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 for x slightly less than 1 (say x = 0.9+)

When 0.9 < x < 1: [x] = 0

Denominator: 2x - [x] = 2x - 0 = 2x > 0

Numerator: [x] = 0

Argument of sgn: [x]/(2x-[x]) = 0/(2x) = 0

Therefore: sgn(0) = 0, so f(x) = cot⁻¹(0) = π/2

∴ lim(x→1⁻) f(x) = π/2

Step 2: Analyze for x slightly greater than 1 (say x = 1.1+)

When x > 1: [x] = 1

Denominator: 2x - [x] = 2x - 1 > 0 (for x > 1)

Numerator: [x] = 1

Argument of sgn: 1/(2x-1) > 0

Therefore: sgn(positive) = 1, so f(x) = cot⁻¹(1) = π/4

∴ 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), the left and right limits are unequal.

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

Step 4: Check differentiability at x = 1

A function must be continuous at a point to be differentiable there. Since f is discontinuous at x = 1, it cannot be differentiable at x = 1.

Statement-2 is TRUE: f(x) is non-differentiable at x = 1 (as a consequence of discontinuity)

∴ Answer: C (Both statements are correct)

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