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.

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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.

All (I), (II) and (III) are TRUE.

Only (II) and (III) are TRUE.

Only (I) and (III) are TRUE.

Only (I) is TRUE.

The correct answer is

Only (I) and (III) are TRUE.

Differentiability Analysis of Function $f(x)$

The function is given by $f(x) = |\log_e x| - |x - 1|$ for $x \in (0, \infty)$. We analyze its differentiability, focusing on the point $x=1$ where the absolute value arguments become zero.

First, we express $f(x)$ piecewise:

For $0 < x < 1$: $\log_e x < 0$ and $x - 1 < 0$. So, $f(x) = -(\log_e x) - (-(x - 1)) = x - \log_e x - 1$.
For $x = 1$: $f(1) = |\log_e 1| - |1 - 1| = |0| - |0| = 0$.
For $x > 1$: $\log_e x > 0$ and $x - 1 > 0$. So, $f(x) = (\log_e x) - (x - 1) = \log_e x - x + 1$.

Statement (I): $f$ is differentiable at all $x > 0$.

We check differentiability at $x=1$ by comparing the left-hand derivative (LHD) and the right-hand derivative (RHD).

LHD at $x=1$:

Using the derivative for $0 < x < 1$, $f'(x) = 1 - \frac{1}{x}$.

LHD = $\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} \left(1 - \frac{1}{x}\right) = 1 - 1 = 0$.

RHD at $x=1$:

Using the derivative for $x > 1$, $f'(x) = \frac{1}{x} - 1$.

RHD = $\lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} \left(\frac{1}{x} - 1\right) = 1 - 1 = 0$.

Since LHD = RHD = 0, the function $f$ is differentiable at $x=1$. For $x \neq 1$, the function is composed of standard functions whose absolute values are differentiable away from zero. Therefore, $f$ is differentiable for all $x > 0$.

Conclusion for Statement (I): Statement (I) is TRUE.

Monotonicity Analysis in $(0, 1)$

Statement (II): $f$ is increasing in $(0, 1)$.

For $0 < x < 1$, $f(x) = x - \log_e x - 1$.

The derivative is $f'(x) = 1 - \frac{1}{x}$.

In the interval $(0, 1)$, we have $0 < x < 1$, which means $\frac{1}{x} > 1$.

Therefore, $f'(x) = 1 - \frac{1}{x} < 0$ for $x \in (0, 1)$.

A negative derivative indicates the function is decreasing.

Conclusion for Statement (II): Statement (II) is FALSE.

Monotonicity Analysis in $(1, \infty)$

Statement (III): $f$ is decreasing in $(1, \infty)$.

For $x > 1$, $f(x) = \log_e x - x + 1$.

The derivative is $f'(x) = \frac{1}{x} - 1$.

In the interval $(1, \infty)$, we have $x > 1$, which means $0 < \frac{1}{x} < 1$.

Therefore, $f'(x) = \frac{1}{x} - 1 < 0$ for $x \in (1, \infty)$.

A negative derivative indicates the function is decreasing.

Conclusion for Statement (III): Statement (III) is TRUE.

Overall Conclusion

Summarizing the findings:

Statement (I) is TRUE.
Statement (II) is FALSE.
Statement (III) is TRUE.

Thus, only statements (I) and (III) are TRUE.

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Important Questions from Differential Calculus

Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$, $x > 2$, is ______
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If the domain of the function $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right)$ is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to
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Let $[t]$ denote the greatest integer less than or equal to $t$. If the function
$f(x) = \begin{cases} b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] \right), & x < 0 \\ \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases}$
is continuous at $x = 0$, then $a^2 + b^2$ is equal to
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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in m, has two equal roots for every $x \in \mathbb{R}$. If $f(0) = 1, f'(0) = 2$, and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to ____________.
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Let $f: \mathbf{R} \to \mathbf{R}$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in \mathbf{R}$ and $f'(a-1) = 0$, where $a$ is a real number. Let $g(x) = f(\tan^2 x - 2\tan x + a)$, $0 < x < \frac{\pi}{2}$.
Consider the following two statements :
(I) $g$ is increasing in $\left(0, \frac{\pi}{4}\right)$
(II) $g$ is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
Then,
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