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(B) Given $(f \circ g)(x) = x$, $g(x)$ is the inverse function of $f(x)$.Let $g(1 + (2n - 1) \pi) = x_n$. Then $f(x_n) = 1 + (2n - 1) \pi$.Substitutin

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$(99)^2 \frac{\pi}{2}$

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(B) Given $(f \circ g)(x) = x$, $g(x)$ is the inverse function of $f(x)$.Let $g(1 + (2n - 1) \pi) = x_n$. Then $f(x_n) = 1 + (2n - 1) \pi$.Substituting $f(x) = 2x + \cos x + \sin^2 x$:$2x_n + \cos x_n + \sin^2 x_n = 1 + (2n - 1) \pi$$2x_n + \cos x_n + 1 - \cos^2 x_n = 1 + (2n - 1) \pi$$2x_n + \cos x_n - \cos^2 x_n = (2n - 1) \pi$.For $x_n = (2n - 1) \frac{\pi}{2}$, we have $\cos x_n = 0$.Substituting this into the equation: $2((2n - 1) \frac{\pi}{2}) + 0 - 0 = (2n - 1) \pi$, which holds true.Thus, $g(1 + (2n - 1) \pi) = (2n - 1) \frac{\pi}{2}$.Now, $\sum_{n=1}^{99} g(1 + (2n - 1) \pi) = \sum_{n=1}^{99} (2n - 1) \frac{\pi}{2} = \frac{\pi}{2} \sum_{n=1}^{99} (2n - 1)$.The sum of the first $99$ odd numbers is $99^2$.Therefore, the sum is $(99)^2 \frac{\pi}{2}$.

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be differentiable functions such that $(f \circ g)(x) = x$. If $f(x) = 2x + \cos x + \sin^2 x$, then the value of $\sum_{n=1}^{99} g(1 + (2n - 1) \pi)$ is

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