Differential Equations
Separable Equations
Grade 12
Question:
<p>If <span class='latex'>f(x)</span> be a positive, continuous and differentiable on the interval <span class='latex'>(a, b)</span>. If <span class='latex'>\lim_{x \to a^+} f(x) = 1</span> and <span class='latex'>\lim_{x \to b^-} f(x) = 31/4</span>. Also <span class='latex'>f'(x) = f^3(x) + \frac{1}{f(x)}</span>, then</p>
<p>(A) <span class='latex'>b - a \geq \pi/4</span></p>
<p>(B) <span class='latex'>b - a \leq \pi/4</span></p>
<p>(C) <span class='latex'>b - a \geq \pi/24</span></p>
<p>(D) None of these</p>
Step-by-Step Solution
Key Concept: Separate variables and integrate the differential equation using standard antiderivative formulas.
<p><strong>Solution:</strong> From the differential equation <span class='latex'>f'(x) = f^3(x) + \frac{1}{f(x)}</span>, we can write:</p><p><span class='latex'>\frac{df}{dx} = f^3(x) + \frac{1}{f(x)} = \frac{f^4(x) + 1}{f(x)}</span></p><p>Separating variables: <span class='latex'>\frac{f(x)df}{f^4(x) + 1} = dx</span></p><p>Integrating both sides from <span class='latex'>x = a</span> to <span class='latex'>x = b</span>:</p><p><span class='latex'>\int_1^{31/4} \frac{f\,df}{f^4 + 1} = b - a</span></p><p>Since <span class='latex'>\int \frac{f\,df}{f^4 + 1} = \frac{1}{2}\tan^{-1}(f^2)</span>, we have:</p><p><span class='latex'>\frac{1}{2}[\tan^{-1}(961/16) - \tan^{-1}(1)] = b - a</span></p><p>Since <span class='latex'>\tan^{-1}(961/16) \leq \pi/2</span>, we get <span class='latex'>b - a \geq \pi/4</span></p>
Correct Answer: A