Question

Let $f(x)=|\log_e x|-|x-1|+5$. Statement 1: $f(x)$ is differentiable for all $x\in(0,\infty)$
Statement 2: $f(x)$ is increasing in $(1,\infty)$
Statement 3: $f(x)$ is decreasing in $(0,1)$
Which of the following is correct?

• A. Statement 1
• B. Statement 2
• C. Statement 3
• D. Statement 4

Hint:

Always remove modulus by breaking the function into intervals before checking monotonicity or differentiability.

Answer 1

The Correct Option is A, C

To analyze the differentiability and behavior of the function \( f(x) = |\log_e x| - |x-1| + 5 \), we evaluate each statement step by step.

Step 1: Differentiability

The function contains absolute value terms \( |\log_e x| \) and \( |x-1| \), both of which change their form at \( x = 1 \).

• \(|\log_e x|\) changes sign at \( x = 1 \).
• \(|x-1|\) changes form at \( x = 1 \).

For \( x > 1 \):

\(|\log_e x| = \log_e x\), \( |x-1| = x-1 \)

\[ f'(x) = \frac{1}{x} - 1 \]

For \( 0 < x < 1 \):

\(|\log_e x| = -\log_e x\), \( |x-1| = 1-x \)

\[ f'(x) = -\frac{1}{x} + 1 \]

At \( x = 1 \):

Left derivative \( = -1 + 1 = 0 \)

Right derivative \( = 1 - 1 = 0 \)

Since both derivatives are equal, \( f(x) \) is differentiable for all \( x \in (0,\infty) \).

Step 2: Behavior on \( (1,\infty) \)

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

So, \( f(x) \) is not increasing on \( (1,\infty) \).

Step 3: Behavior on \( (0,1) \)

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

Hence, \( f(x) \) is decreasing on \( (0,1) \).

Conclusion

• Statement 1 is correct.
• Statement 3 is correct.

Correct answers: Statement 1 and Statement 3.