Relations & Functions
Properties of Functions
Grade 12
Question:
<p>Let \(f(x)\) and \(g(x)\) are functions defined in the real domain and co-domain, such that \(\sqrt{1 - f^2(x)} = g(x)\), then which of the following statements are necessarily true?</p><p>(a) If \(g(x)\) is periodic with period 1, then \(f(x)\) is periodic with period half.</p><p>(b) If \(f'(c) = -f(c) = 0.5\), then \(\dfrac{g'(c)}{g(c)} = \dfrac{1}{3}\).</p><p>(c) If \(g(x)\) is an even function, then \(f(x)\) is odd.</p><p>(d) If \(g(x)\) is continuous function then \(f(x)\) is also continuous in their respective domains.</p>
<p>If \(g(x)\) is periodic with period 1, then \(f(x)\) is periodic with period half.</p>
<p>If \(f'(c) = -f(c) = 0.5\), then \(\dfrac{g'(c)}{g(c)} = \dfrac{1}{3}\).</p>
<p>If \(g(x)\) is an even function, then \(f(x)\) is odd.</p>
<p>If \(g(x)\) is continuous function then \(f(x)\) is also continuous in their respective domains.</p>
Step-by-Step Solution
Key Concept: From √(1 - f²(x)) = g(x), we get f²(x) + g²(x) = 1, which means f and g are constrained on a unit circle. Differentiating this constraint and analyzing even/odd properties of g reveals what must be true about f.
<p><strong>Step 1: Establish the constraint</strong></p><p>From √(1 - f²(x)) = g(x), squaring gives: f²(x) + g²(x) = 1. Also, g(x) ≥ 0 always.</p><p><strong>Step 2: Check statement (a)</strong></p><p>If g has period 1, f need not have period 1/2. For example, f(x) = sin(πx) and g(x) = |cos(πx)| doesn't satisfy g's periodicity claim. <strong>FALSE</strong></p><p><strong>Step 3: Check statement (b)</strong></p><p>Differentiate f²(x) + g²(x) = 1: 2f(x)f'(x) + 2g(x)g'(x) = 0</p><p>So g'(x) = -f(x)f'(x)/g(x)</p><p>Given f'(c) = -0.5 and f(c) = 0.5:</p><p>g'(c)/g(c) = -f(c)f'(c)/g²(c) = -(0.5)(-0.5)/g²(c) = 0.25/g²(c)</p><p>Since f²(c) + g²(c) = 1: (0.5)² + g²(c) = 1 → g²(c) = 0.75</p><p>g'(c)/g(c) = 0.25/0.75 = 1/3 <strong>TRUE</strong></p><p><strong>Step 4: Check statement (c)</strong></p><p>If g is even: g(-x) = g(x) ≥ 0. Then 1 - f²(-x) = 1 - f²(x), so f²(-x) = f²(x).</p><p>This means f(-x) = ±f(x), so f could be even or odd. Not necessarily odd. <strong>FALSE</strong></p><p><strong>Step 5: Check statement (d)</strong></p><p>g(x) = √(1 - f²(x)). If g is continuous, then 1 - f²(x) is continuous (composition of continuous functions with square root of non-negative continuous function is continuous).</p><p>Therefore f²(x) is continuous, which implies f(x) is continuous. <strong>TRUE</strong></p><p><strong>∴ Answer: BD</strong></p>
Correct Answer: BD