<p>If <span>\(g(1) = g(2)\)</span>, then <span>\(\int_1^2 \frac{[f\{g(x)\}]^{-1} f'\{g(x)\} g'(x)}{f^2(x)} dx\)</span> is equal to</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand can be rewritten as a derivative of a composite function using substitution u = g(x), and the condition g(1) = g(2) makes the definite integral zero because the antiderivative evaluates to the same value at both limits.
<p><strong>Step 1:</strong> Let u = g(x), then du = g'(x)dx. The integral becomes:</p><p>$$\int_1^2 \frac{[f\{g(x)\}]^{-1} f'\{g(x)\} g'(x)}{f^2(x)} dx = \int_{g(1)}^{g(2)} \frac{[f(u)]^{-1} f'(u)}{f^2(x)} du$$</p><p><strong>Step 2:</strong> Observe that the integrand can be written as a derivative. Note that:</p><p>$$\frac{d}{du}\left[-\frac{1}{f(u)}\right] = -\left(-\frac{f'(u)}{f^2(u)}\right) = \frac{f'(u)}{f^2(u)}$$</p><p><strong>Step 3:</strong> The integral becomes:</p><p>$$\int_{g(1)}^{g(2)} \frac{[f(u)]^{-1} f'(u)}{f^2(u)} du = \int_{g(1)}^{g(2)} \frac{f'(u)}{f^2(u)} du = \left[-\frac{1}{f(u)}\right]_{g(1)}^{g(2)}$$</p><p><strong>Step 4:</strong> Apply the Fundamental Theorem of Calculus:</p><p>$$\left[-\frac{1}{f(g(2))}\right] - \left[-\frac{1}{f(g(1))}\right] = -\frac{1}{f(g(2))} + \frac{1}{f(g(1))}$$</p><p><strong>Step 5:</strong> Since g(1) = g(2) (given condition), we have f(g(1)) = f(g(2)):</p><p>$$-\frac{1}{f(g(2))} + \frac{1}{f(g(1))} = -\frac{1}{f(g(1))} + \frac{1}{f(g(1))} = 0$$</p><p><strong>∴ Answer:</strong> c</p>
Correct Answer: c