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

### Related Formula textFor x geq 1, quad ln x geq 0 text and x-1 geq 0$$\text{For } x \geq 1, \quad \ln x \geq 0 \text{ and } x-1 \geq 0$$ textFor 0 < x < 1, quad ln x < 0 text and x-1 < 0$$\text{For } 0 < x < 1, \quad \ln x < 0 \text{ and } x-1 < 0$$ ### Core Logic Break the function into piecewise intervals based on the critical point x=1$x=1$. f(x) = begincases ln x - (x - 1) & x geq 1 \\ -ln x - (-(x - 1)) & 0 < x < 1 endcases$$f(x) = \begin{cases} \ln x - (x - 1) & x \geq 1 \\ -\ln x - (-(x - 1)) & 0 < x < 1 \end{cases}$$ f(x) = begincases ln x - x + 1 & x geq 1 \\ -ln x + x - 1 & 0 < x < 1 endcases$$f(x) = \begin{cases} \ln x - x + 1 & x \geq 1 \\ -\ln x + x - 1 & 0 < x < 1 \end{cases}$$ ### Step 1: Checking Differentiability (Statement I) Find the derivative f'(x)$f'(x)$: f'(x) = begincases frac1x - 1 & x > 1 \\ -frac1x + 1 & 0 < x < 1 endcases$$f'(x) = \begin{cases} \frac{1}{x} - 1 & x > 1 \\ -\frac{1}{x} + 1 & 0 < x < 1 \end{cases}$$ Check left-hand derivative (LHD) and right-hand derivative (RHD) at x = 1$x = 1$: textRHD = f'(1^+) = frac11 - 1 = 0$$\text{RHD} = f'(1^+) = \frac{1}{1} - 1 = 0$$ textLHD = f'(1^-) = -frac11 + 1 = 0$$\text{LHD} = f'(1^-) = -\frac{1}{1} + 1 = 0$$ Since textLHD = textRHD$\text{LHD} = \text{RHD}$, f(x)$f(x)$ is differentiable at x=1$x=1$ (and clearly everywhere else in its domain). Thus, Statement (I) is TRUE. ### Step 2: Checking Monotonicity (Statements II and III) Check for x > 1$x > 1$: f'(x) = frac1x - 1 < 0 text for all x > 1$f'(x) = \frac{1}{x} - 1 < 0 \text{ for all } x > 1$. Therefore, f$f$ is decreasing in (1, infty)$(1, \infty)$. Statement (III) is TRUE. Check for 0 < x < 1$0 < x < 1$: f'(x) = 1 - frac1x = fracx-1x < 0 text for all 0 < x < 1$f'(x) = 1 - \frac{1}{x} = \frac{x-1}{x} < 0 \text{ for all } 0 < x < 1$. Therefore, f$f$ is also decreasing in (0, 1)$(0, 1)$. Statement (II) is FALSE. Since f$f$ is decreasing in both intervals, it is monotonically decreasing across (0, infty)$(0, \infty)$. ### Step 3: Final Conclusion Only Statement (I) and Statement (III) are true. ### Pattern Recognition Adding or subtracting two modulus functions that change sign at the exact same critical point often smooths out the sharp corner, leading to a perfectly differentiable curve at that node. Always compute LHD and RHD to confirm. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives Class 12 Maths: Limits, Continuity and Differentiability