Limits, Continuity & Differentiability
Continuous Functions and IVT
Grade 12
Question:
<p>For every pair of continuous function <i>f</i>, <i>g</i> : [0, 1] → ℝ such that max{<i>f</i>(<i>x</i>) : <i>x</i> ∈ [0, 1]} = max{<i>g</i>(<i>x</i>) : <i>x</i> ∈ [0, 1]}. The correct statement(s) is/are</p>
<p>(a) [<i>f</i>(<i>c</i>)]<sup>2</sup> + 3<i>f</i>(<i>c</i>) = [<i>g</i>(<i>c</i>)]<sup>2</sup> + 3<i>g</i>(<i>c</i>) for some <i>c</i> ∈ [0, 1]</p>
<p>(b) [<i>f</i>(<i>c</i>)]<sup>2</sup> + <i>f</i>(<i>c</i>) = [<i>g</i>(<i>c</i>)]<sup>2</sup> + 3<i>g</i>(<i>c</i>) for some <i>c</i> ∈ [0, 1]</p>
<p>(c) [<i>f</i>(<i>c</i>)]<sup>2</sup> + 3<i>f</i>(<i>c</i>) = [<i>g</i>(<i>c</i>)]<sup>2</sup> + <i>g</i>(<i>c</i>) for some <i>c</i> ∈ [0, 1]</p>
<p>(d) [<i>f</i>(<i>c</i>)]<sup>2</sup> = [<i>g</i>(<i>c</i>)]<sup>2</sup> for some <i>c</i> ∈ [0, 1]</p>
Step-by-Step Solution
Key Concept: Define h(x) = [f(x)]² + 3f(x) - [g(x)]² - 3g(x) and use the Intermediate Value Theorem. Since both f and g attain the same maximum value M on [0,1], we can construct functions that make h continuous and force it to have a zero.
<p><strong>Step 1: Setup the problem</strong></p><p>Given: f, g : [0,1] → ℝ are continuous with max{f(x)} = max{g(x)} = M (say).</p><p>We need to find which statement holds for EVERY such pair (f,g).</p><p><strong>Step 2: Analyze Option (a)</strong></p><p>Define h(x) = [f(x)]² + 3f(x) - [g(x)]² - 3g(x) = φ(f(x)) - φ(g(x)), where φ(t) = t² + 3t.</p><p><strong>Step 3: Key observation about φ(t)</strong></p><p>Note that φ(t) = t² + 3t is strictly increasing on [0,∞) since φ'(t) = 2t + 3 > 0 for t ≥ 0.</p><p><strong>Step 4: Evaluate h at points where extrema occur</strong></p><p>Since max{f} = M, there exists c₁ ∈ [0,1] where f(c₁) = M.</p><p>Since max{g} = M, there exists c₂ ∈ [0,1] where g(c₂) = M.</p><p>Therefore: h(c₁) = φ(M) - φ(g(c₁)) and h(c₂) = φ(f(c₂)) - φ(M).</p><p><strong>Step 5: Sign analysis</strong></p><p>At c₁: Since g(c₁) ≤ M and φ is strictly increasing, h(c₁) = φ(M) - φ(g(c₁)) ≥ 0.</p><p>At c₂: Since f(c₂) ≤ M and φ is strictly increasing, h(c₂) = φ(f(c₂)) - φ(M) ≤ 0.</p><p><strong>Step 6: Apply Intermediate Value Theorem</strong></p><p>Since h is continuous on [0,1] and h has opposite signs at c₁ and c₂ (or equals 0 at one of them), by IVT there exists c ∈ [0,1] where h(c) = 0.</p><p>This gives: [f(c)]² + 3f(c) = [g(c)]² + 3g(c).</p><p><strong>Step 7: Check why other options fail</strong></p><p>For option (b): φ₁(t) = t² + t is not the same on both sides, and the asymmetry in coefficients (1 vs 3) breaks the required sign balance from the equal maximum condition.</p><p>For options (c) and (d): Similar asymmetries prevent the IVT argument from working universally for all pairs (f,g).</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A