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 the constraint √(1 - f²(x)) = g(x), we have f²(x) + g²(x) = 1, which means f(x) and g(x) are related like sine and cosine. Use this relationship to test each statement by analyzing what properties one function inherits from the other.
<p><strong>Step 1: Establish the fundamental constraint</strong></p><p>From √(1 - f²(x)) = g(x), we have: f²(x) + g²(x) = 1, where g(x) ≥ 0 always.</p><p><strong>Step 2: Analyze statement (a)</strong></p><p>If g(x) has period 1: g(x+1) = g(x). Then f²(x+1) + g²(x+1) = 1 and f²(x) + g²(x) = 1, so f²(x+1) = f²(x). This means |f(x+1)| = |f(x)|, NOT that f(x+1) = f(x). The sign can change, so f(x) need not be periodic with any period. <strong>Statement (a) is FALSE.</strong></p><p><strong>Step 3: Analyze statement (b)</strong></p><p>Given: f'(c) = -f(c) = 0.5, so f(c) = -0.5 and f'(c) = 0.5.</p><p>From f²(x) + g²(x) = 1, differentiate: 2f(x)f'(x) + 2g(x)g'(x) = 0</p><p>Therefore: g(x)g'(x) = -f(x)f'(x)</p><p>At x = c: g(c)g'(c) = -(-0.5)(0.5) = 0.25</p><p>Also, f²(c) + g²(c) = 1 gives g²(c) = 1 - 0.25 = 0.75, so g(c) = √(0.75) = (√3)/2</p><p>Thus: g'(c) = 0.25/g(c) = 0.25/(√3/2) = 1/(2√3) = √3/6</p><p>Therefore: g'(c)/g(c) = (√3/6)/(√3/2) = (√3/6) · (2/√3) = 2/6 = 1/3 <strong>✓ Statement (b) is TRUE.</strong></p><p><strong>Step 4: Analyze statement (c)</strong></p><p>If g(x) is even: g(-x) = g(x). Then f²(-x) + g²(-x) = f²(-x) + g²(x) = 1. This gives f²(-x) = f²(x), so |f(-x)| = |f(x)|. For f to be odd, we need f(-x) = -f(x), which requires f(-x) = -f(x) for all x. This is not necessarily true; f could satisfy f(-x) = f(x) (even) or f(-x) = -f(x) (odd). <strong>Statement (c) is FALSE.</strong></p><p><strong>Step 5: Analyze statement (d)</strong></p><p>If g(x) is continuous, then g²(x) is continuous. From f²(x) = 1 - g²(x), we have f²(x) is continuous. Therefore, √(f²(x)) = |f(x)| is continuous. Since f(x) = ±√(1 - g²(x)) at each point (the sign may vary), and 1 - g²(x) is continuous, the continuous square root function ensures f(x) can be chosen continuously in its domain. More rigorously, in any connected domain, if we define f(x) = ±√(1 - g²(x)) continuously, it must have constant sign by the intermediate value theorem, making f(x) continuous. <strong>Statement (d) is TRUE.</strong></p><p><strong>∴ Answer: BD</strong></p>
Correct Answer: BD