Limits, Continuity & Differentiability Question - None

Question

Let $ f(x) = \cot^{-1}\left(\text{sgn}\left(\dfrac{[x]}{2x - [x]}\right)\right) $:

Statement-1: $ f(x) $ is discontinuous at $ x = 1 $.

Statement-2: $ f(x) $ is non-differentiable at $ x = 1 $.

Which of the following option is correct?

[Note: $[k]$ denotes greatest integer function less than or equal to $k$.]

Accepted Answer

Step 1: Analyze behavior as x→1⁻

For x→1⁻: [x]=0, so [x]/(2x-[x]) = 0/(2x) = 0. Therefore sgn(0) = 0, and cot⁻¹(0) = π/2. Thus lim(x→1⁻) f(x) = π/2

Step 2: Analyze behavior as x→1⁺

For x→1⁺: [x]=1, so [x]/(2x-[x]) = 1/(2x-1). As x→1⁺, this → 1/(2-1) = 1 > 0. Therefore sgn(1) = 1, and cot⁻¹(1) = π/4. Thus lim(x→1⁺) f(x) = π/4

Step 3: Check continuity at x=1

Since lim(x→1⁻) f(x) = π/2 ≠ π/4 = lim(x→1⁺) f(x), function is discontinuous at x=1. ✓ Statement-1 is TRUE

Step 4: Check differentiability at x=1

A function discontinuous at a point cannot be differentiable there. Therefore f(x) is non-differentiable at x=1. ✓ Statement-2 is TRUE

Step 5: Identify correct option

Both statements are true AND Statement-2 follows logically from Statement-1 (discontinuity ⟹ non-differentiability).

∴ Answer: C (Both statements true; Statement-2 is correct explanation of Statement-1)

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