Consider the following three statements for the function f : (0, ∞) - R defined by f(x) = |logₑ x|

Q: Wait, isn't a modulus function always non-differentiable at the point where the inside is zero?

That is a very common trap! While $|x|$ is not differentiable at $x=0$, the combination of two modulus functions can sometimes 'cancel out' the sharp corner. Always check the LHD and RHD explicitly.

Q: Why did we get a negative derivative in $(0, 1)$? I thought log functions were increasing.

The natural log function $\ln x$ is indeed increasing, but look closely at our piecewise definition: $f(x) = -\ln x + x - 1$. The negative sign in front of the log flips the behavior!

Q: Can I just graph it to see if it's differentiable?

Graphing is a great way to visualize, but in JEE Advanced, you need the analytical proof. The LHD and RHD calculation is the only way to be 100% sure about the smoothness at $x=1$.

Question

Consider the following three statements for the function $f : (0, \infty) 	o \mathbb{R}$ 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.

Options:

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

(B) Only (II) and (III) are TRUE.

(C) Only (I) is TRUE.

(D) Only (I) and (III) are TRUE.

Sigma Percentile · Accepted Answer

### Analyzing the Setup The function under investigation is defined as $f(x) = |\log_e x| - |x - 1|$. This function represents a landscape where the modulus operators act as mirrors, potentially creating sharp corners or "cusps" at transition points. To understand the behavior of $f(x)$, we must identify the critical point where the expressions inside the modulus bars change sign. Both $\log_e x$ and $x - 1$ equal zero at $x = 1$. ### The Critical Point For the interval $0 < x < 1$, the natural logarithm $\log_e x$ is negative, and $x - 1$ is also negative. Consequently, the modulus definitions are $|\log_e x| = -\log_e x$ and $|x - 1| = -(x - 1)$. Substituting these into our function, we obtain: $$f(x) = -\log_e x - (-(x - 1)) = -\log_e x + x - 1$$ To determine the behavior, we calculate the derivative: $$f'(x) = -\frac{1}{x} + 1 = \frac{x - 1}{x}$$ In the interval $(0, 1)$, $x$ is positive and $x - 1$ is negative. Since the derivative $f'(x) < 0$, the function is **strictly decreasing** in this interval. ### The Second Act Now, consider the territory where $x > 1$. Here, $\log_e x$ is positive and $x - 1$ is positive, so $|\log_e x| = \log_e x$ and $|x - 1| = x - 1$. Our function simplifies to: $$f(x) = \log_e x - (x - 1) = \log_e x - x + 1$$ The derivative in this region is: $$f'(x) = \frac{1}{x} - 1 = \frac{1 - x}{x}$$ Since $x > 1$, the numerator $(1 - x)$ is negative and the denominator $x$ is positive. Thus, $f'(x) < 0$, confirming that the function is **strictly decreasing** for $x > 1$ as well. ### The Differentiability Test We now address whether the function is differentiable at the transition point $x = 1$. We compare the Left Hand Derivative (LHD) and the Right Hand Derivative (RHD). The LHD is calculated as: $$\lim_{x o 1^-} f'(x) = \lim_{x o 1^-} \frac{x - 1}{x} = 0$$ The RHD is calculated as: $$\lim_{x o 1^+} f'(x) = \lim_{x o 1^+} \frac{1 - x}{x} = 0$$ Because the LHD equals the RHD, the function is **differentiable at $x = 1$**. Despite the presence of modulus bars, the function is smooth everywhere. ### Final Conclusion Based on our rigorous analysis: 1. **Statement (I) is true**: The function is differentiable everywhere. 2. **Statement (II) is false**: The function is decreasing, not increasing, in the interval $(0, 1)$. 3. **Statement (III) is true**: The function is decreasing in the interval $(1, \infty)$. The correct conclusion is that **only (I) and (III) are true**. Always remember that in JEE Advanced, the most reliable tool is the cold, hard logic of calculus rather than intuition alone.

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Visualized Solution

Understanding the Function $f (x)$

Case 1: $0 < x < 1$

Differentiating for $0 < x < 1$

Monotonicity in $(0, 1)$

Case 2: $x > 1$

Differentiating for $x > 1$

Monotonicity in $(1, \infty)$

Differentiability at $x = 1$

Calculating LHD and RHD

Final Verdict on Statement (I)

Conclusion and Final Answer

The Sigma Insight: Monotonicity

Analyzing the Setup

The Critical Point

The Second Act

The Differentiability Test

Final Conclusion

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Visualized Solution

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0<x<1

Differentiating for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0<x<1

Monotonicity in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,1)

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>1

Differentiating for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>1

Monotonicity in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,∞)

Differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1