Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function such that \( f(0) = \frac{1}{\pi} \) and \( f(x) = \frac{x}{e^x - 1} \) for \( x \ne 0 \). Then:

• A. \( f(x) \) is not continuous at \( x = 0 \)
• B. \( f(x) \) is continuous but not differentiable at \( x = 0 \)
• C. \( f(x) \) is differentiable at \( x = 0 \) and \( f'(0) = \frac{\pi}{2} \)
• D. None of these

Hint:

When analyzing continuity and differentiability: - A function must be continuous before it can be differentiable at a point. - To check continuity at a point, verify if the limit of the function as \( x \to a \) equals the function value at \( a \). - To check differentiability at a point, check if the function is continuous and the derivative exists at that point.

Answer 1

The Correct Option is D

Step 1: Analyze the behavior of \( f(x) = \frac{x}{e^x - 1} \) near \( x = 0 \).
Use the Taylor expansion:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \Rightarrow e^x - 1 = x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \] \[ \Rightarrow \frac{x}{e^x - 1} = \frac{1}{1 + \frac{x}{2} + \frac{x^2}{6} + \cdots} = 1 - \frac{x}{2} + \frac{x^2}{12} + \cdots \] Step 2: Take the limit as \( x \to 0 \):
\[ \lim_{x \to 0} \frac{x}{e^x - 1} = 1 \] But:
\[ f(0) = \frac{1}{\pi} \ne 1 \] So, \( f(x) \) is not continuous at \( x = 0 \).

Step 3: Evaluate differentiability.
If a function is not continuous at a point, it is not differentiable there.
Hence, options (2) and (3) are ruled out.

Step 4: Check Option (1):
Option (1) says “not continuous at \( x = 0 \)” which seems correct. But let’s be precise.
The limit exists:
\[ \lim_{x \to 0} f(x) = 1 \] But the function value is:
\[ f(0) = \frac{1}{\pi} \ne 1 \] Hence, discontinuous.

So option (1) is correct.

Conclusion:
The function \( f(x) \) is not continuous at \( x = 0 \), so the correct answer is Option (1).

If a source stated that the correct answer is Option (4), that is **likely a mistake** based on this logical analysis.