Limits, Continuity & Differentiability
Continuity and Differentiation of Integrals
Grade 12
Question:
<p>Let \(f : (-1, 1) \to R\) be a continuous function. If \(\int_0^{\sin x} f(t)\,dt = \dfrac{\sqrt{3}}{2}\,x\), then \(f\!\left(\dfrac{\sqrt{3}}{2}\right)\) is equal to</p>
<p>\(\dfrac{\sqrt{3}}{2}\)</p>
<p>\(\sqrt{3}\)</p>
<p>\(\sqrt{\dfrac{3}{2}}\)</p>
<p>\(\dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: Differentiate both sides of the integral equation with respect to x using Leibniz rule to find a relationship, then use the given functional equation to evaluate f at the specific point.
<p><strong>Step 1: Differentiate both sides with respect to x</strong></p><p>Given: $\int_0^{\sin x} f(t)\,dt = \frac{\sqrt{3}}{2}x$</p><p>Using Leibniz rule: $\frac{d}{dx}\int_0^{\sin x} f(t)\,dt = f(\sin x) \cdot \cos x$</p><p><strong>Step 2: Differentiate the right side</strong></p><p>$\frac{d}{dx}\left(\frac{\sqrt{3}}{2}x\right) = \frac{\sqrt{3}}{2}$</p><p><strong>Step 3: Equate both sides</strong></p><p>$f(\sin x) \cdot \cos x = \frac{\sqrt{3}}{2}$</p><p>Therefore: $f(\sin x) = \frac{\sqrt{3}}{2\cos x}$</p><p><strong>Step 4: Find the required value</strong></p><p>Set $\sin x = \frac{\sqrt{3}}{2}$, which gives $x = \frac{\pi}{3}$</p><p>Then $\cos\frac{\pi}{3} = \frac{1}{2}$</p><p>$f\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2 \cdot \frac{1}{2}} = \sqrt{3}$</p><p>∴ Answer: A</p>
Correct Answer: A