Question

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

• A. All (I), (II) and (III) are TRUE.
• B. Only (II) and (III) are TRUE.
• C. Only (I) and (III) are TRUE.
• D. Only (I) is TRUE.

Hint:

When absolute values are involved, always split the domain and analyze each interval separately.

Answer 1

The Correct Option is C

To determine which of the statements (I), (II), and (III) are true for the function \( f(x) = |\log_e x| - |x - 1| \), we will analyze the function step by step:

Step 1: Differentiability of the function \( f(x) \)

The function \( f(x) = |\log_e x| - |x - 1| \) is composed of the absolute value functions. The differentiability of such a function can be analyzed by considering the points where the absolute value functions change their behavior.

For the function \( |\log_e x| \), it changes at \( x = 1 \) since \( \log_e(x) \) is zero at this point. Similarly, \( |x - 1| \) changes at \( x = 1 \).

Examining both components:

- For \( x > 1 \), \( |\log_e x| = \log_e x \) and \( |x - 1| = x - 1 \)
- For \( x < 1 \), \( |\log_e x| = -\log_e x \) and \( |x - 1| = 1 - x \)
 

The critical point to check is \( x = 1 \).

• For \( x > 1 \), \(\frac{d}{dx}[\log_e x - (x - 1)] = \frac{1}{x} - 1\)
• For \( x < 1 \), \(\frac{d}{dx}[-\log_e x - (1 - x)] = -\frac{1}{x} + 1\)

At \( x = 1 \), both derivatives approach \( 0 \). Therefore, \( f \) is differentiable for all \( x > 0 \), confirming Statement (I) is true.

Step 2: Behavior of \( f(x) \) over intervals

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

• For \( x \in (0, 1) \), the derivative is \(-\frac{1}{x} + 1\). This is positive only if \( x < 1 \), but generally becomes negative as \( x \to 0 \).

Thus, Statement (II) is false.

Statement (III): f is decreasing in (1, ∞)

• For \( x \in (1, \infty) \), the derivative is \(\frac{1}{x} - 1\), which is negative for all \( x > 1 \).

Thus, \( f \) is indeed decreasing for \( x > 1 \), confirming Statement (III) is true.

Therefore, the correct answer is:

Only (I) and (III) are TRUE.