In [1,3], the function <span class="math-inline">\( [x^2 + 1][.] \)</span> denoting the greatest integer function, is continuous -
for all x
for all x except at nine points
for all x except at seven points
for all x except at eight points
Step-by-Step Solution
Key Concept: The greatest integer function [x²+1] is discontinuous at points where x²+1 is an integer. We need to find all integers k in the range of x²+1 for x∈[1,3], then solve x²+1=k to find discontinuity points.
<p><strong>Step 1: Find the range of x²+1 on [1,3]</strong></p><p>For x∈[1,3]:</p><p>• When x=1: x²+1=2</p><p>• When x=3: x²+1=10</p><p>• Since x²+1 is continuous and increasing on [1,3], the range is [2,10]</p><p><strong>Step 2: Identify integer values in the range</strong></p><p>The integers in [2,10] are: 2, 3, 4, 5, 6, 7, 8, 9, 10</p><p>That's 9 integers total.</p><p><strong>Step 3: Find discontinuity points</strong></p><p>The function [x²+1] is discontinuous where x²+1 equals each of these integers.</p><p>For each integer k, solve x²+1=k for x∈[1,3]:</p><p>• x²+1=2 ⟹ x=1 (boundary point, check separately)</p><p>• x²+1=3 ⟹ x=√2 ≈1.414</p><p>• x²+1=4 ⟹ x=√3 ≈1.732</p><p>• x²+1=5 ⟹ x=2</p><p>• x²+1=6 ⟹ x=√5 ≈2.236</p><p>• x²+1=7 ⟹ x=√6 ≈2.449</p><p>• x²+1=8 ⟹ x=√7 ≈2.646</p><p>• x²+1=9 ⟹ x=√8 ≈2.828</p><p>• x²+1=10 ⟹ x=3 (boundary point, check separately)</p><p><strong>Step 4: Check boundary points</strong></p><p>At x=1: [x²+1]=[2]=2. For x slightly greater than 1, x²+1 is slightly greater than 2 but less than 3, so [x²+1]=2. Continuous at x=1.</p><p>At x=3: [x²+1]=[10]=10. For x slightly less than 3, x²+1 is slightly less than 10, so [x²+1]=9. Discontinuous at x=3.</p><p><strong>Step 5: Count interior discontinuity points</strong></p><p>Interior points (x∈(1,3)) where discontinuity occurs: √2, √3, 2, √5, √6, √7, √8</p><p>That's 7 points in the interior.</p><p>Adding the right boundary point x=3: Total = 7 points of discontinuity in [1,3]</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C