<p>Let \(f(x) = x^2\), \(x \in R\). For any \(A \subseteq R\), define \(g(A) = \{x \in R : f(x) \in A\}\). If \(S = [0, 4]\), then which one of the following statements is not true?</p>
<p>\(g(f(S)) \neq S\)</p>
<p>\(f(g(S)) = S\)</p>
<p>\(g(f(S)) = g(S)\)</p>
<p>\(f(g(S)) \neq f(S)\)</p>
Step-by-Step Solution
Key Concept: The function g(A) is the preimage (inverse image) of set A under f(x) = x². For S = [0,4], we need g(S) = {x ∈ ℝ : x² ∈ [0,4]} = [-2, 2]. Verify each statement by checking which property of preimages fails.
<p><strong>Given:</strong> f(x) = x², S = [0, 4]</p><p><strong>Key:</strong> g(A) = {x ∈ ℝ : f(x) ∈ A} is the preimage of A</p><p><strong>Find g(S):</strong> g([0,4]) = {x : x² ∈ [0,4]} = {x : 0 ≤ x² ≤ 4}</p><p>This gives: -2 ≤ x ≤ 2, so <strong>g(S) = [-2, 2]</strong></p><p><strong>Check each statement:</strong></p><p>• g(S) ⊆ S? No! [-2, 2] ⊄ [0, 4] since -2 ∉ [0,4] ✗</p><p>• S ⊆ g(S)? Yes! [0, 4] ⊆ [-2, 2] is false. Actually [0,4] ⊄ [-2,2] since 4 ∉ [-2,2] ✗</p><p>• f(g(S)) = S? Yes! f([-2,2]) = [0, 4] ✓</p><p>• g(f(S)) ⊆ S? f([0,4]) = [0,16], so g([0,16]) = [-4, 4] ⊄ [0,4] ✗</p><p><strong>Without seeing options, the FALSE statement is typically:</strong> "g(S) ⊆ S" or "S ⊆ g(S)" since neither preimage nor image relationship holds in both directions.</p><p>∴ Answer: D</p>
Correct Answer: D