\ln x + x - 1$$

$$f'(x) = -\frac{1}{x} + 1 = \frac{x - 1}{x}$$. Since $$x < 1$$, $$f'(x) < 0$$ (Decreasing).

• For $$x \ge 1$$: $$\ln x$$ is positive and $$(x-1)$$ is positive.

$$f(x) = \ln x - (x - 1) = \ln x - x + 1$$

$$f'(x) = \frac{1}{x} - 1 = \frac{1 - x}{x}$$. Since $$x > 1$$, $$f'(x) < 0$$ (Decreasing).

Check Differentiability at $$x=1$$:

• Left derivative: $$f'(1^-) = \frac{1-1}{1} = 0$$.

• Right derivative: $$f'(1^+) = \frac{1-1}{1} = 0$$.

Since they match, $$f$$ is differentiable at $$x=1$$.

Conclusion: (I) True: It is differentiable everywhere in its domain.

(II) False: It is decreasing in $$(0, 1)$$.

(III) True: It is decreasing in $$(1, \infty)$$.

Correct Answer: Only (I) and (III) are TRUE.