If f(x) = frac{2 - x cos x}{2 + x cos x} and | vedclass

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(A) Let $I = \int_{-\pi/4}^{\pi/4} g(f(x)) dx$. Since $f(-x) = \frac{2 - (-x) \cos(-x)}{2 + (-x) \cos(-x)} = \frac{2 + x \cos x}{2 - x \cos x} = \frac

Vedclass · Accepted Answer

$\ln 1$

Vedclass · Answer

(A) Let $I = \int_{-\pi/4}^{\pi/4} g(f(x)) dx$. Since $f(-x) = \frac{2 - (-x) \cos(-x)}{2 + (-x) \cos(-x)} = \frac{2 + x \cos x}{2 - x \cos x} = \frac{1}{f(x)}$,we have $g(f(-x)) = \ln(f(-x)) = \ln(1/f(x)) = -\ln(f(x)) = -g(f(x))$. Thus,$g(f(x))$ is an odd function. For any odd function $h(x)$,the integral $\int_{-a}^{a} h(x) dx = 0$. Therefore,$I = 0$. Since $\ln 1 = 0$,the correct option is $\ln 1$.

If $f(x) = \frac{2 - x \cos x}{2 + x \cos x}$ and $g(x) = \ln x$ for $x > 0$,then the value of the integral $\int_{-\pi/4}^{\pi/4} g(f(x)) dx$ is

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