Applications of Derivatives
Periodic Functions and Derivatives
Grade 12
Question:
<p><strong>Ex. 16:</strong> <strong>Statement I</strong> If differentiable function \(f(x)\) satisfies the relation \(f(x) + f(x+2) = 0\), \(\forall x \in \mathbb{R}\), and if \(\frac{1}{d}\frac{d}{dx}f(x)\bigg|_{x=a} = 6\), then \(\frac{1}{d}\frac{d}{dx}f(x)\bigg|_{x=a+4000} = 6\)</p><p><strong>Statement II</strong> \(f(x)\) is a periodic function with period 4.</p>
Step-by-Step Solution
Key Concept: Use the given functional equation to show that f has period 4, which implies the derivatives at points differing by 4000 are equal.
<p><strong>Solution:</strong> We have $f(x) + f(x+2) = 0$ ... (i)</p><p>Replace $x$ by $x+2$ in Eq. (i): $f(x+2) + f(x+4) = 0$ ... (ii)</p><p>From Eqs. (i) and (ii): $f(x) = f(x+4) = 0$ ... (iii)</p><p>From Eq. (iii): $f(x+4) = f(x)$</p><p>Therefore $f(x)$ is periodic with period 4.</p><p>Since $f(x+4000) = f(x)$, we have $\frac{d}{dx}f(x)\bigg|_{x=a+4000} = \frac{d}{dx}f(x)\bigg|_{x=a} = 6$</p><p>Hence, both statements are true and Statement II is the correct explanation of Statement I.</p>
Correct Answer: A