Relations & Functions
Greatest Integer Function
Grade 12
Question:
<p>Let \(f(x) = [x]^2 + [x+1] - 3\), where \([x]\) = the greatest integer \(\leq x\). Then</p>
<p>(a) \(f(x)\) is a many-one and into function</p>
<p>(b) \(f(x) = 0\) for infinite number of values of \(x\)</p>
<p>(c) \(f(x) = 0\) for only two real values</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: For any real x, [x] and [x+1] differ by exactly 1 (since [x+1] = [x] + 1), so substitute [x] = n to express f(x) in terms of n and the fractional part of x.
<p><strong>Step 1:</strong> Let [x] = n where n is an integer. Then x ∈ [n, n+1) and [x+1] = [x] + 1 = n + 1.</p><p><strong>Step 2:</strong> Substitute into f(x): f(x) = n² + (n+1) - 3 = n² + n - 2 = (n+1)(n-2).</p><p><strong>Step 3:</strong> For x ∈ [n, n+1), f(x) is constant and equals (n+1)(n-2), so f is a step function that jumps at each integer.</p><p><strong>Step 4:</strong> Find when f(x) = 0: (n+1)(n-2) = 0 gives n = -1 or n = 2.</p><p>• When n = -1: f(x) = 0 for x ∈ [-1, 0)</p><p>• When n = 2: f(x) = 0 for x ∈ [2, 3)</p><p><strong>Step 5:</strong> f is not one-one (multiple intervals give same value). f is onto ℤ since (n+1)(n-2) covers all integers as n varies. f is strictly increasing on intervals where [x] increases (jumps).</p><p>∴ Answer: A (f is onto), B (f is not one-one)</p>
Correct Answer: A,B