Limits, Continuity & Differentiability Question - None

Question

The number of values of $x$, $x \in [-2, 3]$ where $f(x) = [x^2]\sin(\pi x)$ is discontinuous is (where $[\cdot]$ denotes greatest integer function)

Accepted Answer

Step 1: Identify where [x²] is discontinuous.

The greatest integer function [x²] is discontinuous when x² equals a non-negative integer. We need x² ∈ {0, 1, 2, 3, 4, ...} where x ∈ [-2, 3].

Since x ∈ [-2, 3], we have x² ∈ [0, 9].

So x² ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

Step 2: Find all x values where x² equals these integers.

• x² = 0 ⟹ x = 0

• x² = 1 ⟹ x = ±1

• x² = 2 ⟹ x = ±√2 ≈ ±1.414

• x² = 3 ⟹ x = ±√3 ≈ ±1.732

• x² = 4 ⟹ x = ±2

• x² = 5 ⟹ x = ±√5 ≈ ±2.236

• x² = 6 ⟹ x = ±√6 ≈ ±2.449

• x² = 7 ⟹ x = ±√7 ≈ ±2.646

• x² = 8 ⟹ x = ±√8 ≈ ±2.828

• x² = 9 ⟹ x = ±3

Step 3: Filter for x ∈ [-2, 3].

Valid x values: x = -2, -√8, -√7, -√6, -√5, -2, -√3, -√2, -1, 0, 1, √2, √3, √5, √6, √7, √8, 2, 3

Listing distinct values in order:

x ∈ {-2, -√8, -√7, -√6, -√5, -√3, -√2, -1, 0, 1, √2, √3, √5, √6, √7, √8, 2, 3}

Step 4: Count the discontinuities.

From negative side: -2, -√8, -√7, -√6, -√5, -√3, -√2, -1 (8 points)

At origin: 0 (1 point)

From positive side: 1, √2, √3, √5, √6, √7, √8, 2, 3 (9 points)

Total: 8 + 1 + 9 = 18 points

Step 5: Verify these are actual discontinuities.

At each point where x² = n (integer), [x²] jumps from n-1 to n. Since sin(πx) is continuous and non-zero at most of these points, the product f(x) = [x²]sin(πx) is indeed discontinuous at each of these 18 points.

∴ Answer: 18

Question:

Step-by-Step Solution

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