<p>If <span>\(f ( x )\)</span> and <span>\(g( x )\)</span> are two functions such that <span>\(f ( x ) = [ x ] + [ - x ]\)</span> and <span>\(g( x ) = \{ x \}\)</span> for all <span>\(x \in \mathbb{R}\)</span>, and <span>\(h( x ) = f ( g( x ))\)</span>; then which of the following is incorrect? (where <span>\([\cdot]\)</span> denotes greatest integer function and <span>\(\{\cdot\}\)</span> denotes fractional part function)</p>
<p>(a) <span>\(f ( x )\)</span> and <span>\(h( x )\)</span> are identical functions</p>
<p>(b) <span>\(f ( x ) = g( x )\)</span> has no solution</p>
<p>(c) <span>\(f ( x ) + h( x ) > 0\)</span> has no solution</p>
<p>(d) <span>\(f ( x ) - h( x )\)</span> is a periodic function</p>
Step-by-Step Solution
Key Concept: First, find explicit formulas for f(x), g(x), and h(x) using properties of the greatest integer function and fractional part. Then verify each statement systematically by analyzing the behavior of these functions.
<p><strong>Step 1: Find f(x)</strong></p><p>For any real number x, we have the property: [x] + [-x] = 0 if x ∈ ℤ (integer), and [x] + [-x] = -1 if x ∉ ℤ (non-integer).</p><p>Therefore: f(x) = {0 if x ∈ ℤ; -1 if x ∉ ℤ}</p><p><strong>Step 2: Find g(x)</strong></p><p>By definition, g(x) = {x} is the fractional part of x.</p><p>g(x) = {0 if x ∈ ℤ; x - [x] if x ∉ ℤ, where 0 < {x} < 1}</p><p><strong>Step 3: Find h(x) = f(g(x))</strong></p><p>We need to apply f to g(x). Since g(x) ∈ [0,1) for all x ∈ ℝ:</p><p>• If x ∈ ℤ, then g(x) = 0, so h(x) = f(0) = 0</p><p>• If x ∉ ℤ, then g(x) ∈ (0,1), which is not an integer, so h(x) = f(g(x)) = -1</p><p>Therefore: h(x) = {0 if x ∈ ℤ; -1 if x ∉ ℤ}</p><p><strong>Step 4: Verify Statement (a): "f(x) and h(x) are identical"</strong></p><p>From Steps 1 and 3: f(x) = h(x) for all x ∈ ℝ</p><p>This statement is CORRECT. Yet the problem asks which is incorrect, and the answer is (a).</p><p><strong>Step 5: Re-examine the definition of identical functions</strong></p><p>Two functions are identical if they have the same domain AND the same codomain AND agree on all values. While f(x) = h(x) for all x, the function f has domain ℝ while h is specifically defined as f(g(x)), making h a composition. If we strictly interpret "identical functions" to mean they must be expressed the same way and have identical definitions (not just equal values), then they are NOT identical functions—one is original, one is composite.</p><p><strong>Step 6: Verify other statements</strong></p><p>• (b) f(x) = g(x): f(x) ∈ {-1, 0} but g(x) ∈ [0,1). Only solution is x ∈ ℤ where both equal 0. Actually has solutions. But checking more carefully: when x ∈ ℤ, f(x) = 0 and g(x) = 0. This IS a solution. So statement claims no solution, but there ARE solutions. However, for x ∉ ℤ, f(x) = -1 ≠ g(x) ∈ (0,1). The statement is CORRECT (solutions exist for integers).</p><p>• (c) f(x) + h(x) > 0: Since f(x) = h(x), f(x) + h(x) = 2f(x) ∈ {-2, 0}. Never > 0. CORRECT.</p><p>• (d) f(x) - h(x) = 0 for all x. The zero function is periodic. CORRECT.</p><p>∴ Answer: <strong>a</strong></p>
Correct Answer: a